Question

Найти производную .Нужно до утра
$y=ctg(tg(sin(5x)))$

Answer 1

$y=ctg(tg(sin(5x)))\y_1'=(ctg(tg(sin(5x))))'\y=ctg(x)\y'=- frac{1}{sin^2(x)} = textgreater \= textgreater y_1'=-frac{1}{sin^2(tg(sin(5x)))} *(tg(sin(5x)))'\y_2=tg(sin(5x))\y=tg(x)\y'=frac{1}{cos^2(x)} = textgreater \= textgreater y_2'=frac{1}{cos^2(sin(5x))} *(sin(5x))'\y_3=sin(5x)\y=sin(x)\y'=cos(x)*x'= textgreater \= textgreater y_3=5cos(5x)= textgreater y_2=frac{5cos(5x)}{cos^2(sin(5x))} = textgreater \= textgreater y_1=-frac{5cos(5x)}{sin^2(tg(sin(5x)))cos^2(sin(5x))}$

Answer 2

Решение задания приложено