Question

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

Answer 1

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