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.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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