Jeśli funkcja f ma pochodną w punkcie x oraz c jest dowolną stałą, to: ${\left(c\cdot f\left(x\right)\right)}^{{}^{\prime }}=c\cdot f^{{}^{\prime }}\left(x\right)$

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a)

$f\left(x\right)=12x^3$  

Wyznaczamy pochodną funkcji f:

$f^{{}^{\prime }}\left(x\right)={\left(12x^3\right)}^{{}^{\prime }}=12\cdot {\left(x^3\right)}^{{}^{\prime }}=12\cdot 3x^2=36x^2$  

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b) 

$f\left(x\right)=0,5x^6$  

$f^{{}^{\prime }}\left(x\right)={\left(0,5x^6\right)}^{{}^{\prime }}=0,5\cdot {\left(x^6\right)}^{{}^{\prime }}=0,5\cdot 6x^5=3x^5$  

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c)

$f\left(x\right)=4x^7$  

$f^{{}^{\prime }}\left(x\right)={\left(4x^7\right)}^{{}^{\prime }}=4\cdot {\left(x^7\right)}^{{}^{\prime }}=4\cdot 7x^6=28x^6$  

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d)

$f\left(x\right)=6x^{-1},x\ne 0$  

$f^{{}^{\prime }}\left(x\right)={\left(6x^{-1}\right)}^{{}^{\prime }}=6\cdot {\left(x^{-1}\right)}^{{}^{\prime }}=6\cdot \left(-1\right)\cdot x^{-2}=$

$=-6x^{-2}$   

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e)

$f\left(x\right)=2x^{-6},x\ne 0$  

$f^{{}^{\prime }}\left(x\right)={\left(2x^{-6}\right)}^{{}^{\prime }}=2\cdot {\left(x^{-6}\right)}^{{}^{\prime }}=2\cdot \left(-6\right)x^{-7}=$

$=-12x^{-7}$   

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f) 

$f\left(x\right)=\frac{1}{2}\sqrt[4]{x},x\ge 0$  

$f^{{}^{\prime }}\left(x\right)={\left(\frac{1}{2}\sqrt[4]{x}\right)}^{{}^{\prime }}=\frac{1}{2}\cdot {\left(\sqrt[4]{x}\right)}^{{}^{\prime }}=$

$=\frac{1}{2}\cdot {\left(x^{\frac{1}{4}}\right)}^{{}^{\prime }}=\frac{1}{2}\cdot \frac{1}{4}{x}^{-\frac{3}{4}}=$   

$=\frac{1}{8}{x}^{-\frac{3}{4}},x>0$