Найти производную .Нужно до утра

Решите задачу:

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