Question

Să se calculeze derivatele funcțiilor :

Answer 1

$a)f(x) = {x}^{3} + 3x + 1$

$f'(x) = ( {x}^{3} + 3x + 1)'$

$f'(x) = ( {x}^{3} )' + (3x)' + 1'$

$f'(x) = 3 {x}^{3 - 1} + 3x' + 0$

$f'(x) = 3 {x}^{2} + 3 \times 1$

$f'(x) = 3 {x}^{2} + 3$

$b)f(x) = 2x - {x}^{4}$

$f'(x) = (2x - {x}^{4} )'$

$f'(x) = (2x)' - ( {x}^{4} )'$

$f'(x) = 2x' - 4 {x}^{4 - 1}$

$f'(x) = 2 \times 1 - 4 {x}^{3}$

$f'(x) = 2 - 4 {x}^{3}$

$c)f(x) = x + 2 \sqrt{x}$

$f'(x) = (x + 2 \sqrt{x} )'$

$f'(x) = x' + (2 \sqrt{x} )'$

$f'(x) = 1 + 2\times ( \sqrt{x} )'$

$f'(x) = 1 + 2 \times \frac{1}{2 \sqrt{x} }$

$f'(x) = 1 + \frac{1}{ \sqrt{x} }$

$f'(x) = 1 + \frac{ \sqrt{x} }{x}$

$f'(x) = \frac{x + \sqrt{x} }{x}$

$d)f(x) = {x}^{3} + sinx + cosx$

$f'(x) = ( {x}^{3} + sinx + cosx)'$

$f'(x) = ( {x}^{3} )' + (sinx)' + (cosx) '$

$f'(x) = 3 {x}^{3 - 1} + cosx + ( - sinx)$

$f'(x) = 3 {x}^{2} + cosx - sinx$

$e)f(x) = 2 {x}^{3} + lnx$

$f'(x) = (2 {x}^{3} + lnx)'$

$f'(x) = (2 {x}^{3} )' + (lnx)'$

$f'(x) = 2( {x}^{3} )' + \frac{1}{x}$

$f'(x) = 2 {x}^{3 - 1} + \frac{1}{x}$

$f'(x) = 2 {x}^{2} + \frac{1}{x}$

$f'(x) = \frac{2 {x}^{3} + 1}{x}$

$f)f(x) = {2}^{x} + {3}^{x} - x$

$f'(x) = ( {2}^{x} + {3}^{x} - x)'$

$f'(x) = ({2}^{x})' + ( {3}^{x} )' - x'$

$f'(x) = {2}^{x} ln2 + {3}^{x} ln3 - 1$