1.4 KiB
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