# Na podstawie definicji obliczyć pochodną funkcji f w podanym punkcie $x_0$.
## 1
$f(x)=2x-x^{2},gdzie\  x_0=1$
$f'(x)=2-2x$
$f'(x_{0})=0$
## 2
$f(x)=x^{2}-7x,gdzie\ x_{0}=0$
$f'(x)=2x-7$
$f'(x_{0})=-7$
## 3
$f(x)=x^{3},x_{0}=1$
$f'(x)=3x^2$
$f'(x_{0})=3$
## 4
$f(x)=-2x^{3}+x, x_{0}=-1$
$f'(x)=-6x^{2}+1$
$f'(x_{0})=-5$
## 5
$f(x)=-\sqrt{x+2},\ x_{0}=2$
$f(x)=-(x+2)^{\frac{1}{2}},\ x_{0}=2$
$f'(x)=-\frac{1}{2\sqrt{x+2}}$
$f'(x_{0})=-\frac{1}{4}$
## 6
$f(x)=\sqrt{1+2x},\ x_{0}=4$
$f(x)=(1+2x)^{\frac{1}{2}}$
$f'(x)=\cfrac{\frac{1}{2\sqrt{1+2x}}}{2}$
$f'(x_{0})=\frac{1}{3}$
## 7
$f(x)=3+2\sqrt{4x-1},\ x_{0}=2\frac{1}{2}$
$f'(x)=2\cdot \frac{1}{2\sqrt{4x-1}}\cdot 4$
$f'(x_{0})=\frac{4}{3}$
## 8
$f(x)=\frac{1}{x^{2}},\ x_{0}=1$
$f'(x)=-\frac{2}{x^3}$
$f'(x_{0})=-2$ 
## 9
$f(x)=\frac{2}{x^{3}},\ x_{0}=-1$
$f(x)=2(x^{-3})$
$f'(x)=-\frac{6}{x^{4}}$
$f'(x_{0})=-6$
## 10
$f(x)=\sqrt{x^{2}-1},\ x_{0}=-3$
$f'(x)=\cfrac{1}{2\sqrt{x^{2}-1}} \cdot 2x$
$f'(x_{0})=\frac{-6}{4\sqrt{2}}=\frac{-3}{2\sqrt{2}}$
## 11 
$f(x)=\sqrt{2x^{2}+1},\ x_{0}=2$
$f'(x)=\cfrac{1}{2\sqrt{2x^{2}+1}} \cdot 4x$
$f'(x_{0})=\frac{4}{3}$
## 12
$f(x)=\sin{x},\ x_{0}=\frac{\pi}{2}$
$f'(x)=\cos x$
$f'(x_{0})=0$ 
## 13
$f(x)=\cos{x},\ x_{0}=\frac{\pi}{2}$
$f'(x)=-\sin x$
$f'(x_{0})=-1$ 
## 14
$f(x)=\sin{x^2},\ x_{0}=0$
$f'(x)=\cos x^{2}\cdot 2x$
$f'(x_{0})=0$
## 15
$f(x)=\sin\sqrt{x},\ x_{0}=0$
$f'(x)=\cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}}$
$f'(x_{0})=1\cdot\frac{1}{0}=brak\ rozwiązania$