Rumus Turunan dan Sifat Turunan Pertama
$\begin{aligned}&\\ \textrm{Untuk}:&\\ &\begin{cases} a & \in \mathbb{R} \\ n& \in \mathbb{Q}\\ c& \textrm{konstanta}\\ U&=g(x)\\ V&=h(x) \end{cases}\\ & \end{aligned}$.
$\begin{array}{|l|}\hline \color{blue}\textbf{Sifat-Sifat}\\\hline \begin{aligned}y&=c\rightarrow \qquad{y}'=0\\ y&=c.U\rightarrow \quad{y}'=c.{U}'\\ y&=U\pm V\rightarrow {y}'={U}'\pm {V}'\\ y&=U.V\rightarrow \: \: \: \: {y}'={U}'.V+U.{V}'\\ y&=\displaystyle \frac{U}{V}\rightarrow \quad\: \: \: {y}'=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}\\\hline \color{blue}\textbf{Fungsi Aljabar}\\\hline \begin{aligned}y&=a.x^{n}\rightarrow {y}\: '=n.a.x^{(n-1)}\\ y&=a.U^{n}\rightarrow {y}\: '=n.a.U^{(n-1)}.{U}'\end{aligned}\\\hline \color{blue}\textbf{Fungsi Trigonometri}\\\hline \begin{aligned}y&=a\sin U\rightarrow & {y}'&=\left (a\cos U \right ).{U}'\\ y&=a\cos U\rightarrow &{y}'&=\left (-a\sin U \right ).{U}'\\ y&=a\tan U\rightarrow &{y}'&=\left (a\sec ^{2}U \right ).{U}'\end{aligned}\\\hline \color{blue}\textbf{Aturan rantai }\\\hline \begin{aligned} &\textrm{pada turunan untuk} \: \: y=f(u),\: \: \textrm{jika}\\ &\textrm{untuk}\: \: u\: \: \textrm{merupakan fungsi}\: \: x,\: \: \textrm{maka}:\\ &\color{red}{y}\: '={f}\: '(x).{u}'\: \: \color{black}\textrm{atau}\: \: \: \: \color{red}\displaystyle \frac{dy}{dx}=\displaystyle \frac{dy}{du}.\frac{du}{dx}\\ & \end{aligned} \\\hline \end{array}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Dengan menggunakan rumus turunan}\\ &{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h},\\ &\textrm{tunjukkan bahwa turunan}\: \: \\ &\textrm{a}.\quad f(x)=ax^{n}\: \: \textrm{adalah}\: \: {f}\, '(x)=n.ax^{n-1}\\ &\textrm{b}.\quad f(x)=u(x)+v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x)+{v}\,'(x)\\ &\textrm{c}.\quad f(x)=u(x).v(x)\: \: \textrm{adalah}\: \: {f}\, '(x)= {u}\,'(x).v(x)+u(x).{v}\,'(x)\\ &\textrm{d}.\quad f(x)=\displaystyle \frac{u(x)}{v(x)}\: \: \textrm{adalah}\: \: {f}\, '(x)=\displaystyle \frac{{u}\,'(x).v(x)+u(x).{v}\,'(x)}{(v(x))^{2}} \\ &\textrm{e}.\quad f(x)=\sin x\: \: \textrm{adalah}\: \: {f}\, '(x)=\cos x \\ &\textrm{f}.\quad f(x)=\cos x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\sin x \\ &\textrm{g}.\quad f(x)=\tan x\: \: \textrm{adalah}\: \: {f}\, '(x)=\sec^{2} x \\ &\textrm{h}.\quad f(x)=\cot x\: \: \textrm{adalah}\: \: {f}\, '(x)=-\csc^{2} x \\ \end{array}$.
$\begin{aligned}&\textbf{Bukti}\\ &\begin{array}{ll}\\ 1.\: \textrm{a}&{f}\, '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\, '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a(x+h)^{n}-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left ( x^{n}+\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )-ax^{n}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{a\left (\binom{n}{1}x^{n-1}.h+\binom{n}{2}x^{n-2}.h^{2}+\binom{n}{3}x^{n-3}.h^{3}+\cdots +\binom{n}{n-1}x.h^{n-1}+h^{n}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{ah\left (\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}.h+\binom{n}{3}x^{n-3}.h^{2}+\cdots +\binom{n}{n-1}x.h^{n-2}+h^{n-1}\right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle a\binom{n}{1}x^{n-1}+a\binom{n}{2}x^{n-2}.h+a\binom{n}{3}x^{n-3}.h^{2}+\cdots +ah^{n-1}\\ &=a\binom{n}{1}x^{n-1}+0+0+...+0\\ &=a.\displaystyle \frac{n!}{(n-1)!.1!}x^{n-1}\\ &=a.\displaystyle \frac{n.(n-1)!}{(n-1)!}x^{n-1}\\ &=a.n.x^{n-1}\qquad\quad \blacksquare \end{aligned} \end{array} \end{aligned}$.
$\begin{array}{ll}\\ 1.\: \textrm{b}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (u(x+h)+v(x+h) \right )-\left ( u(x)+v(x) \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \left (\displaystyle \frac{u(x+h)-u(x)}{h}+\frac{v(x+h)-v(x)}{h} \right )\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)-u(x)}{h}+ \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{v(x+h)-v(x)}{h}\\ &=u'(x)+v'(x)\qquad\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 1.\: \textrm{c}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\left (u(x+h)\times v(x+h) \right )-\left ( u(x)\times v(x) \right )}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)\times v(x+h)-u(x+h)\times v(x)+u(x+h)\times v(x)-u(x)\times v(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \left (u(x+h)\times \displaystyle \frac{v(x+h)-v(x)}{h}+v(x)\times \displaystyle \frac{u(x+h)-u(x)}{h} \right )\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: u(x+h)\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{v(x+h)-v(x)}{h}+\underset{h\rightarrow 0}{\textrm{Lim}}\: \: v(x)\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{u(x+h)-u(x)}{h}\\ &=u(x)\times v'(x)+v(x)\times u'(x)\\ &=u'(x)\times v(x)+u(x)\times v'(x)\qquad\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 1.\: \textrm{d}&\textrm{Misalkan}\: \: p(x)=\displaystyle \frac{u(x)}{v(x)}\\ &\textrm{Sebelumnya telah diketahui dari no. 1. c}\\ &\: {u}\: (x)=p(x)\times v(x)\\ &\begin{aligned}{u}\: '(x)&=p'(x)\times v(x)+p(x)\times v'(x)\\ \textrm{Sekar}&\textrm{ang kita substitusikan pemisalan}\\ \textrm{di at}&\textrm{as, yaitu}: \end{aligned}\\ &\begin{aligned}\: p'(x)&\times v(x)=u'(x)-p(x)\times v'(x)\\ &=u'(x)-\displaystyle \frac{u(x)}{v(x)}\times v'(x)\\ &=\displaystyle \frac{u'(x)\times v(x)-u(x)\times v'(x)}{v(x)}\\ p'(x)&=\displaystyle \frac{u'(x)\times v(x)-u(x)\times v'(x)}{v^{2}(x)}\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 1.\: \textrm{e}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{2\cos \displaystyle \frac{1}{2}(2x+h)\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: 2\cos \displaystyle \frac{1}{2}(2x+h).\displaystyle \frac{\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle 2\cos \displaystyle \frac{1}{2}(2x+h)\times \displaystyle \frac{1}{2}\\ &=2\cos \displaystyle \frac{1}{2}(2x+0)\times \displaystyle \frac{1}{2}\\ &=\cos \displaystyle \frac{1}{2}(2x)\\ &=\cos x\qquad\quad \blacksquare \end{aligned} \end{array}$ .
$\begin{array}{ll}\\ 1.\: \textrm{f}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos (x+h)-\cos x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2\sin \displaystyle \frac{1}{2}(2x+h)\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle -2\sin \displaystyle \frac{1}{2}(2x+h).\frac{\sin \displaystyle \frac{1}{2}h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle -2\sin \displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}\\ &=-2\sin \displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}\\ &=-\sin \displaystyle \frac{1}{2}(2x)\\ &=-\sin x\qquad\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 1.\: \textrm{g}&{f}\: '(x)=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\begin{aligned}{f}\: '(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h}{1-\tan x.\tan h}-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{\tan x+\tan h-\tan x+\tan^{2}x.\tan h}{1-\tan x.\tan h}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h\left ( 1+\tan^{2}x \right )}{h\left ( 1-\tan x.\tan h \right )}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan h}{h}\times \underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{1+\tan^{2}x}{1-\tan x.\tan h}\\ &=1 \times \displaystyle \frac{1+\tan^{2}x}{1-0}\\ &=1+\tan^{2}x\\ &=\sec ^{2}x\qquad\quad \blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah turunannya}\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=x^{6}&\textrm{i}.&f(x)=(x+1)(x-2)&\textrm{q}.&f(x)=\displaystyle \frac{2}{x^{3}}\\ \textrm{b}.&f(x)=\displaystyle \frac{1}{2}x^{2}&\textrm{j}.&f(x)=(3-x)(5-x)&\textrm{r}.&f(x)=\displaystyle \frac{1}{2\sqrt{x}}\\ \textrm{c}.&f(x)=-2x^{5}&\textrm{k}.&f(x)=(2x+3)^{2}&\textrm{s}.&f(x)=\displaystyle \frac{1}{3x^{5}}\\ \textrm{d}.&f(x)=ax^{3}&\textrm{l}.&f(x)=(x-2)^{3}&\textrm{t}.&f(x)=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ \textrm{e}.&f(x)=4x^{4}-x^{2}+2017&\textrm{m}.&f(x)=(4x+1)(4x-1)&\textrm{u}.&f(x)=\displaystyle \frac{x}{2}+\frac{2}{x}\\ \textrm{f}.&f(x)=6-x-3x^{2}&\textrm{n}.&f(x)=(x-1)(x+1)(x+2)&\textrm{v}.&f(x)=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ \textrm{g}.&f(x)=(x-3)^{2}&\textrm{o}.&f(x)=\displaystyle \frac{1}{2}x^{-4}&\textrm{w}.&f(x)=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ \textrm{h}.&f(x)=\left ( x^{3}-2 \right )^{2}&\textrm{p}.&f(x)=x^{-5}&\textrm{x}.&f(x)=\displaystyle \frac{2+4x}{x^{2}}\\ &&&&\textrm{y}.&f(x)=\left ( x+1+\displaystyle \frac{1}{x} \right )\left ( x+1-\displaystyle \frac{1}{x} \right ) \end{array} \end{array}$.
$\begin{aligned}&\textrm{Untuk}\: \: y=f(x)\: \: \textrm{maka}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}&\textrm{c}\\\hline \begin{aligned}y&=x^{6}\\ {y}\: '&=6x^{6-1}\\ &=6x^{5}\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{2}\\ {y}\: '&=2.\displaystyle \frac{1}{2}x^{2-1}\\ &=x \end{aligned}&\begin{aligned}y&=-2x^{5}\\ {y}\: '&=-5.2x^{5-1}\\ &=-10x^{4}\\ & \end{aligned}\\\hline \textrm{d}&\textrm{e}&\textrm{f}\\\hline \begin{aligned}y&=ax^{3}\\ {y}\: '&=3.a.x^{3-1}\\ &=3.a.x^{2} \end{aligned}&\begin{aligned}y&=4x^{4}-x^{2}+2017\\ {y}\: '&=4.4x^{4-1}-2.x^{2-1}+0\\ &=16x^{3}-2x \end{aligned}&\begin{aligned}y&=6-x-3x^{2}\\ {y}\: '&=0-1.x^{1-1}-2.3x^{2-1}\\ &=-1-6x \end{aligned}\\\hline \end{array} \end{aligned}$.
$\begin{array}{|c|c|}\hline \textrm{g}&\textrm{h}\\\hline \begin{aligned}y&=(x-3)^{2}\\ {y}\: '&=2.(x-3)^{2-1}.1\\ &=2(x-3)\\ &\end{aligned}&\begin{aligned}y&=\left ( x^{3}-2 \right )^{2}\\ {y}\: '&=2.\left ( x^{3}-2 \right )^{2-1}.3x^{2}\\ &=6\left ( x^{3}-2 \right )x^{2}\\ &=6x^{5}-12x^{2} \end{aligned}\\\hline \textrm{i}&\textrm{j}\\\hline \begin{aligned}y&=(x+1)(x-2)\\ {y}\: '&=1.(x+2)+(x+1).1\\ &=2x+3 \end{aligned}&\begin{aligned}y&=(3-x)(5-x)\\ {y}\: '&=-1.(5-x)+(3-x).-1\\ &=2x-8 \end{aligned}\\\hline \end{array}$ .
$\begin{array}{|c|c|c|c|}\hline \textrm{k}&\textrm{l}&\textrm{m}_{1}&\textrm{m}_{2}\\\hline \begin{aligned}y&=(2x+3)^{2}\\ {y}\: '&=2.(2x+3)^{2-1}.2\\ &=4(2x+3)^{1}\\ &=8x+12\\ &\\ & \end{aligned}&\begin{aligned}y&=(x-2)^{3}\\ {y}\: '&=3(x-2)^{3-1}.1\\ &=3(x-2)^{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 1}\\ {y}\: '&=4(4x-1)+(4x+1).4\\ &=16x-4+16x+4\\ &=32x\\ & \end{aligned}&\begin{aligned}y&=(4x+1)(4x-1)\\ \textrm{c}&\textrm{ara 2}\\ y&=16x^{2}-1\\ {y}\: '&=2.16x^{2-1}-0\\ &=32x^{1}\\ &=32x \end{aligned}\\\hline \textrm{n}&\textrm{o}&\textrm{p}&\textrm{q}\\\hline \begin{aligned}y&=(x-1)(x+1)(x+2)\\ &=(x^{2}-1)(x+2)\\ &=x^{3}+2x^{2}-x-2\\ {y}\: '&=3x^{3-1}+2.2x^{2-1}-x^{1-1}-0\\ &=3x^{2}+4x-1\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{2}x^{-4}\\ {y}\: '&=-4.\displaystyle \frac{1}{2}x^{-4-1}\\ &=-2x^{-5}\\ &=-\displaystyle \frac{2}{x^{5}}\\ & \end{aligned}&\begin{aligned}y&=x^{-5}\\ {y}'\: &=-5.x^{-5-1}\\ &=-5x^{-6}\\ &=-\displaystyle \frac{5}{x^{6}}\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2}{x^{3}}\\ &=2x^{-3}\\ {y}\: '&=-3.2x^{-3-1}\\ &=-6x^{-4}\\ &=-\displaystyle \frac{6}{x^{4}} \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|c|c|c|}\hline \textrm{r}&\textrm{s}&\textrm{t}&\textrm{u}\\\hline \begin{aligned}y&=\displaystyle \frac{1}{2\sqrt{x}}\\ &=\displaystyle \frac{1}{2x^{\frac{1}{2}}}\\ &=\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}}\\ {y}\: '&=-\displaystyle \frac{1}{2}.\frac{1}{2}x^{^{-\frac{1}{2}-1}}\\ &=-\displaystyle \frac{1}{4}x^{^{-\frac{3}{2}}}\\ &=-\displaystyle \frac{1}{4x^{^{\frac{3}{2}}}}\\ &=-\displaystyle \frac{1}{4}\sqrt{x^{3}} \end{aligned}&\begin{aligned}y&=\displaystyle \frac{1}{3x^{5}}\\ &=\displaystyle \frac{1}{3}x^{-5}\\ {y}\: '&=-5.\displaystyle \frac{1}{3}x^{-5-1}\\ &=-\displaystyle \frac{5}{3}x^{-6}\\ &=-\displaystyle \frac{5}{3x^{6}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=2x^{4}+\displaystyle \frac{1}{2x^{3}}\\ &=2x^{4}+\displaystyle \frac{1}{2}x^{-3}\\ {y}\: '&=4.2x^{4-1}+(-3).\displaystyle \frac{1}{2}x^{-3-1}\\ &=8x^{3}-\displaystyle \frac{3}{2}x^{-4}\\ &=8x^{3}-\displaystyle \frac{3}{2x^{4}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{x}{2}+\frac{2}{x}\\ &=\displaystyle \frac{1}{2}x+2x^{-1}\\ {y}\: '&=\displaystyle \frac{1}{2}x^{1-1}+(-1).2x^{-1-1}\\ &=\displaystyle \frac{1}{2}x^{0}-2x^{-2}\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{2}{x^{2}}\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|c|c|c|}\hline \textrm{v}&\textrm{w}\\\hline \begin{aligned}y&=\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )^{2}\\ &=\left ( 2x^{3}+x^{-1} \right )^{2}\\ {y}\: '&=2.\left ( 2x^{3}+x^{-1} \right )^{2-1}.\left ( 3.2x^{3-1}+(-1)x^{-1-1} \right )\\ &=2.\left ( 2x^{3}+x^{-1} \right )^{1}.\left ( 6x^{2}-x^{-2} \right )\\ &=2\left ( 2x^{3}+\displaystyle \frac{1}{x} \right )\left ( 6x^{2}-\displaystyle \frac{1}{x^{2}} \right )\\ &=2\left ( 12x^{5}-2x+6x-\displaystyle \frac{1}{x^{3}} \right )\\ &=24x^{5}+8x-\displaystyle \frac{2}{x^{3}}\\ & \end{aligned}&\begin{aligned}y&=x^{2}\left ( 1+\sqrt{x} \right )^{2}\\ &=x^{2}\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}\\ {y}\: '&=2x.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2}+x^{2}.2.\left ( 1+x^{^{\frac{1}{2}}} \right )^{2-1}.\left ( 0+\displaystyle \frac{1}{2}.x^{^{\frac{1}{2}-1}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( 1+\sqrt{x} \right ).\left ( \displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+2x^{2}.\left ( \displaystyle \frac{1}{2x^{^{\frac{1}{2}}}} \right ).\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{2-\frac{1}{2}}}.\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+x^{^{\frac{3}{2}}}\left ( 1+\sqrt{x} \right )\\ &=2x\left ( 1+\sqrt{x} \right )^{2}+\left ( x\sqrt{x}+x^{2} \right ) \end{aligned}\\\hline \end{array}$.
$\begin{array}{|c|c|}\hline \textrm{x}_{1}&\textrm{x}_{2}\\\hline \begin{aligned}y&=\displaystyle \frac{U}{V}\\ {y}\: '&=\displaystyle \frac{{U}'.V-U.{V}'}{V^{2}} \end{aligned}&\begin{aligned}y&=UV\\ {y}\: '&={U}'.V+U.{V}' \end{aligned}\\\hline \begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}&\begin{aligned}&&&\\ U&=2+4x&\rightarrow {U}'&=4\\ V&=x^{2}&\rightarrow {V}'&=2x\\ &&& \end{aligned}\\\hline \begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ {y}\: '&=\displaystyle \frac{(4)(x^{2})-(2+4x).(2x)}{\left ( x^{2} \right )^{2}}\\ &= \displaystyle \frac{4x^{2}-4x-8x^{2}}{x^{4}}\\ &=\displaystyle \frac{-4x^{2}-4x}{x^{4}}\\ &=\displaystyle \frac{-4x-4}{x^{3}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}y&=\displaystyle \frac{2+4x}{x^{2}}\\ &=(2+4x).x^{-2}\\ {y}\: '&=(4).x^{-2}+(2+4x).-2x^{-2-1}\\ &=4x^{-2}-(4+8x).x^{-3}\\ &=4x^{-2}-4x^{-3}-8x^{-2}\\ &=-4x^{-3}-4x^{-2}\\ &=\displaystyle \frac{-4}{x^{3}}+\displaystyle \frac{-4}{x^{2}}\\ &=\displaystyle \frac{-4-4x}{x^{3}}\\ &=\displaystyle \frac{-4x-4}{x^{3}} \end{aligned}\\\hline \end{array}$.
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