Podstawowe wzory na pochodne funkcji:

Funkcja	Pochodna funkcji
$f\left(x\right)=c,c\in \mathbb{R}$	$f{}^{\prime }\left(x\right)=0$
$f\left(x\right)=x$	$f{}^{\prime }\left(x\right)=1$
$f\left(x\right)=x^n,n\in \mathbb{N}$	$f{}^{\prime }\left(x\right)=n\cdot x^{n-1}$
$f\left(x\right)=\frac{1}{x}$	$f{}^{\prime }\left(x\right)=-\frac{1}{x^2}$
$f\left(x\right)=\sqrt{x}$	$f{}^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$
$f\left(x\right)=x^a,a\in \mathbb{R}$	$f{}^{\prime }\left(x\right)=a\cdot x^{a-1}$

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Twierdzenie  Jeśli funkcje f i g są różniczkowalne w zbiorze D, to dla dowolnej liczby x, x ∈ D: ${\left[f\left(x\right)+g\left(x\right)\right]}^{{}^{\prime }}=f{}^{\prime }\left(x\right)+g{}^{\prime }\left(x\right)$   ${\left[f\left(x\right)-g\left(x\right)\right]}^{{}^{\prime }}=f{}^{\prime }\left(x\right)-g{}^{\prime }\left(x\right)$   ${\left[f\left(x\right)\cdot g\left(x\right)\right]}^{{}^{\prime }}=f{}^{\prime }\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g{}^{\prime }\left(x\right)$   ${\left[c\cdot f\left(x\right)\right]}^{{}^{\prime }}=c\cdot f{}^{\prime }\left(x\right),c\in \mathbb{R}$   ${\left[\frac{f\left(x\right)}{g\left(x\right)}\right]}^{{}^{\prime }}=\frac{f{}^{\prime }\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g{}^{\prime }\left(x\right)}{{\left[g\left(x\right)\right]}^2},g\left(x\right)\ne 0$

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a)

Dana jest funkcja 

$f\left(x\right)=\frac{x^3-2}{3x^2+x},x\in \mathbb{R}/\left\{-\frac{1}{3},0\right\}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

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

$=\frac{3x^2\cdot \left(3x^2+x\right)-\left(x^3-2\right)\cdot \left(3\cdot 2x+1\right)}{{\left(3x^2+x\right)}^2}=\frac{9x^4+3x^3-\left(x^3-2\right)\cdot \left(6x+1\right)}{{\left(3x^2+x\right)}^2}=$  

$=\frac{9x^4+3x^3-\left(6x^4+x^3-12x-2\right)}{{\left(3x^2+x\right)}^2}=\frac{3x^4+2x^3+12x+2}{{\left(3x^2+x\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)==\frac{3x^4+2x^3+12x+2}{{\left(3x^2+x\right)}^2},x\in \mathbb{R}/\left\{-\frac{1}{3},0\right\}$  

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b)

Dana jest funkcja 

$f\left(x\right)=\frac{4x^2-5x}{x^2+x+2},x\in \mathbb{R}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

$f{}^{\prime }\left(x\right)=\left(\frac{4x^2-5x}{x^2+x+2}\right){}^{\prime }=\frac{\left(4x^2-5x\right){}^{\prime }\cdot \left(x^2+x+2\right)-\left(4x^2-5x\right)\cdot \left(x^2+x+2\right){}^{\prime }}{{\left(x^2+x+2\right)}^2}=$  

$=\frac{\left(4\cdot 2x-5\right)\left(x^2+x+2\right)-\left(4x^2-5x\right)\left(2x+1\right)}{{\left(x^2+x+2\right)}^2}=$  

$=\frac{\left(8x-5\right)\left(x^2+x+2\right)-\left(8x^3+4x^2-10x^2-5x\right)}{{\left(x^2+x+2\right)}^2}=$  

$=\frac{8x^3+8x^2+16x-5x^2-5x-10-8x^3+6x^2+5x}{{\left(x^2+x+2\right)}^2}=$  

$=\frac{9x^2+16x-10}{{\left(x^2+x+2\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)=\frac{9x^2+16x-10}{{\left(x^2+x+2\right)}^2},x\in \mathbb{R}$  

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c)

Dana jest funkcja 

$f\left(x\right)=\frac{x^2+8}{x+2},x\in \mathbb{R}/\left\{-2\right\}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

$f{}^{\prime }\left(x\right)=\left(\frac{x^2+8}{x+2}\right){}^{\prime }=\frac{\left(x^2+8\right){}^{\prime }\cdot \left(x+2\right)-\left(x^2+8\right)\cdot \left(x+2\right){}^{\prime }}{{\left(x+2\right)}^2}=$  

$=\frac{2x\cdot \left(x+2\right)-\left(x^2+8\right)\cdot 1}{{\left(x+2\right)}^2}=\frac{2x^2+4x-x^2-8}{{\left(x+2\right)}^2}=\frac{x^2+4x-8}{{\left(x+2\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)=\frac{x^2+4x-8}{{\left(x+2\right)}^2},x\in \mathbb{R}/\left\{-2\right\}$  

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d)

Dana jest funkcja 

$f\left(x\right)=\frac{5}{x^4+x^3+1},x\in \mathbb{R}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

$f{}^{\prime }\left(x\right)=\left(\frac{5}{x^4+x^3+1}\right){}^{\prime }=\frac{\left(5\right){}^{\prime }\cdot \left(x^4+x^3+1\right)-5\cdot \left(x^4+x^3+1\right){}^{\prime }}{{\left(x^4+x^3+1\right)}^2}=$  

$=\frac{0\cdot \left(x^4+x^3+1\right)-5\cdot \left(4x^3+3x^2\right)}{{\left(x^4+x^3+1\right)}^2}=\frac{-20x^3-15x^2}{{\left(x^4+x^3+1\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)=\frac{-20x^3-15x^2}{{\left(x^4+x^3+1\right)}^2},x\in \mathbb{R}$  

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e)

Dana jest funkcja 

$f\left(x\right)=\frac{x^4+3x+2}{x^3+1},x\in \mathbb{R}/\left\{-1\right\}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

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

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

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

$=\frac{4x^6+4x^3+3x^3+3-3x^6-9x^3-6x^2}{{\left(x^3+1\right)}^2}=\frac{x^6-2x^3-6x^2+3}{{\left(x^3+1\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)=\frac{x^6-2x^3-6x^2+3}{{\left(x^3+1\right)}^2},x\in \mathbb{R}/\left\{-1\right\}$  

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f)

Dana jest funkcja 

$f\left(x\right)=\frac{x}{x^5-1},x\in \mathbb{R}/\left\{1\right\}$  

Stosujemy wzór na pochodną ilorazu i otrzymujemy 

$f{}^{\prime }\left(x\right)=\left(\frac{x}{x^5-1}\right){}^{\prime }=\frac{\left(x\right){}^{\prime }\cdot \left(x^5-1\right)-x\cdot \left(x^5-1\right){}^{\prime }}{{\left(x^5-1\right)}^2}=$  

$=\frac{1\cdot \left(x^5-1\right)-x\cdot 5x^4}{{\left(x^5-1\right)}^2}=\frac{x^5-1-5x^5}{{\left(x^5-5\right)}^2}=\frac{-4x^5-1}{{\left(x^5-1\right)}^2}$  

czyli 

$f{}^{\prime }\left(x\right)=\frac{-4x^5-1}{{\left(x^5-1\right)}^2},x\in \mathbb{R}/\left\{1\right\}$