Ingat bentuk turunan $f(u)=\sqrt{u}$$\rightarrow f'(u)=\frac{u’}{2\sqrt{u}}$

$\frac{dr}{d\alpha}=r’=\frac{cos\alpha}{2\sqrt{sin\alpha}}$

$y=cos^{4}x$

Misalkan $y=u^{4}\Rightarrow\frac{dy}{du}=4u^{3}$

$u=cosx\Rightarrow\frac{du}{dx}=-sinx$

$\begin{aligned}\frac{dy}{dx} & =\frac{dy}{du}\cdot\frac{du}{dx}\\
& =4u^{3}\cdot-sinx\\
& =-4\cdot cos^{3}x\cdot sinx
\end{aligned}
$

$\begin{aligned}f(x) & =5\cdot sinx\cdot cosx\\
& =\frac{5}{2}\cdot2\cdot sinx\cdot cosx\\
& =\frac{5}{2}\cdot sin2x
\end{aligned}
$

$\begin{aligned}f'(x) & =\frac{5}{2}\cdot2\cdot cos2x\\
& =5\cdot cos2x
\end{aligned}
$

$\begin{aligned}y & =\frac{sin2x}{1+cos2x}\\
& =\frac{2\cdot sinx\cdot cosx}{2\cdot cos^{2}x}\\
& =\frac{sinx}{cosx}\\
& =tan
\end{aligned}
$

Jadi turunan dari $y=\frac{sin2x}{1+cos2x}$ adalah $y’=sec^{2}x$

$\begin{aligned}f(x) & =-\left(cos^{2}x-sin^{2}x\right)\\
& =-cos2x
\end{aligned}
$

$\begin{aligned}f'(x) & =2\cdot sin2x\\
& =2\cdot2\cdot sinx\cdot cosx\\
& =4\cdot sinx\cdot cosx
\end{aligned}
$

Misalkan $u=(3x-2)$$\Rightarrow u’=3$

$v=sin(2x+1)$$\Rightarrow v’=2.cos(2x+1)$

$f'(x)=u’v+uv’$

$f'(x)=3\cdot sin(2x+1)+(3x-2)\cdot2cos(2x+1)$

$=3\cdot sin(2x+1)+(6x-4)\cdot cos(2x+1)$

$f(x)=x\cdot cosx$

Misalkan $u=x$ dan $v=cosx$,

$u^{1}=1$ dan $v^{1}=-\sin x$

$\begin{aligned}f'(x) & =u’v+uv’\\
& =1(cosx)+x(-sinx)
\end{aligned}
$

$f'(x+\frac{\pi}{2})=1(cos(x+\frac{\pi}{2}))$$+(x+\frac{\pi}{2})\left\{ -sin(x+\frac{\pi}{2})\right\} $

$\begin{aligned}f'(x+\frac{\pi}{2}) & =-sinx-(x+\frac{\pi}{2})\left\{ cosx)\right\} \\
& =-sinx-x\cdot cosx-\frac{\pi}{2}cosx
\end{aligned}
$

$f(x)=a\cdot tanx+bx$

$f'(x)=a\cdot sec^{2}x+b$

$\begin{aligned}f'(\frac{\pi}{4}) & =a\cdot sec^{2}\left(\frac{\pi}{4}\right)+b\\
& =2a+b\\
& =3
\end{aligned}
$…..(1)

$\begin{aligned}f'(\frac{\pi}{3}) & =a\cdot sec^{2}\left(\frac{\pi}{3}\right)+b\\
& =4a+b\\
& =9
\end{aligned}
$…..(2)

Persamaan (2) kurangkan dengan pers (1) sehingga diperoleh :

$2a=6\rightarrow a=3$

Jika $a=3,$ maka $b=-3$

Jadi $\begin{aligned}a+b & =3+(-3)\\
& =0.
\end{aligned}
$

Misalkan $u=e^{2x}$$\Rightarrow u’=2\cdot e^{2x}$

$v=sinx$$\Rightarrow v’=cosx$

$f'(x)=u’v+uv’$

$\begin{aligned}f'(x) & =2e^{2x}\cdot sinx+e^{2x}\cdot cosx\\
& =e^{2x}\left(2sinx+cosx\right)
\end{aligned}
$

$\begin{aligned}f'(0) & =e^{0}\left(2\cdot sin0+cos0\right)\\
& =1(0+1)\\
& =1
\end{aligned}
$

$f(x)=\frac{sin^{2}x}{1+cos^{2}x}$

Misalkan;

$\begin{aligned}u & =sin^{2}x\\
\Rightarrow u’ & =2\cdot sinx\cdot cosx\\
& =sin2x
\end{aligned}
$

$\begin{aligned}v & =1+cos^{2}x\\
\Rightarrow v’ & =2\cdot cosx\cdot-sinx\\
& =-sin2x
\end{aligned}
$

$f'(x)=\frac{u’v-uv’}{v^{2}}$

$\begin{aligned}f'(x) & =\frac{sin2x\cdot(1+cos^{2}x)-sin^{2}x\left(-sin2x\right)}{(1+cos^{2}x)^{2}}\\
& =\frac{sin2x(1+cos^{2}x+sin^{2}x)}{(1+cos^{2}x)^{2}}\\
& =\frac{sin2x(1+1)}{(1+cos^{2}x)^{2}}\\
& =\frac{2\cdot sin2x}{(1+cos^{2}x)^{2}}
\end{aligned}
$

$\begin{aligned}f’\left(\frac{\pi}{4}\right) & =\frac{2\cdot sin2\left(\frac{\pi}{4}\right)}{(1+cos^{2}\left(\frac{\pi}{4}\right))^{2}}\\
& =\frac{2.(1)}{\left(1+\frac{1}{2}\right)^{2}}\\
& =\frac{2}{\left(\frac{3}{2}\right)^{2}}\\
& =\frac{8}{9}
\end{aligned}
$

Ini adalah turunan ekplisit :

$\frac{dy}{dx}\left(cos3y\right)=\frac{dy}{dx}\left(tan2x\right)$

$-3.sin3y\cdot\frac{dy}{dx}=2\cdot sec^{2}2x$

$\frac{dy}{dx}=-\frac{2\cdot sec^{2}2x}{3\cdot sin3y}$

$\frac{dy}{dx}\left(a\cdot cos^{2}(x+y)\right)=\frac{dy}{dx}(b)$

$2a\cdot cos\left(x+y\right)\cdot\left\{ -sin(x+y)\right\} $$\cdot(1+1\cdot\frac{dy}{dx})=0$

$(1+1\cdot\frac{dy}{dx})=0$

$\frac{dy}{dx}=y’=-1$

Ingat bentuk $F(x)=\ln g(x)$ memiliki turunan $F'(x)=\frac{g(x)’}{g(x)}$

$g(x)=\frac{1+sinx}{cosx}$

misalkan $u=1+sinx$$\Rightarrow u’=cosx$

$\begin{aligned}v & =cosx\\
\Rightarrow v’ & =-sinx
\end{aligned}
$

$\begin{aligned}g'(x) & =\frac{u’v-uv’}{v^{2}}\\
& =\frac{cosx(cosx)-(1+sinx)(-sinx)}{cos^{2}x}\\
& =\frac{cos^{2}x+sin^{2}x+sinx}{cos^{2}x}\\
& =\frac{1+sinx}{cos^{2}x}
\end{aligned}
$

Jadi turunan

$\begin{aligned}F'(x) & =\frac{\frac{1+sinx}{cos^{2}x}}{\frac{1+sinx}{cosx}}\\
& =\frac{1}{cosx}\\
& =secx
\end{aligned}
$

$\begin{aligned}f(x) & =sin^{2}x\\
\Rightarrow f'(x) & =2\cdot sinx\cdot cosx\\
& =sin2x
\end{aligned}
$

$\begin{aligned}f\left(x+\frac{1}{k}\right) & =sin^{2}\left(x+\frac{1}{k}\right)\\
\Rightarrow f’\left(x+\frac{1}{k}\right) & =2\cdot sin\left(x+\frac{1}{k}\right)\cdot cos\left(x+\frac{1}{k}\right)\\
& =sin\left(2x+\frac{2}{k}\right)
\end{aligned}
$

$\underset{k\rightarrow\infty}{lim}k.\left\{ f’\left(x+\frac{1}{k}\right)-f'(x)\right\} $$=\underset{k\rightarrow\infty}{lim}k\cdot\left\{ sin\left(2x+\frac{2}{k}\right)-sin2x)\right\} $

$=\underset{k\rightarrow\infty}{lim}k$$\cdot\left\{ 2\cdot cos\frac{1}{2}\left(2x+\frac{2}{k}+2x\right)\cdot sin\frac{1}{2}\left(2x+\frac{2}{k}-2x\right)\right\} $

$=\underset{k\rightarrow\infty}{lim}k\cdot2\cdot cos2x\cdot sin\frac{1}{k}$

$=\underset{k\rightarrow\infty}{lim}2\cdot cos2x\cdot\frac{sin\frac{1}{k}}{\frac{1}{k}}$

$=\underset{k\rightarrow\infty}{lim}2\cdot cos2x\cdot\underset{k\rightarrow\infty}{lim}\frac{sin\frac{1}{k}}{\frac{1}{k}}$

$=2\cdot cos2x\cdot1$

$=2\cdot cos2x$

$\frac{dy}{dx}\left(cosxy-sinxy\right)=\frac{dy}{dx}(2x-y^{2})$

$-sinxy\cdot(1y+x\cdot y’)-cosxy\cdot$$(1\cdot y+x\cdot y’)=2-2y\cdot y’$

$-y\cdot sinxy-xy’sinxy-y\cdot cosxy$$-x\cdot y’cosxy+2y\cdot y’=2$

$-xy’sinxy-x\cdot y’cosxy$$+2y\cdot y’=2+y\cdot sinxy+y\cdot cosxy$

$y'(-x\cdot sinxy-x\cdot cosxy+2y)=2+y\cdot sinxy$$+y\cdot cosxy$

$y’=\frac{2+y\cdot sinxy+y\cdot cosxy}{2y-x\cdot sinxy-x\cdot cosxy}$