${\left(x^3\right)}^{{}^{\prime }}=3x^2$ ${\left(\frac{1}{x}\right)}^{{}^{\prime }}=-\frac{1}{x^2},x\ne 0$ ${\left(\sqrt{x}\right)}^{{}^{\prime }}=\frac{1}{2\sqrt{x}},x>0$

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a)

$f\left(x\right)=x^3,x_0=-5$

Wyznaczamy funkcję pochodną: 

$f{}^{\prime }\left(x\right)=\left(x^3\right){}^{\prime }=3x^2$

wobec tego:

$f{}^{\prime }\left(x_0\right)=f{}^{\prime }\left(-5\right)=3\cdot {\left(-5\right)}^2=3\cdot 25=75$     

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b)

$f\left(x\right)=\frac{1}{x},x_0=\frac{\sqrt{2}}{2}$

Wyznaczamy funkcję pochodną:

$f{}^{\prime }\left(x\right)=\left(\frac{1}{x}\right){}^{\prime }=-\frac{1}{x^2}$

wobec tego:

$f{}^{\prime }\left(x_0\right)=f{}^{\prime }\left(\frac{\sqrt{2}}{2}\right)=-\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}=$   

$=-\frac{1}{\frac{2}{4}}=-\frac{1}{\frac{1}{2}}=-1\cdot 2=-2$  

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c)

$f\left(x\right)=\sqrt{x},x_0=1\frac{9}{16}=\frac{25}{16}$

Wyznaczamy funkcję pochodną:

$f{}^{\prime }\left(x\right)=\left(\sqrt{x}\right){}^{\prime }=\frac{1}{2\sqrt{x}}$

wobec tego:

$f{}^{\prime }\left(x_0\right)=f{}^{\prime }\left(\frac{25}{16}\right)=\frac{1}{2\sqrt{\frac{25}{16}}}=$   

$=\frac{1}{2\cdot \frac{5}{4}}=\frac{1}{\frac{5}{2}}=1\cdot \frac{2}{5}=\frac{2}{5}$