Bilangan Bentuk Pangkat ( Lanjutan Kelas X KTSP )

Silahkan review kembali ulasan yang sudah ada di

sekedar mengingatkan

\LARGE a^{m}=\displaystyle \underset{m\: \: \: \textbf{faktor}}{\underbrace{a\times a\times a\times a\times a\times a\times a\times a\times a\: \: ...\: \times a}}.

maka,

\begin{aligned}\textbf{b}^{\textbf{m}}\: \times \: \textbf{b}^{\textbf{n}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\textbf{b}\times \textbf{b}\times \textbf{b}\times \textbf{b}\: \: ...\: \times \textbf{b}}}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}\\\\ &=\textbf{b}^{^{\textbf{...}}} \end{aligned}.

Sebagai misal

\begin{aligned}3^{1010}\: \times \: 3^{1007}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{3\times 3\times 3\times 3\: \: ...\: \times 3}}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}\\\\ &=...^{\: ^{...}} \end{aligned}.

untuk

\begin{aligned}\textbf{b}^{\textbf{m}}\: : \: \textbf{b}^{\textbf{n}}&=\displaystyle \frac{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad\: \begin{matrix} ...\\ \end{matrix}\: \qquad}}\: \: \times \: \: \overset{...\: \: \: \textbf{faktor}}{\overbrace{\qquad \: \begin{matrix} ...\\ \end{matrix}\: \qquad}}}}}{\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad}}\\\\ &=\textbf{...}^{^{\textbf{...}}} \end{aligned}.

Silahkan dicoba bentuk berikut

\begin{aligned}\textbf{5}^{\textbf{2034}}\: : \: \textbf{5}^{\textbf{17}}&=\displaystyle \frac{\overset{2034\: \: \: \textbf{faktor}}{\overbrace{\overset{2017\: \: \: \textbf{faktor}}{\overbrace{\qquad\: \begin{matrix} ...\\ \end{matrix}\: \qquad}}\: \: \times \: \: \overset{17\: \: \: \textbf{faktor}}{\overbrace{\qquad \: \begin{matrix} ...\\ \end{matrix}\: \qquad}}}}}{\displaystyle \underset{17\: \: \: \textbf{faktor}}{\underbrace{\qquad\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad\qquad}}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\qquad\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: \qquad}}\\\\ &=\textbf{...}^{^{\textbf{...}}} \end{aligned}.

Bentuk selanjutnya adalah

\begin{aligned}\left ( \textbf{b}^{\textbf{m}} \right )^{\textbf{n}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\times \left ( \textbf{b}^{\textbf{m}} \right )\: \: ...\: \times \left ( \textbf{b}^{\textbf{m}} \right )}}\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times ...\times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \\ \end{matrix}\qquad\: \: }}\\\\ &=\textbf{b}^{^{\textbf{...}}} \end{aligned}.

Silahkan coba soal berikut

\begin{aligned}\left ( \textbf{4}^{\textbf{5}} \right )^{\textbf{2017}}&=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\times \left ( \textbf{4}^{\textbf{5}} \right )\: \: ...\: \times \left ( \textbf{4}^{\textbf{5}} \right )}}\\\\ &=\displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \times ...\times \: \: \: \displaystyle \underset{...\: \: \: \textbf{faktor}}{\underbrace{\: \: \qquad\begin{matrix} .\: .\: .\\ \end{matrix}\qquad\: \: }}\: \: \: \\ \end{matrix}\qquad\: \: }}\\\\ &=\textbf{b}^{^{\textbf{...}}} \end{aligned}.

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Selamat Hari Raya Idul Fitri 1438 H ( Tahun 2017 )

Saya mengucapkan Selamat Hari Raya Idul Fitri 1438 H tahun 2017.

Mohon maaf lahir dan batin.

“Taqobballahu minna waminkum. Ja’alanallahu waiyyakum minal ‘aidin wal faizina. Kullu ‘amin wa antum bikhoir.”

Ahmad Thohir sekeluarga.

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Lanjutan Contoh Soal Turunan Fungsi

\begin{array}{ll}\\ \fbox{11}.&\textrm{Jika}\: \: \: f(x)=\displaystyle (x^{2}+2)\sqrt{x^{2}+x+3} \: ,\: \textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 13 \\\\ \textrm{b}.\quad \displaystyle 17 \quad &\textrm{c}.\quad \displaystyle 15 \quad &\textrm{e}.\quad 12 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}f(x)&=\displaystyle (x^{2}+2)\sqrt{x^{2}+x+3} \\ {f}\, '(x)&=(2x)\sqrt{x^{2}+x+3}+(x^{2}+2).\displaystyle \frac{1}{2}\left ( x^{2}+x+3 \right )^{^{-\frac{1}{2}}}.\left ( 2x+1 \right ) \\ &=2x\sqrt{x^{2}+x+3}+\displaystyle \frac{(x^{2}+2)(2x+1)}{2\sqrt{x^{2}+x+3}} \\ {f}\, '(2)&=2(2)\sqrt{(2)^{2}+(2)+3}+\displaystyle \frac{((2)^{2}+2)(2(2)+1)}{2\sqrt{(2)^{2}+(2)+3}} \\ &=4\sqrt{9}+\displaystyle \frac{6.5}{2\sqrt{9}} \\ &=4.3+\displaystyle \frac{6.5}{2.3}\\ &=12+5\\ &=17 \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika rusuk suatu kubus bertambah panjang dengan laju 7}\: \: cm/detik\: , \: \textrm{maka laju}\\ &\textrm{bertambahnya volume pada saat rusuk panjangnya 15}\: \: cm\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 675 \: \: \: \: cm^{3}/detik &&\textrm{d}.\quad 4725 \: \: \: \: cm^{3}/detik \\\\ \textrm{b}.\quad \displaystyle 1575 \: \: \: \: cm^{3}/detik \quad &\textrm{c}.\quad \displaystyle 3375 \: \: \: \: cm^{3}/detik \quad &\textrm{e}.\quad 23625 \: \: \: \: cm^{3}/detik \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{d}\\\\ \begin{aligned}\textrm{Laju }&\textrm{pertambahan volumenya}:\\\\ &\begin{cases} \bullet \: \: \: \frac{\mathrm{d} s}{\mathrm{d} t} & = 7\: \: cm/detik\\ \bullet \: \: \: V & =s^{3} \rightarrow \textrm{d}V=3s^{2}\: \: \textrm{d}s\: \: \: \textrm{atau}\\ \, \: \: \: \: \frac{\mathrm{d} V}{\mathrm{d} s} & =3s^{2}\: \: cm^{3}/cm\rightarrow s=15\: \: cm \end{cases}\\\\ \frac{\mathrm{d} V}{\mathrm{d} t}&=\frac{\mathrm{d} V}{\mathrm{d} t}\\ &=\frac{\mathrm{d} V}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}\\ &=3s^{2} \: \: \: \: cm^{3}/cm \times 7\: \: \: \: cm/detik \\ &=3(15)^{2}\times 7\: \: \: \: cm^{3}/detik\\ &=4725\: \: \: \: cm^{3}/detik \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{persamaan garis singgung di}\: \: x=1\: \: \textrm{pada kurva}\: \: y=x^{3}-3x^{2}+1\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad y=-3x+2&&\textrm{d}.\quad y=3x-2\\ \textrm{b}.\quad y=-3x+4\quad &\textrm{c}.\quad y=3x-4\quad &\textrm{e}.\quad y=-3x+3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{array}{|c|c|c|}\hline \textrm{Titik singgung di}\: \: x=1&\textrm{Gradien garis singgung di}\: \: x=1&\textrm{Persamaan garis singgung}\\\hline \begin{aligned}&\\ y&=x^{3}-3x^{2}+1\\ y&=(1)^{3}-3(1)^{2}+1\\ &=1-3+1\\ &=-1\\ &\\ \textrm{titik}&\: (a,b)=(1,-1) \end{aligned}&\begin{aligned}&\\ {y}\, '=m\, _{_{x=1}}&=3x^{2}-6x\\ &=3(1)^{2}-6(1)\\ &=3-6\\ &=-3\\ &\\ & \end{aligned}&\begin{aligned}y&=m(x-a)+b\\ &=-3(x-1)+(-1)\\ &=-3x+3-1\\ &=-3x+2\\ & \end{aligned}\\\hline \end{array}.

Berikut ilustrasi gambar kurva dan garis singgungnya

\begin{array}{ll}\\ \fbox{14}.&\textrm{Suatu kurva}\: \: y=x^{3}+2ax^{2}+b\: .\: \: \textrm{Sebuah garis}\: \: y=-9x-2\: \: \textrm{menyinggung}\\ &\textrm{kurva di titik dengan} \: \: x=1,\: \textrm{maka nilai}\: \: a \: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -3&&\textrm{d}.\quad 3\\ \textrm{b}.\quad -\displaystyle \frac{1}{3} \quad &\textrm{c}.\quad \displaystyle \frac{1}{3} \quad &\textrm{e}.\quad 8 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}\textrm{Gradien garis singgung}\: \: \: y\: _{_{_{di\: \: x=1}}}&\begin{cases} y & =x^{3}+2ax^{2}+b \rightarrow m={y}\, '=3x^{2}+4ax\\ y & =-9x-2 \rightarrow m=-9 \end{cases}\\ \textrm{Sehingga}\:, &\\ m&=m={y}\, '\\ -9&=3x^{2}+4ax\\ -9&=3(1)^{2}+4a(1)\\ -9&=3+4a\\ -4a&=3+9\\ a&=\displaystyle \frac{12}{-4}\\ &=-3\\ & \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika grafik fungsi}\: \: f(x)=x^{3}+ax^{2}+bx+c\: \: \textrm{hanya turun untuk interval}\: \: -1< x< 5\, ,\\ &\textrm{maka nilai}\: \: a+b \: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -21&&\textrm{d}.\quad 21\\ \textrm{b}.\quad -\displaystyle 9 \quad &\textrm{c}.\quad \displaystyle 9 \quad &\textrm{e}.\quad 24 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}\textrm{Diketahui bahwa}:\quad\qquad &\\ f(x)&=x^{3}+ax^{2}+bx+c\\ \textrm{grafik fungsi turun}\: &\: \textrm{berarti}:\: \: \: {f}\, '(x)< 0\\ {f}\, '(x)&< 0\\ 3x^{2}+2ax+b&< 0\\ (x+1)(x-5)&< 0 \: \quad \textrm{ini maksud pada interval}\: \: -1< x< 5\: \: \: \textrm{pada soal}\\ x^{2}-4x-5&< 0\: \quad \textrm{dikalikan dengan 3 supaya terjadi persamaan}\\ 3x^{2}-3.4x-3.5&=3x^{2}+2ax+b< 0\\ 3x^{2}+2(-6)x+(-15)&=3x^{2}+2ax+b< 0\\ &\begin{cases} a & =-6 \\ b & = -15 \end{cases}\\ \textrm{Sehingga}\: ,&\\ a+b&=-6+(-15)=-21\\ & \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Jika grafik fungsi}\: \: f(x)=5+15x+9x^{2}+x^{3}\: \: \textrm{naik untuk}\: \: x\: \: \textrm{yang memenuhi}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x< 1\: \: \textrm{atau}\: \: x> 5 &&\textrm{d}.\quad x< -5\: \: \textrm{atau}\: \: x> -1 \\ \textrm{b}.\quad -\displaystyle 1< x< 5 \quad &\textrm{c}.\quad \displaystyle -5< x< -1 \quad &\textrm{e}.\quad -5< x< 1 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{d}\\\\ \begin{aligned}\textrm{Diketa}&\textrm{hui bahwa}:\\ f(x)&=x^{3}+9x^{2}+15x+5.\\ \textrm{Dika}&\textrm{takan fungsi \textit{f} naik, maka}\\ {f}\, '(x)&> 0\\ 3x^{2}+&18x+15> 0,\qquad \textrm{masing-masing ruas dibagi 3}\\ x^{2}+&6x+5> 0\\ (x+1&)(x+5)> 0\\ &\\ \textrm{Jika}&\: \textrm{dibuatkan garis bilangan adalah sebagai berikut}\\ &\begin{array}{ll|llll|lll}\\ \multicolumn{8}{r}{.}\\\cline{1-2}\cline{7-8} &+&&-&-&&+&\\ &&&&&&&&X\\\cline{1-8} &\multicolumn{3}{l}{-5}&&\multicolumn{3}{l}{-1} \end{array} \end{aligned}.

Berikut ilustrasi gambar grafiknya

\begin{array}{ll}\\ \fbox{17}.&\textrm{Sebuah bola dilempar ke atas secara vertikal. Jika lintasan bola pada saat}\: \: t\\ &\textrm{detik adalah}\: \: \: h(t)=-\displaystyle \frac{1}{4}t^{4}+\frac{2}{3}t^{3}+4t^{2}+5\: \: \textrm{m},\: \: \textrm{maka tinggi maksimum yang} \\ &\textrm{dicapai oleh bola tersebut adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{67}{3} &&\textrm{d}.\quad \displaystyle \frac{133}{3} \\\\ \textrm{b}.\quad \displaystyle \frac{123}{3} \quad &\textrm{c}.\quad \displaystyle \frac{128}{3} \quad &\textrm{e}.\quad \displaystyle \frac{143}{3} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}\textrm{Perh}&\textrm{atikan bahwa lintasan bola saat dilempar vertikal dituliskan dengan fungsi}\\ h(t)&=-\displaystyle \frac{1}{4}t^{4}+\frac{2}{3}t^{3}+4t^{2}+5.\\ \textrm{mak}&\textrm{a tinggi maksimum akan dicapai bola saat}\: \: \: {h}\, '(t)=0,\\ \textrm{Sela}&\textrm{njutnya},\\ {h}\, '(t)&=0\\ -t^{3}+&2t^{2}+8t=0\\ -t(t^{2}&-2t-8)=0\\ t(t-&4)(t+2)=0\begin{cases} t & =0 \\ t & =4 \\ t & =-2 \end{cases}\\ \textrm{kita}&\: \textrm{ambil yang bernilai positif untuk \textit{t} yaitu}\: \: t=4.\\ t=4&\rightarrow h(4)=-\displaystyle \frac{1}{4}(4)^{4}+\frac{2}{3}(4)^{3}+4(4)^{2}+5\\ h(4)&=-64+\displaystyle \frac{128}{3}+64+5\\ &=\displaystyle \frac{143}{3}\: \: m \end{aligned}.

Berikut untuk ilustrasi gambarnya

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
  2. Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: Bumi Aksara.
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Alhamdulillah …. Marhaban Ya Ramadhan

Alhamdulillah kita masih dipertemukan dengan bulan yang penuh maghfiroh ini yakni bulan Ramadhan 1438 H semoga kita dapat mengisinya dengan berbagai amal solih dan semoga kita juga diberikan kesempatan untuk dapat memperbanyak amal solih tersebut. Dan juga semoga kita diberikan kekuatan oleh Allah untuk  dapat terhindar dari perbuatan-perbuatan maksiat. amiin

 

Marhaban ya ramadhan

alhamdulillah

 

 

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Contoh Soal Turunan Fungsi

\begin{array}{ll}\\ \fbox{1}.&\textrm{Diketahui}\: \: {f}\, '(2)=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2}, \textrm{maka fungsi}\: \: f(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 12&&\textrm{d}.\quad x^{3}\\ \textrm{b}.\quad 2\quad&\textrm{c}.\quad x\quad&\textrm{e}.\quad x-8 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\qquad \textbf{d}\\ \begin{array}{l}\\ \multicolumn{1}{c}{\begin{aligned}&\\ \textrm{Turunan fungsi f di}\: \: x&=c\: \: \textrm{adalah}\\ {f}\, '(c)&=\underset{x\rightarrow c}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)-f(c)}{x-c},\: \: \textrm{maka}\\ \textrm{turunan fungsi f di}\: \: x&=2\: \: \textrm{adalah}\\ {f}\, '(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{x^{3}-8}{x-2}\\ \textrm{sehingga akan didapa}&\textrm{tkan fungsi}\: \: f\: \: \textrm{nya yaitu}\\ f(x)&=x^{3}\\ & \end{aligned}}\\ \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Jika}\: \: a\neq 0\, ,\: \textrm{maka nilai dari}\: \: \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3a\displaystyle \sqrt[3]{a}&&\textrm{d}.\quad \displaystyle \frac{1}{2a}\sqrt[3]{a} \\\\ \textrm{b}.\quad 2a\displaystyle \sqrt[3]{a} \quad&\textrm{c}.\quad 0\quad&\textrm{e}.\quad \displaystyle \frac{1}{3a}\sqrt[3]{a} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\qquad \textbf{e}\\.

Alternatif 1

\begin{aligned}{f}\, '(a)&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{\left ( \sqrt[3]{x}-\sqrt[3]{a} \right )\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{\left ( \sqrt[3]{x^{2}}+\sqrt[3]{xa}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{\left ( \sqrt[3]{a^{2}}+\sqrt[3]{a.a}+\sqrt[3]{a^{2}} \right )}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\\ &=\displaystyle \frac{1}{3\sqrt[3]{a^{2}}}\times \displaystyle \frac{\sqrt[3]{a}}{\sqrt[3]{a}}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned}.

Alternatif 2 (dengan aturan L’Hopital) 

Jawaban ini sekaligus sebagai contoh penggunaan aturan L’Hopital pada bahasan sebelumnya di sini

\begin{aligned}\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x) }{{g}\, '(x) }\\ \underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \sqrt[3]{x}-\displaystyle \sqrt[3]{a}}{x-a}&=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{x^{^{\frac{1}{3}}}-a^{^{\frac{1}{3}}} }{x-a}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{\frac{1}{3}-1}}\: -0}{1-0}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{3}x^{^{-\frac{2}{3}}}}{1}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3\sqrt[3]{x^{2}}}\times \frac{\sqrt[3]{x}}{\sqrt[3]{x}}\\ &=\underset{x\rightarrow a}{\textrm{Lim}}\: \: \displaystyle \frac{1}{3x}\sqrt[3]{x}\\ &=\displaystyle \frac{1}{3a}\sqrt[3]{a} \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{1}{\sqrt{x}}\, ,\: \textrm{maka nilai dari}\: \: -2{f}\, '(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{x\sqrt{x}} &&\textrm{d}.\quad \displaystyle -\frac{1}{2x\sqrt{x}} \\\\ \textrm{b}.\quad \displaystyle x\sqrt{x} \quad&\textrm{c}.\quad \displaystyle -\frac{1}{2\sqrt{x}} \quad&\textrm{e}.\quad \displaystyle -2x\sqrt{x} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\qquad \textbf{d}\\\\ \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ f(x)&=\displaystyle \frac{1}{\sqrt{x}}\\ &=\displaystyle \frac{1}{x^{^{\frac{1}{2}}}}\\ &=x^{^{-\frac{1}{2}}}\\ & \end{aligned}}\\\hline y=ax^{n}\rightarrow {y}\, '=nax^{n-1}&\begin{aligned}&\\ y&=\displaystyle \frac{U}{V}\rightarrow {y}\, '=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}}\\ & \end{aligned}\\\hline \begin{aligned} {f}\, '(x)&=-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{2}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{2x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{2x^{1}.x^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}&\begin{aligned} {f}\, '(x)&=\displaystyle \frac{0.\sqrt{x}-1.\frac{1}{2}x^{^{\frac{1}{2}-1}}}{\left ( \sqrt{x}\right )^{2}}\\ &=\displaystyle \frac{-\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}}{x}\\ &=-\displaystyle \frac{1}{2xx^{^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{1}{2x\sqrt{x}} \end{aligned}\\\hline  \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Turunan pertama dari}\: \: y=\displaystyle \sqrt[n]{x}\: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{n}x^{^{\frac{1}{n}}}&&\textrm{d}.\quad (n-1)\sqrt[n-1]{x}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{n}x^{^{\frac{1-n}{n}}}\quad &\textrm{c}.\quad \displaystyle \frac{1}{n-1}x^{^{n-1}}\quad &\textrm{e}.\quad \sqrt[n-1]{x} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}y&=\sqrt[n]{x}\\ &=\displaystyle x^{^{\frac{1}{n}}}\\ {y}\, '&=\displaystyle \frac{1}{n}x^{^{\frac{1}{n}-1}}\\ &=\displaystyle \frac{1}{n}x^{^{\frac{1-n}{n}}} \end{aligned}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Turunan ke}-n\: \: \textrm{dari}\: \: \: y=\displaystyle \frac{1}{x} \: \: \textrm{adalah}.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle n!\: .\: x^{-(n+1)} &&\textrm{d}.\quad (-1)^{n+1}\: n!\: .\: x^{-(n+1)} \\\\ \textrm{b}.\quad \displaystyle (n+1)!\: .\: x^{-(n+1)} \quad &\textrm{c}.\quad \displaystyle (-1)^{n}\: n!\: .\: x^{-(n+1)} \quad &\textrm{e}.\quad (-1)^{n+1}\: (n+1)!\: .\: x^{-(n+1)} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{c}\\\\ \begin{array}{|c|c|c|c|c|c|c|}\hline &\multicolumn{6}{|c|}{\textrm{Turunan}}\\\cline{2-7} \raisebox{1.5ex}[0cm][0cm]{Fungsi}&\textrm{pertama}&\textrm{kedua}&\textrm{ketiga}&\textrm{keempat}&\cdots &\textrm{ke}-n\\\hline &{y}\, '&{y}\, ''&{y}\, '''&{y}\, ''''&\cdots &y\, ^{n}\\\cline{2-7} y=\displaystyle \frac{1}{x}&-x^{-2}&2x^{-3}&-6x^{-4}&24x^{-5}&\cdots &\\\cline{2-5} \begin{aligned}&\\ &\textrm{atau}\\ & \end{aligned}&-1.x^{-(1+1)}&-2.-1.x^{-(2+1)}&-3.-2.-1.x^{-(3+1)}&-4.-3.-2.-1.x^{-(4+1)}&&\\\cline{2-5} \begin{aligned}&\\ y&=x^{-1}\\ & \end{aligned} &(-1)^{1}.1!.x^{-(1+1)}&(-1)^{2}.2!.x^{-(2+1)}&(-1)^{3}.3!.x^{-(3+1)}&(-1)^{4}.4!.x^{-(4+1)}&&(-1)^{n}.n!.x^{-(n+1)}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: W=\sin 2t\: ,\: \textrm{maka}\: \: \displaystyle \frac{dW}{dt}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos 2t&&\textrm{d}.\quad 2t\cos 2t+\sin 2t\\ \textrm{b}.\quad 2\cos 2t\quad &\textrm{c}.\quad \sin 2t+t\cos 2t\quad &\textrm{e}.\quad \sin 2t-t\cos 2t \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{b}\\\\ \begin{aligned}\textrm{Diketahui bahwa}&:\\ &\\ W&=\sin 2t\\ W&=\sin u\, ,\qquad\quad \textrm{dengan}\: \: u=2t\\ \displaystyle \frac{dW}{dt}&=\displaystyle \frac{dW}{du}.\frac{du}{dt}\\ &=\cos u.2\\ &=2\cos u\\ &=2\cos 2t \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{2x+4}{1+\sqrt{x}}\: ,\: \textrm{maka}\: \: \displaystyle {f}\, '(4) =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4} &&\textrm{d}.\quad 1 \\\\ \textrm{b}.\quad \displaystyle \frac{3}{7} \quad &\textrm{c}.\quad \displaystyle \frac{3}{5} \quad &\textrm{e}.\quad 4 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{a}\\\\ \begin{aligned}f(x)&=\displaystyle \frac{2x+4}{1+\sqrt{x}}\\ &=\displaystyle \frac{U}{V}\\ {f}\, '(x)&=\displaystyle \frac{{U}\, '.V-U.{V}\, '}{V^{2}}\\ &=\displaystyle \frac{(2)\left ( 1+\sqrt{x} \right )-(2x+4).\left ( 1.\left ( 1+\sqrt{x} \right )^{0} .\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )}{\left ( 1+\sqrt{x} \right )^{2}} \, ,\qquad\quad \textrm{ingat}\: \: \sqrt{x}=x^{^{\frac{1}{2}}} \\ {f}\, '(4) &=\displaystyle \frac{2\left ( 1+\sqrt{4} \right )-(2.4+4).\frac{1}{2}. 4^{^{-\frac{1}{2}}} }{\left ( 1+\sqrt{4} \right )^{2}}\\ &=\displaystyle \frac{2(1+2)-(12).\frac{1}{2}. \frac{1}{2}}{(1+2)^{2}}\, ,\qquad\quad \textrm{ingat juga}\: \: 4^{^{-\frac{1}{2}}}=\left ( 2^{2} \right )^{^{-\frac{1}{2}}}=2^{^{-1}}=\displaystyle \frac{1}{2^{1}}=\frac{1}{2}\\ &=\displaystyle \frac{6-3}{9}\\ &=\displaystyle \frac{1}{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Jika}\: \: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x} \: ,\: \textrm{maka}\: \: {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2} &&\textrm{d}.\quad 2 \\\\ \textrm{b}.\quad \displaystyle -\frac{1}{2} \quad &\textrm{c}.\quad \displaystyle \frac{3}{4} \quad &\textrm{e}.\quad -2 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{d}\\\\ \begin{aligned}f(x)&=\displaystyle \frac{1+\sin x}{\cos x}\\ {f}\, '(x)&=\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x +\sin ^{2}x}{\cos ^{2}x}\: ,\qquad \textrm{ingat bahwa}\: \: \sin ^{2}x+\cos ^{2}x=1\\ &=\displaystyle \frac{1+\sin x }{\cos ^{2}x}\\ {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )&=\displaystyle \frac{1+\sin \left ( \displaystyle \frac{1}{6}\pi \right )}{\cos ^{2}\left ( \displaystyle \frac{1}{6}\pi \right )}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{2} }{\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{3}{2}\times \frac{4}{3}\\ &=2 \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Jika}\: \: \: f(x)=\displaystyle 3x^{2}-2ax+7\: \: \textrm{dan}\: \: {f}\, '(1)=0 \: ,\: \textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1 &&\textrm{d}.\quad 6 \\\\ \textrm{b}.\quad \displaystyle 2 \quad &\textrm{c}.\quad \displaystyle 4 \quad &\textrm{e}.\quad 8 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{d}\\\\ \begin{aligned}f(x)&=\displaystyle 3x^{2}-2ax+7\\ {f}\, '(x)&=6x-2a\\ {f}\, '(1)&=0\\ 6(1)-2a&=0\\ 6&=2a\\ 3&=a\\ \textrm{sehingga}\, &\: \\ {f}\, '(x)&=6x-6\\ \textrm{maka}\, ,\: \quad &\\ {f}\, '(2)&=6.2-6\\ &=12-6\\ &=6 \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Jika}\: \: \: f(x)=\displaystyle (6x-3)^{3}(2x-1)\: ,\: \textrm{maka}\: \: {f}\, '(1)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 162 \\\\ \textrm{b}.\quad \displaystyle 24 \quad &\textrm{c}.\quad \displaystyle 54 \quad &\textrm{e}.\quad 216 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\quad \textbf{e}\\\\ \begin{aligned}f(x)&=\displaystyle (6x-3)^{3}(2x-1)\\ &=(3.(2x-1))^{3}(2x-1)^{1}\\ &=3^{3}.(2x-1)^{3+1}\\ &=27(2x-1)^{4}\\ {f}\, '(x)&=4.27(2x-1)^{4-1}.2\\ &=216.(2x-1)^{3}\\ {f}\, '(1)&=216.(2.1-1)^{3}\\ &=216.1\\ &=216 \end{aligned}.

Catatan:

Pada penyelesaian soal semisal pada No. 10, cara penyelesaian tidak harus seperti yang dicontohkan. Kebetulan pada nomor tersebut fungsi tersebut dapat disederhanakan dan sebagai alternatif cara penyelesaian yang lain Anda dapat menggunakan rumus U. V dan hasil akhirnya akan sama. Untuk penggunan rumus U.V, Anda dapat perhatikan pembahasan No. 11  pada halaman berikutnya

 

 

 

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Penggunaan Turunan Fungsi (Lanjutan Turunan Fungsi)

Terdapat pada :

  • Persamaan garis singgung
  • Fungsi naik dan fungsi turun
  • menggambar grafik fungsi aljabar
  • Maksimum dan minimum fungsi
  • Teorema L’Hopital  (dibaca: Lopital)
  • Nilai stasioner
  • Titik belok
  • Kecepatan dan percepatan

Perhatikanlah tabel berikut!

\begin{array}{|c|c|l|c|}\hline \multicolumn{3}{|c|}{\textrm{Turunan Pertama}}&\textrm{Turunan Kedua}\\\hline 1.&\textrm{\textrm{Gradien garis singgung}}&m={f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}=\displaystyle \frac{f(x+h)-f(x)}{h}&\textrm{Belok}\\\cline{1-3} 2.&\textrm{Fungsi naik dan turun}&y=f(x)\begin{cases} {f}'(x)> 0, & \text{ fungsi naik } \\ {f}'(x)< 0, & \text{ fungsi turun } \end{cases}&\textrm{Percepatan}\\\cline{1-3} 3.&\textrm{Jarak, kec, percepatan}&y=s(t)\begin{cases} s(t) & \text{ jarak} \\ {s}\, '(t) & \text{ kecepatan } \\ {s}\, ''(t) & \text{ percepatan} \end{cases}&\textrm{Maksimum}\\\cline{1-3} &&\begin{aligned}\textrm{Maksimum}:&\\ \rightarrow {f}&\, ''(k)< 0\\ \textrm{titik mak}&\: \left ( k, f(k)\right ) \end{aligned}&\textrm{Minimum}\\\cline{3-3} 4.&\textrm{Stasioner}&\begin{aligned}\textrm{Minimum}:&\\ \rightarrow {f}&\, ''(k) > 0\\ \textrm{titik min}&\: \left ( k, f(k)\right ) \end{aligned}&\\\cline{3-3} &{f}\, '(x)=0\rightarrow x=k&\begin{aligned}\textrm{Belok}:&\\ \rightarrow {f}&\, ''(k)= 0\\ \textrm{titik belok}&\: \left ( k, f(k)\right ) \end{aligned}&\\\cline{1-3} 5.&\begin{aligned}&\textrm{Limit fungsi}\\ &\textrm{bentuk tak tentu} \end{aligned}&\begin{aligned}&\textrm{Aturan L'Hopital}\\ &\\ &\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{f(x)}{g(x)}=\underset{x\rightarrow h}{\textrm{Lim}}\: \: \displaystyle \frac{{f}\, '(x)}{g\,'(x)}\\ &\\ &\textrm{untuk hasil limit}\\ &\textrm{bentuk}\: \: \frac{0}{0}\: \: \textrm{atau}\: \: \frac{\infty }{\infty }\end{aligned}&\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah persamaan garis singgung pada kurva}\: \: y=x^{2}+2x-8\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: 1 \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Diketahui},\: &\textrm{persamaan sebuah kurva adalah:}\\ &y=x^{2}+2x-8\\ & \end{aligned}}\\\hline \textrm{Titik singgung}&\textrm{Gradien}\: ,\: x=1&\textrm{Persamaan garis singgung}\\\hline \begin{aligned}x=1\rightarrow y&=(1)^{2}+2(1)-8\\ &=1+2-8\\ &=-5\\ &\\ \textrm{di titik}&\: \: (a,b)=(1,-5)\end{aligned}&\begin{aligned}\displaystyle \frac{dy}{dx}=m&=2x+2\\ &=2(1)+2\\ &=4\\ &\\ & \end{aligned}&\begin{aligned}y&=m(x-a)+b\\ &=4(x-1)+(-5)\\ &=4x-4-5\\ &=4x-9\\ & \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Jadi},\: &\textrm{persamaan garis singgungnya adalah:}\\ &y=4x-9\Leftrightarrow y-4x+9=0\\ & \end{aligned}}\\\hline \end{array}.

Berikut ilustasi gambar dari persoalan di atas

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah interval di mana kurva fungsi}\: \: f(x)=x^{3}+3x^{2}-9x+5\\ &\textrm{a}.\quad \textrm{naik}\\ &\textrm{b}.\quad \textrm{turun} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ f(x)&=x^{3}+3x^{2}-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ &=3(x+3)(x-1)\\ & \end{aligned}}\\\hline \textrm{naik}\: ;\: \: {f}\, '(x)> 0&\textrm{turun}\: ;\: \: {f}\, '(x)< 0\\\hline \begin{aligned}&\\ &3(x+3)(x-1)> 0\\ & \end{aligned}&\begin{aligned}&\\ &3(x+3)(x-1) < 0\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{array}{ll|lll|lll}\\ &\multicolumn{3}{c}{.}&\multicolumn{3}{c}{.}&\\\cline{1-2}\cline{6-7} +&+&-&-&-&+&+&\\ &&&&&&\\\hline &\multicolumn{2}{l}{-3}&&\multicolumn{2}{c}{1}&&\\ \end{array} }\\\hline \begin{aligned}&\\ \textrm{naik}\: ,&\: x< -3\: \: \textrm{atau}\: \: x> 1\\ & \end{aligned}&\begin{aligned}&\\ \textrm{turun}\: ,&\: -3< x< 1\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai stasioner fungsi}\: \: f(x)=x^{3}+3x^{2}-9x+5 \: \: \textrm{dan tentukan pula jenisnya}.\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\\ f(x)&=x^{3}+3x^{2}-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ &=3(x+3)(x-1)\\ &\textrm{dan}\\ &\begin{array}{ll|lll|lll}\\ &\multicolumn{3}{c}{.}&\multicolumn{3}{c}{.}&\\\cline{1-2}\cline{6-7} +&+&-&-&-&+&+&\\ &&&&&&\\\hline &\multicolumn{2}{l}{-3}&&\multicolumn{2}{c}{1}&&\\ \end{array} \\ &\textrm{untuk nilai stasionernya} \\ f(-3)&=(-3)^{3}+3(-3)^{2}-9(-3)+5=32&\rightarrow \left ( -3,32 \right )\: \textrm{adalah titik balik maksimum}\\ f(1)&=(1)^{3}+3(1)^{2}-9(1)+5=0&\rightarrow \left ( 1,0 \right )\: \textrm{adalah titik balik minimum} \end{aligned}.

Berikut ilustrasi gambar kurvanya baik untuk jawaban No. 2 maupun No. 3

\begin{array}{ll}\\ 4.&\textrm{Masih sama dengan soal seperti pada No. sebelumnya, yaitu fungsi}\: \: f(x)=x^3+3x^2-9x+5\: .\\ & \textrm{Tentukanlah koordinat titik beloknya}\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|c|l|c|}\hline \multicolumn{6}{|c|}{\begin{aligned}&\\ f(x)&=x^3+3x^2-9x+5\\ {f}\, '(x)&=3x^{2}+6x-9\\ {f}\, ''(x)&=6x+6\\ \textrm{Proses}\: &\textrm{mencari beloknya}\\ {f}\, ''(x)&=0\\ 6x+6&=0\\ 6x&=-6\\ x&=-1\\ & \end{aligned}}\\\hline \textrm{Interval}&f(x)&{f}\, '(x)&{f}\, ''(x)&\: \: \qquad \textrm{Keterangan}&\textrm{Koordinat titik beloknya}\\\hline x< -1&&&-&\textrm{grafik cekung ke bawah}&\\\cline{1-5} x= -1&16&-12&0&\textrm{grafik memiliki titik belok}&(-1,16)\\\cline{1-5} x > -1&&&+&\textrm{grafik cekung ke atas}&\\\hline \end{array}.

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Turunan Fungsi (KTSP MA/SMA Kelas XI)


A. Turunan Fungsi

A. 1  Pendahuluan

mengenal laju perubahan untuk nilai fungsi

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Laju Perubahan}}\\\hline \textrm{Laju erubahan rata-rata}&\textrm{Laju perubahan sesaat}\\\hline \begin{aligned}&\\ &\displaystyle \frac{\Delta y}{\Delta x}=\displaystyle \frac{f\left ( x_{2} \right )-f\left ( x_{1} \right )}{x_{2}-x_{1}}\\ & \end{aligned}&\begin{aligned}&\\ &\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(a+h)-f(a)}{h}\\ & \end{aligned}\\\hline \end{array}.

A. 2  Pengertian

Perhatikanlah ilustrasi berikut!

Misalkan diketahui fungsi   y=f(x)   terdefinisi untuk semua harga x di sekitar  x=k . Jika  \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}  ada, maka bentuk  \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}  disebut turunan dari fungsi  f(x)  saat  x=k.

A. 3  Notasi

  • Notasi turunan fungsi dilambangkan dengan  {f}'(k)   dengan  {f}'(k)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(k+h)-f(k)}{h}.
  • Lambang  {f}'(k)  dibaca  f  aksen  k  desebut  turunan atau derivatif  untuk fungsi  f(x)   terhadap  x   saat   x=k.
  • Jika limitnya ada, dapat dikatakan fungsi  f(x)   diferensiabel(dapat didiferebsialkan)  saat  x=k   dan bentuk limitnya selanjutna dilambangkan dengan  {f}'(k).
  • Misalkan fungsi  f(x)  memiliki turunan  {f}'(x) . Jika   {f}'(k)   tidak terdefinisi  maka  f(x)   tidak diferensiabel  di   x=k .

A. 4  Bentuk Umum (Turunan Pertama)

Bentuk umum turunan fungsi ang selanjutnya disebut juga turunan pertama  fungsi  y   terhadap  x  dapat dinotasikan dengan berbagai bentuk berikut yaitu:

{y}\: '=\displaystyle \frac{dy}{dx}=\displaystyle \frac{d\left ( f(x) \right )}{dx}={f}'(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Jika} \: \: g(x)=3x-5, \textrm{hitunglah laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{g(x)=3x-5}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned} g(x)&=3x-5\\ g(2)&=3.2-5=1\\ {g}'(2)&=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{g(x)-g(2)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{(3x-5)-(1)}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle \frac{3x-6}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{Lim}}\: \: \displaystyle 3\\ &=3 \end{aligned}&\begin{aligned}g(2)\: \,&=1\\ g(2+&h)=3(2+h)-5=3h+1\\ g(2+&h)-g(2)=3h\\ {g}'(2)\: &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{g(2+h)-g(2)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle\frac{3h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 3\\ &=3\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Jadi, laju perubahan fungsi}\: \: g\: \: \textrm{di}\: \: x=2\: \: \textrm{adalah 3}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: f(x)=2017x^{2},\: \: \textrm{tentukanlah}\: \: {f}'(x)\: \: \textrm{dan}\: \: {f}'(1)\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|c|c|}\hline {f}'(x)&{f}'(1)\\\hline \begin{aligned}f(x)&=2017x^{2}\\ f(x+h)&=2017(x+h)^{2}\\ &=2017\left ( x^{2}+2xh+h^{2} \right )\\ &=2017x^{2}+4034xh+2017h^{2}\\ {f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (2017x^{2}+4034xh+2017h^{2} \right )-\left (2017x^{2} \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{4034xh+2017h^{2}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 4034x+2017h\\ &=4034x \end{aligned}&\begin{aligned}{f}'(x)&=4034x\\ \textrm{maka},&\\ {f}'(1)&=4034.1\\ &=4034\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}\end{array}.

\begin{array}{ll}\\ 3.&\textrm{Diketahui bahwa fungsi}\: \: f(x)=\displaystyle \frac{1}{x^{2}}\: \: \textrm{dengan daerah asal}\: \: \textrm{D}_{f}=\left \{ x\: |\: x\in \mathbb{R},\: \: \textrm{dan}\: \: x\neq 0 \right \}.\\ &\textrm{a}.\quad \textrm{Tunjukkan bahwa}\: \: {f}'(a)=\displaystyle -\frac{2}{a^{3}}\\ &\textrm{b}.\quad \textrm{jelaskanlah mengapa}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\textrm{Jawab}:\\\\ &\begin{array}{|l|l|}\hline \begin{aligned}{f}'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{1}{(x+h)^{2}}-\displaystyle \frac{1}{x^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{x^{2}-(x+h)^{2}}{\left (x(x+h) \right )^{2}}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{x^{2}-\left ( x^{2}+2xh+h^{2} \right )}{h\left ( x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2xh-h^{2}}{h\left (x(x+h) \right )^{2}}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: -\displaystyle \frac{2x+h}{\left (x(x+h) \right )^{2}}\\ &=-\displaystyle \frac{2x}{x^{4}}\\ &=-\displaystyle \frac{2}{x^{3}} \end{aligned}&\begin{aligned}\textrm{Untuk}&\: \textrm{jawaban poin a dan b adalah sebagai berikut}\\ {f}'(x)&=-\displaystyle \frac{2}{x^{3}}\\ {f}'(a)&=-\displaystyle \frac{2}{a^{3}}\\ \textrm{maka},&\\ {f}'(0)&=-\displaystyle \frac{2}{0^{3}}\\ &=-\displaystyle \frac{2}{0}\\ &\\ &\textrm{karena penyebut berupa bilangan}\: \: 0\\ &\textrm{maka}\: \: {f}'(0)\: \: \textrm{tidak terdefinisi}\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Selanjutnya berikut beberapa hasil turunan dari berbagai fungsi aljabar}\\ &\\ &\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|}\hline y&\cdots &x^{-2}&x^{-1}&C&x&x^{2}&x^{3}&x^{4}&x^{5}&\cdots \\\hline \multicolumn{11}{|c|}{\begin{aligned}&\\ &\textrm{dan hasilnya}\\ & \end{aligned}}\\\hline {y}\: '&\cdots &-\displaystyle \frac{2}{x^{3}}&-\displaystyle \frac{1}{x^{2}}&0&1&2x&3x^{2}&4x^{3}&5x^{4}&\cdots \\\hline \end{array} \end{array}.

B. Rumus Turunan

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Untuk}:&\\ &\begin{cases} a & \in \mathbb{R} \\ n& \in \mathbb{Q}\\ c& \textrm{konstanta}\\ U&=g(x)\\ V&=h(x) \end{cases}\\ & \end{aligned}}\\\hline \textrm{Sifat-Sifat}&\textrm{Fungsi Aljabar}&\textrm{Fungsi Trigonometri}\\\hline \begin{aligned}y&=c\rightarrow &{y}'&=0\\ y&=c.U\rightarrow &{y}'&=c.{U}'\\ y&=U\pm V\rightarrow &{y}'&={U}'\pm {V}' \end{aligned}&\begin{aligned}y&=a.x^{n}\rightarrow {y}\: '=n.a.x^{(n-1)}\end{aligned}&\begin{aligned}y&=a\sin x\rightarrow & {y}'&=a\cos x\\ y&=a\cos x\rightarrow &{y}'&=-a\sin x\\ y&=a\tan x\rightarrow &{y}'&=a\sec ^{2}x\end{aligned}\\\cline{2-3} \begin{aligned}y&=U.V\rightarrow &{y}'&={U}'.V+U.{V}'\\ y&=\displaystyle \frac{U}{V}\rightarrow &{y}'&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=a.U^{n}\rightarrow {y}\: '=n.a.U^{(n-1)}.{U}'\end{aligned}&\begin{aligned}y&=a\sin U\rightarrow & {y}'&=\left (a\cos U \right ).{U}'\\ y&=a\cos U\rightarrow &{y}'&=\left (-a\sin U \right ).{U}'\\ y&=a\tan U\rightarrow &{y}'&=\left (a\sec ^{2}U \right ).{U}'\end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Aturan}&\: \: \textrm{rantai untuk turunan pada} \: \: y=f(u),\: \: \textrm{maka}\\ \textrm{untuk}\: \: &u\: \: \textrm{merupakan fungsi}\: \: x,\: \: \textrm{adalah}:\\ &\\ {y}\: '&={f}\: '(x).{u}'\\ &\textrm{atau}\\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ & \end{aligned}}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Dengan menggunakan rumus turunan}\: \: {f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h},\: \: \textrm{tunjukkan bahwa turunan}\: \: \\ &\textrm{a}.\quad f(x)=ax^{n}\: \: \textrm{adalah}\: \: {f}\, '(x)=n.ax^{n-1}\\ &\textrm{b}.\quad f(x)=u(x)+v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x)+{v}\,'(x)\\ &\textrm{c}.\quad f(x)=u(x).v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x).v(x)+u(x).{v}\,'(x)\\ &\textrm{d}.\quad f(x)=\displaystyle \frac{u(x)}{v(x)}\: \: \textrm{adalah}\: \: {f}\, '(x)=\displaystyle \frac{{u}\,'(x).v(x)+u(x).{v}\,'(x)}{(v(x))^{2}} \\ &\textrm{e}.\quad f(x)=\sin x\: \: \textrm{adalah}\: \: {f}\, '(x)=\cos x \\ &\textrm{f}.\quad f(x)=\cos x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\sin x \\ &\textrm{g}.\quad f(x)=\tan x\: \: \textrm{adalah}\: \: {f}\, '(x)=\sec^{2} x \\ &\textrm{h}.\quad f(x)=\cot x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\csc^{2} x \\ \end{array}.

Bukti:

\begin{array}{ll}\\ 1.\: \textrm{a}&{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\, '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a(x+h)^{n}-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left ( x^{n}+\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left (\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\left (\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}.h+\binom{n}{3}x^{n-3}.h^{2}+\cdots +\binom{n}{n-1}x.h^{n-2}+h^{n-1}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle a\binom{n}{1}x^{n-1}+a\binom{n}{2}x^{n-2}.h+a\binom{n}{3}x^{n-3}.h^{2}+\cdots +ah^{n-1}\\ &=a\binom{n}{1}x^{n-1}+0+0+...+0\\ &=a.\displaystyle \frac{n!}{(n-1)!.1!}x^{n-1}\\ &=a.\displaystyle \frac{n.(n-1)!}{(n-1)!}x^{n-1}\\ &=a.n.x^{n-1}\qquad\quad \blacksquare \end{aligned} \end{array}.

\begin{array}{ll}\\ 1.\: \textrm{g}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h-\tan x+\tan^{2}x.\tan h}{1-\tan x.\tan h}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h\left ( 1+\tan^{2}x \right )}{h\left ( 1-\tan x.\tan h \right )}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h}{h}\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1+\tan^{2}x}{1-\tan x.\tan h}\\ &=1 \times \displaystyle \frac{1+\tan^{2}x}{1-0}\\ &=1+\tan^{2}x\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned} \end{array}.

Untuk bukti yang lain silahkan dicoba sebagai latihan mandiri.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=x^{6}&\textrm{i}.&f(x)=(x+1)(x-2)&\textrm{q}.&f(x)=\displaystyle \frac{2}{x^{3}}\\ \textrm{b}.&f(x)=\displaystyle \frac{1}{2}x^{2}&\textrm{j}.&f(x)=(3-x)(5-x)&\textrm{r}.&f(x)=\displaystyle \frac{1}{2\sqrt{x}}\\ \textrm{c}.&f(x)=-2x^{5}&\textrm{k}.&f(x)=(2x+3)^{2}&\textrm{s}.&f(x)=\displaystyle \frac{1}{3x^{5}}\\ \textrm{d}.&f(x)=ax^{3}&\textrm{l}.&f(x)=(x-2)^{3}&\textrm{t}.&f(x)=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ \textrm{e}.&f(x)=4x^{4}-x^{2}+2017&\textrm{m}.&f(x)=(4x+1)(4x-1)&\textrm{u}.&f(x)=\displaystyle \frac{x}{2}+\frac{2}{x}\\ \textrm{f}.&f(x)=6-x-3x^{2}&\textrm{n}.&f(x)=(x-1)(x+1)(x+2)&\textrm{v}.&f(x)=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ \textrm{g}.&f(x)=(x-3)^{2}&\textrm{o}.&f(x)=\displaystyle \frac{1}{2}x^{-4}&\textrm{w}.&f(x)=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ \textrm{h}.&f(x)=\left ( x^{3}-2 \right )^{2}&\textrm{p}.&f(x)=x^{-5}&\textrm{x}.&f(x)=\displaystyle \frac{2+4x}{x^{2}}\\ &&&&\textrm{y}.&f(x)=\left ( x+1+\displaystyle \frac{1}{x} \right )\left ( x+1-\displaystyle \frac{1}{x} \right ) \end{array} \end{array}.

Jawab:

Untuk  y=f(x)  maka ,

\begin{array}{|c|c|c|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}&\textrm{d}&\textrm{e}&\textrm{f}\\\hline \begin{aligned}y&=x^{6}\\ {y}\: '&=6x^{6-1}\\ &=6x^{5} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{2}\\ {y}\: '&=2.\displaystyle \frac{1}{2}x^{2-1}\\ &=x \end{aligned}&\begin{aligned}y&=-2x^{5}\\ {y}\: '&=-5.2x^{5-1}\\ &=-10x^{4} \end{aligned}&\begin{aligned}y&=ax^{3}\\ {y}\: '&=3.a.x^{3-1}\\ &=3.a.x^{2} \end{aligned}&\begin{aligned}y&=4x^{4}-x^{2}+2017\\ {y}\: '&=4.4x^{4-1}-2.x^{2-1}+0\\ &=16x^{3}-2x \end{aligned}&\begin{aligned}y&=6-x-3x^{2}\\ {y}\: '&=0-1.x^{1-1}-2.3x^{2-1}\\ &=-1-6x \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{g}&\textrm{h}&\textrm{i}&\textrm{j}\\\hline \begin{aligned}y&=(x-3)^{2}\\ {y}\: '&=2.(x-3)^{2-1}.1\\ &=2(x-3)\\ &\end{aligned}&\begin{aligned}y&=\left ( x^{3}-2 \right )^{2}\\ {y}\: '&=2.\left ( x^{3}-2 \right )^{2-1}.3x^{2}\\ &=6\left ( x^{3}-2 \right )x^{2}\\ &=6x^{5}-12x^{2} \end{aligned}&\begin{aligned}y&=(x+1)(x-2)\\ {y}\: '&=1.(x+2)+(x+1).1\\ &=2x+3\\ & \end{aligned}&\begin{aligned}y&=(3-x)(5-x)\\ {y}\: '&=-1.(5-x)+(3-x).-1\\ &=2x-8 \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{k}&\textrm{l}&\multicolumn{2}{|c|}{\textrm{m}}\\\hline \begin{aligned}y&=(2x+3)^{2}\\ {y}\: '&=2.(2x+3)^{2-1}.2\\ &=4(2x+3)^{1}\\ &=8x+12\\ &\\ & \end{aligned}&\begin{aligned}y&=(x-2)^{3}\\ {y}\: '&=3(x-2)^{3-1}.1\\ &=3(x-2)^{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 1}\\ {y}\: '&=4(4x-1)+(4x+1).4\\ &=16x-4+16x+4\\ &=32x\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 2}\\ y&=16x^{2}-1\\ {y}\: '&=2.16x^{2-1}-0\\ &=32x^{1}\\ &=32x \end{aligned}\\\hline \textrm{n}&\textrm{o}&\textrm{p}&\textrm{q}\\\hline \begin{aligned}y&=(x-1)(x+1)(x+2)\\ &=(x^{2}-1)(x+2)\\ &=x^{3}+2x^{2}-x-2\\ {y}\: '&=3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ &=3x^{2}+4x-1\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{-4}\\ {y}\: '&=-4.\displaystyle \frac{1}{2}x^{-4-1}\\ &=-2x^{-5}\\ &=-\displaystyle \frac{2}{x^{5}}\\ & \end{aligned}&\begin{aligned}y&=x^{-5}\\ {y}'\: &=-5.x^{-5-1}\\ &=-5x^{-6}\\ &=-\displaystyle \frac{5}{x^{6}}\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2}{x^{3}}\\ &=2x^{-3}\\ {y}\: '&=-3.2x^{-3-1}\\ &=-6x^{-4}\\ &=-\displaystyle \frac{6}{x^{4}} \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{r}&\textrm{s}&\textrm{t}&\textrm{u}\\\hline \begin{aligned}y&=\displaystyle \frac{1}{2\sqrt{x}}\\ &=\displaystyle \frac{1}{2x^{\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ {y}\: '&=-\displaystyle \frac{1}{2}.\frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{4}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{4x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{4}\sqrt{x^{3}} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{3x^{5}}\\ &=\displaystyle \frac{1}{3}x^{-5}\\ {y}\: '&=-5.\displaystyle \frac{1}{3}x^{-5-1}\\ &=-\displaystyle \frac{5}{3}x^{-6}\\ &=-\displaystyle \frac{5}{3x^{6}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ &=2x^{4}+\displaystyle \frac{1}{2}x^{-3}\\ {y}\: '&=4.2x^{4-1}+(-3).\displaystyle \frac{1}{2}x^{-3-1}\\ &=8x^{3}-\displaystyle \frac{3}{2}x^{-4}\\ &=8x^{3}-\displaystyle \frac{3}{2x^{4}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{x}{2}+\frac{2}{x}\\ &=\displaystyle \frac{1}{2}x+2x^{-1}\\ {y}\: '&=\displaystyle \frac{1}{2}x^{1-1}+(-1).2x^{-1-1}\\ &=\displaystyle \frac{1}{2}x^{0}-2x^{-2}\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{2}{x^{2}}\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \textrm{v}&\textrm{w}\\\hline \begin{aligned}y&=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ &=\left ( 2x^{3}+x^{-1} \right )^{2}\\ {y}\: '&=2.\left ( 2x^{3}+x^{-1} \right )^{2-1}.\left ( 3.2x^{3-1}+(-1)x^{-1-1} \right )\\ &=2.\left ( 2x^{3}+x^{-1} \right )^{1}.\left ( 6x^{2}-x^{-2} \right )\\ &=2\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )\left ( 6x^{2}-\displaystyle \frac{1}{x^{2}} \right )\\ &=2\left ( 12x^{5}-2x+6x-\displaystyle \frac{1}{x^{3}} \right )\\ &=24x^{5}+8x-\displaystyle \frac{2}{x^{3}}\\ & \end{aligned}&\begin{aligned}y&=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ &=x^{2}\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}\\ {y}\: '&=2x.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}+x^{2}.2.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2-1}.\left ( 0+\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( 1+\sqrt{x} \right ).\left ( \displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( \displaystyle \frac{1}{2x^{^{\frac{1}{2}}}} \right ).\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{2-\frac{1}{2}}}.\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{\frac{3}{2}}}\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+\left ( x\sqrt{x}+x^{2} \right ) \end{aligned}\\\hline \end{array}.

\begin{array}{|c|c|c|c|}\hline \multicolumn{2}{|c|}{\textrm{x}}\\\hline \textrm{Cara Pertama}&\textrm{Cara Kedua}\\\hline \begin{aligned}y&=\displaystyle \frac{U}{V}\\ {y}\: '&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=UV\\ {y}\: '&={U}'.V+U.{V}' \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}}\\\hline \begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ {y}\: '&=\displaystyle \frac{(4)(x^{2})-(2+4x).(2x)}{\left ( x^{2} \right )^{2}}\\ &= \displaystyle \frac{4x^{2}-4x-8x^{2}}{x^{4}}\\ &=\displaystyle \frac{-4x^{2}-4x}{x^{4}}\\ &=\displaystyle \frac{-4x-4}{x^{3}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ &=(2+4x).x^{-2}\\ {y}\: '&=(4).x^{-2}+(2+4x).-2x^{-2-1}\\ &=4x^{-2}-(4+8x).x^{-3}\\ &=4x^{-2}-4x^{-3}-8x^{-2}\\ &=-4x^{-3}-4x^{-2}\\ &=\displaystyle \frac{-4}{x^{3}}+\displaystyle \frac{-4}{x^{2}}\\ &=\displaystyle \frac{-4-4x}{x^{3}}\\ &=\displaystyle \frac{-4x-4}{x^{3}} \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Ruas kiri = Ruas kanan}}\\\hline \end{array}.

\begin{array}{|l|}\hline \begin{aligned}&\\ &\begin{array}{|r|l|}\hline \textbf{Perhatikanlah bukti poin 1. g}&\\\cline{1-1} \multicolumn{2}{|l|}{.}\\ \multicolumn{2}{|c|}{\LARGE\textrm{\textbf{Cara lain :}}}\\ \multicolumn{2}{|r|}{.}\\\cline{2-2} &\begin{aligned}f(x)&=\tan x=\displaystyle \frac{\sin x}{\cos x}\\\\ &=\displaystyle \frac{U}{V}\quad \Rightarrow \quad \begin{cases} U=\sin x & \rightarrow {U}'=\cos x \\ V=\cos x & \rightarrow {V}'=- \sin x \end{cases}\\\\ {f}\: '(x)&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}}\\ &=\displaystyle \frac{\cos x.\cos x-\sin x.(-\sin x)}{\left (\cos x \right )^{2}}\\ &=\displaystyle \frac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned}\\\hline \end{array}\\ & \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=2\sin x\cos x&\textrm{f}.&f(x)=4\sin \left ( 5x+\pi \right )^{3}&\textrm{k}.&f(x)=\sin \left ( \displaystyle \frac{x+3}{x-2} \right )\\ \textrm{b}.&f(x)=x^{2}.\sin x&\textrm{g}.&f(x)=\cos ^{3}(x+5)&\textrm{l}.&f(x)=\cos \left ( x^{2}-6x \right )\\ \textrm{c}.&f(x)=3\cos ^{2}x&\textrm{h}.&f(x)=\sin ^{2}\left ( \pi -3x \right )&\textrm{m}.&f(x)=\displaystyle \frac{1}{\sin x}\\ \textrm{d}.&f(x)=3\sin ^{2}x-x^{3}&\textrm{i}.&f(x)=\displaystyle \frac{\sin ^{2}x}{\cos x}&\textrm{n}.&f(x)=\displaystyle \frac{\sin x}{1+\cos x}\\ \textrm{e}.&f(x)=5\sin ^{4}x&\textrm{j}.&f(x)=\displaystyle \frac{\sin x^{4}}{x^{2}}&\textrm{o}.&f(x)=\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}\\ \end{array} \end{array}.

Jawab:

\begin{array}{ll|l}\\ 3.\: \textrm{a}&f(x)=2\sin x\cos x&\qquad \textbf{atau}\\ &\begin{aligned}f(x)&=\sin 2x\\ {f}\: '(x)&=\cos 2x.(2)\\ &=2\cos 2x\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\sin 2x\quad \rightarrow \quad \begin{cases} y=\sin u & \rightarrow \displaystyle \frac{dy}{du}=\cos u\\\\ u=2x & \rightarrow \displaystyle \frac{du}{dx}=2 \end{cases} \\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ &=\cos u. 2\\ &=2\cos u\\ &=2\cos 2x \end{aligned} \end{array}.

\begin{array}{ll|l}\\ 3.\: \textrm{h}&f(x)=\sin ^{2}\left ( \pi -3x \right )&\textbf{atau}\\ &\begin{aligned}{f}\: '(x)&=2\sin ^{2-1}\left ( \pi -3x \right ).\cos \left ( \pi -3x \right ).(-3)\\ &=-6\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3.2\sin \left ( \pi -3x \right )\cos \left ( \pi -3x \right )\\ &=-3\sin 2\left ( \pi -3x \right )\\ &=-3\sin \left ( 2\pi -6x \right )\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\sin ^{2}\left ( \pi -3x \right )\quad \rightarrow \quad \begin{cases} y=u^{2} &\rightarrow \displaystyle \frac{dy}{du}=2u. \\\\ u=\sin w &\rightarrow \displaystyle \frac{du}{dw}=\cos w \\\\ w=\pi -3x &\rightarrow \displaystyle \frac{dw}{dx}=-3 \end{cases}\\ \displaystyle \frac{dy}{dx}&=\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}\\ &=\left ( 2u \right ).\left ( \cos w \right ).\left ( -3 \right )\\ &= -3.\left ( 2\sin w \right ).\left ( \cos w \right )\\ &=-3\sin 2w\\ &=-3\sin \left ( 2\pi -6x \right )\end{aligned} \end{array}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah nilai dari}\: \: \: \displaystyle \frac{dy}{dx}\\ &\textrm{a}.\quad y=\sqrt{x}\\ &\textrm{b}.\quad y=\sqrt{3+\sqrt{x}}\\ &\textrm{c}.\quad y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ &\textrm{d}.\quad y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}} \end{array}\\\\\\ \textrm{Jawab:}\\\\ \begin{array}{|c|c|}\hline \begin{aligned}\textrm{a}.\qquad y&=\sqrt{x}\\ &=x^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{2\sqrt{x}}\\ &\\ & \end{aligned}&\begin{aligned} \textrm{b}.\qquad y&=\sqrt{3+\sqrt{x}}\\ &=\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}\\ {y}'&=\displaystyle \frac{1}{2}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}-1}}.\displaystyle \frac{1}{2}x^{^{\frac{1}{2}-1}}\\ &=\displaystyle \frac{1}{4}\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{-\frac{1}{2}}}.x^{^{-\frac{1}{2}}}\\ &=\displaystyle \frac{1}{4\left ( 3+x^{^{\frac{1}{2}}} \right )^{^{\frac{1}{2}}}.x^{^{\frac{1}{2}}}}\\ &=\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}\\ & \end{aligned}\\\hline \end{array}.

C. Turunan Kedua

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ \textrm{Suatu fungsi}\qquad &\\ &f\: :\: x\: \rightarrow \: y\\ & \end{aligned}}\\\hline \textrm{Notasi}&\textrm{Proses}&\textrm{Contoh Soal}\\\hline \begin{aligned}&\\ &\begin{cases} {f}\: ' (x)& =\displaystyle \frac{dy}{dx} \\\\ {f}\: ''(x) & =\displaystyle \frac{d^{2}y}{dx^{2}} \end{cases}\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ y&=ax^{n}\\ {y}'&=n.a.x^{n-1}\\ {y}''&=(n-1).n.a.x^{n-2}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\textrm{Misalkan}\: \: y=15x^{3}-4x\\ &\textrm{maka},\\ &\textrm{turunan pertamanya adalah}\quad \displaystyle \frac{dy}{dx}=45x^{2}-4,\: \: \textrm{dan}\\ &\textrm{turunan keduanya adalah}\quad \displaystyle \frac{d^{2}y}{dx^{2}}=90x\\ &\end{aligned}\\\hline \end{array}.

Sebagai tambahan contoh yang lain adalah sebagai berikut:

\begin{array}{|l|}\hline \begin{aligned}&\\ &\begin{aligned}y&=\displaystyle \frac{x^{4}}{3}+\frac{x^{2}}{2}+x+\sqrt{x}+1+\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^{2}}+\frac{3}{x^{4}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan pertama}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{dy}{dx}={y}\, '&=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &=\displaystyle \frac{4x^{3}}{3}+x+1+\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^{2}}-\frac{2}{x^{3}}-\frac{12}{x^{5}}\\ &\qquad\quad \parallel \\ &\textbf{Turunan kedua}\\ &\qquad\quad \Downarrow \\ \displaystyle \frac{d^{2}y}{dx^{2}}={y}\, ''&=4x^{2}+1+0-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}}\\ &=4x^{2}+1-\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^{2}\sqrt{x}}+\frac{2}{x^{3}}+\frac{6}{x^{4}}+\frac{60}{x^{6}} \end{aligned}\\ & \end{aligned}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

  1. Kerjakanlah Soal-soal yang belum dijawab/dibahas pada contoh soal
  2. Carilah turunan kedua dari setiap fungsi pada contoh soal No. 2 di atas
  3. Carilah turunan kedua dari setiap fungsi pada contoh soal No. 3 di atas

Sumber Referensi

  1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (WAJIB). Bandung: SEWU.
  2. Kartini, Suprapto, Subandi, dan Setiyadi, U. (2005). Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
  3. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS.
  4. Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: Bumi Aksara.
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