Найти производную функций: а)y=(5x+1) б)y=cos в)f(x)=

Question 1

Найти производную функций:
а)y=(5x+1) $^{6}$
б)y=cos $x^{2}$ $\frac{x}{3}$
в)f(x)= $\sqrt{4-x}$

Question 2

A) $f^{'}( \sqrt{ (5x+1)^{6} } )$
$f^{'}((5x+1)^{6} )* \frac{1}{ 2 \sqrt{(5x+1)^{6}} }$
$f^{'}((5x+1)(5x+1)^{5} )* \frac{3}{ \sqrt{(5x+1)^{6}} }$
$f^{'}(5x)(5x+1)^{5}* \frac{3}{ \sqrt{(5x+1)^{6}} }$
$f^{'}(x)(5x+1)^{5}* \frac{15}{ \sqrt{(5x+1)^{6}} }$
$(5x+1)^{5} * \frac{15}{ \sqrt{(5x+1)^{6}} }$
$15 (5x+1)^{2}$

б) $f^{'}( cos x^{ \frac{2x}{3}} )=- sin x^{ \frac{2x}{3} }* f^{'} (x^{ \frac{2x}{3}})=- sin x^{ \frac{2x}{3} }* f^{'} (e^{ln(x)* \frac{2x}{3} })=$ $- sin x^{ \frac{2x}{3} }* f^{'} (ln(x)* \frac{2x}{3} })* e^{ln(x)* \frac{2x}{3} } =$ $- sin x^{ \frac{2x}{3} }*e^{ln(x)* \frac{2x}{3} }* (f^{'} (ln(x)* \frac{2x}{3}+ln(x)* f^{'} ( \frac{2x}{3} })) =-$ $- sin x^{ \frac{2x}{3} }*e^{ln(x)* \frac{2x}{3} }* ( \frac{2}{3}+ln(x) \frac{3* f^{'}(2x)+0*x}{9} ) =$ $- sin x^{ \frac{2x}{3} }*e^{ln(x)* \frac{2x}{3} }* ( \frac{2}{3}+ln(x) \frac{2* f^{'}(x)}{3} ) =$ $- sin x^{ \frac{2x}{3} }*( \frac{2}{3}+ \frac{2}{3}ln(x))*e^{ln(x)* \frac{2x}{3} } =$ $- ( \frac{2}{3}+ \frac{2}{3}ln(x))*sin x^{ \frac{2x}{3} }*e^{ln(x)* \frac{2x}{3} } =$ $- ( \frac{2}{3}+ \frac{2}{3}ln(x))*sin x^{ \frac{2x}{3} }*e^{ln(x)* \frac{2x}{3} } =- \frac{2}{3}(ln(x)+1) * sin x^{x* \frac{2}{3} }*e^{x* \frac{2}{3}*ln(x) }$ $=- \frac{2}{3}(ln(x)+1) * x^{x* \frac{2}{3} }*sin^{x* \frac{2}{3}}$

в) $f^{'}( \sqrt{4-x} ) =f^{'}(4-x) \frac{1}{2 \sqrt{4-x} } =f^{'}(-x) \frac{1}{2 \sqrt{4-x} } =-f^{'}(x) \frac{1}{2 \sqrt{4-x} } = \frac{1}{2 \sqrt{4-x} }$