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Să se calculeze derivatele funcțiilor :

Să Se Calculeze Derivatele Funcțiilor class=

Răspuns :

[tex]a)f(x) = {x}^{3} + 3x + 1[/tex]

[tex]f'(x) = ( {x}^{3} + 3x + 1)'[/tex]

[tex]f'(x) = ( {x}^{3} )' + (3x)' + 1'[/tex]

[tex]f'(x) = 3 {x}^{3 - 1} + 3x' + 0[/tex]

[tex]f'(x) = 3 {x}^{2} + 3 \times 1[/tex]

[tex]f'(x) = 3 {x}^{2} + 3[/tex]

[tex]b)f(x) = 2x - {x}^{4} [/tex]

[tex]f'(x) = (2x - {x}^{4} )'[/tex]

[tex]f'(x) = (2x)' - ( {x}^{4} )'[/tex]

[tex]f'(x) = 2x' - 4 {x}^{4 - 1} [/tex]

[tex]f'(x) = 2 \times 1 - 4 {x}^{3} [/tex]

[tex]f'(x) = 2 - 4 {x}^{3} [/tex]

[tex]c)f(x) = x + 2 \sqrt{x} [/tex]

[tex]f'(x) = (x + 2 \sqrt{x} )'[/tex]

[tex]f'(x) = x' + (2 \sqrt{x} )'[/tex]

[tex]f'(x) = 1 + 2\times ( \sqrt{x} )'[/tex]

[tex]f'(x) = 1 + 2 \times \frac{1}{2 \sqrt{x} } [/tex]

[tex]f'(x) = 1 + \frac{1}{ \sqrt{x} } [/tex]

[tex]f'(x) = 1 + \frac{ \sqrt{x} }{x} [/tex]

[tex]f'(x) = \frac{x + \sqrt{x} }{x} [/tex]

[tex]d)f(x) = {x}^{3} + sinx + cosx[/tex]

[tex]f'(x) = ( {x}^{3} + sinx + cosx)'[/tex]

[tex]f'(x) = ( {x}^{3} )' + (sinx)' + (cosx) '[/tex]

[tex]f'(x) = 3 {x}^{3 - 1} + cosx + ( - sinx)[/tex]

[tex]f'(x) = 3 {x}^{2} + cosx - sinx[/tex]

[tex]e)f(x) = 2 {x}^{3} + lnx[/tex]

[tex]f'(x) = (2 {x}^{3} + lnx)'[/tex]

[tex]f'(x) = (2 {x}^{3} )' + (lnx)'[/tex]

[tex]f'(x) = 2( {x}^{3} )' + \frac{1}{x} [/tex]

[tex]f'(x) = 2 {x}^{3 - 1} + \frac{1}{x} [/tex]

[tex]f'(x) = 2 {x}^{2} + \frac{1}{x} [/tex]

[tex]f'(x) = \frac{2 {x}^{3} + 1}{x} [/tex]

[tex]f)f(x) = {2}^{x} + {3}^{x} - x[/tex]

[tex]f'(x) = ( {2}^{x} + {3}^{x} - x)'[/tex]

[tex]f'(x) = ({2}^{x})' + ( {3}^{x} )' - x'[/tex]

[tex]f'(x) = {2}^{x} ln2 + {3}^{x} ln3 - 1[/tex]