$\underset{p\rightarrow0}{lim}\frac{f(x+p)-f(x)}{p}=f'(x)$

$f(x)=x^{2}-1$$\rightarrow f'(x)=2x$

$f(x)=\frac{3x^{2}-5}{x+6}$

$\begin{aligned}f(0) & =\frac{3(0)^{2}-5}{0+6}\\
& =\frac{-5}{6}
\end{aligned}
$

Misalkan $u=3x^{2}-5$ dan $v=x+6,$

$\begin{aligned}f'(x) & =\frac{u’v-uv’}{v^{2}}\\
& =\frac{(6x)(x+6)-(3x^{2}-5)(1)}{(x+6)^{2}}
\end{aligned}
$

$\begin{aligned}f'(0) & =\frac{(6.0)(0+6)-(3(0)^{2}-5)(1)}{(0+6)^{2}}\\
& =\frac{0-(-5)}{36}\\
& =\frac{5}{36}
\end{aligned}
$

Jadi $\begin{aligned}f(0)+6f'(0) & =\frac{-5}{6}+6.\frac{5}{36}\\
& =-\frac{5}{6}+\frac{5}{6}\\
& =0
\end{aligned}
$

Misalkan $y=u^{\frac{3}{4}}$, maka $\frac{dy}{du}=\frac{3}{4.\sqrt[4]{u}}$

$u=2x^{2}-3$, maka $\frac{du}{dx}=4x$

$\begin{aligned}\frac{dy}{dx} & =\frac{dy}{du}\cdot\frac{du}{dx}\\
& =\frac{3}{4.\sqrt[4]{u}}\cdot4x\\
& =\frac{3}{4\cdot\sqrt[4]{2x^{2}-3}}\cdot4x\\
& =\frac{3x}{\sqrt[4]{2x^{2}-3}}
\end{aligned}
$

Misalkan $u=x^{2}-2x+1$$\Rightarrow u’=2x-2$

$v=x-3$$\Rightarrow v’=1$

$\begin{aligned}f'(x) & =\frac{u’v-uv’}{v^{2}}\\
& =\frac{(2x-2)(x-3)-\left(x^{2}-2x+1\right)(1)}{\left(x-3\right)^{2}}
\end{aligned}
$

$\begin{aligned}f'(2) & =\frac{(2(2)-2)(2-3)-\left((2)^{2}-2(2)+1\right)(1)}{\left(2-3\right)^{2}}\\
& =\frac{2(-1)-\left(1\right)}{(-1)^{2}}\\
& =\frac{-3}{1}\\
& =-3
\end{aligned}
$

$y=\frac{1}{3}x^{3}+\frac{5}{2}x^{2}-6x$

syarat stasioner adalah ketika $y’=0$

$x^{2}+5x-6=0$

$\left(x+6\right)(x-1)=0$

$x=1$ atau $x=-6$

untuk $x=1$$\Rightarrow y=\frac{1}{3}(1)^{3}+\frac{5}{2}(1)^{2}-6(1)$$=-\frac{19}{6}$ (minimum)

untuk $x=-6$$\Rightarrow y=\frac{1}{3}(-6)^{3}+\frac{5}{2}(-6)^{2}-6(-6)$$=54$ (maksimum)

Jadi titik balik minimumnya adalah $\left(1,-\frac{19}{6}\right).$

$f(x)=(x-2)(x-1)^{2}$

Misalkan $u=x-2$ dan $v=(x-1)^{2}$

$f'(x)=u’v+uv’$

$f'(x)=(1)$$\left((x-1)^{2}+(x-2)(2(x-1)\right)$

$\begin{aligned}f'(x) & =x^{2}-2x+1+2x^{2}-6x+4\\
& =3x^{2}-8x+5
\end{aligned}
$

Syarat mencapai nilai ekstrim adalah ketika $f'(x)=0$

$3x^{2}-8x+5=0$

$\left(3x-5\right)(x-1)=0$

$x=\frac{5}{3}$ atau $x=1$

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

$v=sinx$$\Rightarrow v’=cosx$

$f(x)=u\cdot v$$\Rightarrow f'(x)=u’v+uv’$

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

Misalkan $u=\frac{1}{5}x^{5}$$\Rightarrow u’=x^{4}$

$v=Inx-\frac{1}{5}$$\Rightarrow v’=\frac{1}{x}$

$\begin{aligned}y’ & =u’v+uv’\\
& =x^{4}(\ln x-5)+\frac{1}{5}x^{5}\cdot\frac{1}{x}\\
& =x^{4}\cdot\ln x-5x^{4}+\frac{1}{5}x^{4}\\
& =x^{4}\left(\ln x-5+\frac{1}{5}\right)\\
& =x^{4}\left(\ln x-\frac{24}{5}\right)
\end{aligned}
$

Turunan berbentuk $\ln g(x)$ adalah $\frac{g'(x)}{g(x)}$

$g(x)=\frac{2+x}{2-x}$

Misalkan $u=2+x$ dan $v=2-x$

$\begin{aligned}g'(x) & =\frac{u’v-uv’}{v^{2}}\\
& =\frac{1(2-x)-(2+x)(-1)}{(2-x)^{2}}\\
& =\frac{4}{(2-x)^{2}}
\end{aligned}
$

Jadi

$\begin{aligned}y’ & =\frac{g'(x)}{g(x)}\\
& =\frac{\frac{4}{(2-x)^{2}}}{\frac{2+x}{2-x}}\\
& =\frac{-2x}{(2-x)(2+x)}\\
& =\frac{4}{4-x^{2}}
\end{aligned}
$

$f(x)=x^{3}-px^{2}-px-1$

syarat stasioner adalah $f'(x)=0$

$\begin{aligned}f'(x) & =3x^{2}-2px-p\\
& =0
\end{aligned}
$

karena $x=p,$ maka :

$3p^{2}-2p^{2}-p=0$

$p^{2}-p=0$

$p\left(p-1\right)=0$

$p=0$ atau $p=1.$

$y=sec^{5}2x^{4}$

Misalkan $y=u^{5}$$\Rightarrow\frac{dy}{du}=5u^{4}$

$u=secv$$\Rightarrow\frac{du}{dv}=secv\cdot tanv$

$v=2x^{4}$$\Rightarrow\frac{dv}{dx}=8x^{3}$

Jadi

$\begin{aligned}\frac{dy}{dx} & =\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}\\
& =5u^{4}\cdot secv\cdot tanv\cdot8x^{3}\\
& =40x^{3}\cdot sec^{5}2x^{4}\cdot tan2x^{4}
\end{aligned}
$

$y=\left(\frac{4-3\cdot sinx}{4+3\cdot sinx}\right)^{7}$

misalkan $y=u^{7}$$\Rightarrow\frac{dy}{du}=7u^{6}$

u = $\frac{4-3.sinx}{4+3.sinx}$$\Rightarrow\frac{du}{dx}$ gunakan aturan turunan pembagian :

$\frac{du}{dx}=$$\frac{\left(-3cosx\right)\left(4+3\cdot sinx\right)-\left(4-3\cdot sinx\right)\left(3\cdot cosx\right)}{\left(4+3\cdot sinx\right)^{2}}$

$=\frac{-12\cdot cosx-9\cdot sinx\cdot cosx-12\cdot cosx+9\cdot sinx\cdot cosx}{\left(4+3\cdot sinx\right)^{2}}$

$=\frac{-24cosx}{\left(4+3\cdot sinx\right)^{2}}$

Jadi

$\frac{dy}{du}=\frac{dy}{du}\cdot\frac{du}{dx}$

$=7u^{6}\cdot\frac{-24cosx}{\left(4+3\cdot sinx\right)^{2}}$

$=7\cdot\left(\frac{4-3\cdot sinx}{4+3\cdot sinx}\right)^{6}\cdot\frac{-24cosx}{\left(4+3\cdot sinx\right)^{2}}$

$=\frac{-128\cdot cosx\cdot\left(4-3.sinx\right)^{6}}{\left(4+3\cdot sinx\right)^{8}}$

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$

$v=cosx$$\Rightarrow v’=-sinx$

$\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}
$

$f(x)=x\cdot g(x)$

gunakan aturan perkalian turunan fungsi

misalkan $u=x$, dan $v=g(x)$

$f'(x)=u’v+uv’$

$f'(x)=1\cdot g(x)+x\cdot g'(x)$

$f'(2)=1\cdot g(2)+2\cdot g'(2)$

$\begin{aligned}f'(2) & =1\cdot2+2\cdot(-1)\\
& =2-2\\
& =0
\end{aligned}
$

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

$=sin\left(2x+\frac{2}{k}\right)$

$\underset{k\rightarrow\infty}{lim}k\cdot\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\} $

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

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

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

$=2\cdot cos2x\cdot1$

$=2\cdot cos2x$