Question

Решите, пожалуйста, что можете
$f(x)=5x^{3} +x^{5}-2\sqrt{x}$
$f(x)=(3+x^{2} )*(7-\frac{3}{x})$
$f(x)=\frac{3-2x}{7x+3}$
$y=\frac{1}{(6x+1)^{3} } + \sqrt{5-9x}$
$y=sin2x+ctg7x-2$
$f(x)=e^{6x} * ln8x$
$y=\frac{e^{-5x}-2 }{3x}$

Answer 1

Ответ:

Объяснение:

$f'(x)=(5x^3+x^5-2\sqrt{x} )=5*3x^{2} +5x^{4}-\frac{2}{2\sqrt{x} } =15x^{2} +5x^{4} -\frac{1}{\sqrt{x} }$

$f'(x)=(21+7x^{2} -\frac{9}{x}-3x)' =7*2x+\frac{9}{x^2}-3=14x+\frac{9}{x^2}-3$

$f'(x)=\frac{(3-2x)'(7x+3)-(3-2x)(7x+3)'}{(7x+3)^2}=\frac{-2(7x+3)-7(3-2x)}{(7x+3)^2}=\\ \\ =\frac{-14x-6-21+14x}{(7x+3)^2} =\frac{-27}{(7x+3)^2}$

$f'(x)=(\frac{1}{6x+3)^2}+\sqrt{5-9x})'=((6x+3)^{-2})'+\frac{(5-9x)'}{2\sqrt{5-9x} }=\\ \\ =-\frac{2}{(6x+1)^3}*(6x+1)'-\frac{9}{2\sqrt{5-9x}}=-\frac{12}{((6x+1)^3}-\frac{9}{\sqrt{5-9x} }$

$f'(x)=(sin2x+ctg7x-2)'=cos2x*2-\frac{1}{sin^27x}*7=2cos2x-\frac{7}{sin^27x}$

$f'(x)=(e^{6x}*ln8x)'=(e^{6x})'*ln8x+e^{6x}*(ln8x)'=6e^{6x}*ln8x+e^{6x}\frac{8}{8x}=\\ \\ =6e^{6x}*ln8x+e^{6x}\frac{1}{x}$

$f'(x)=\frac{(e^{-5x}-2)'3x-(3x)'(e^{-5x}-2)}{9x^2}=\frac{-5e^{-5x}*3x-3e^{-5x}+6}{9x^2}=\\ \\ =\frac{-15xe^{-5x}-3e^{-5x}+6}{9x^2}$