Question

Найти производную функции (arccos4x)(cos(8x-2))

Answer 1

$y=arccos4x\\ y'=(arccos4x)'=-\frac{1}{\sqr{1-(4x)^2}}*4x'=-\frac{4}{\sqrt{1-16x^2}}$

$y=cos(8x-2)\\ y'=(8x-2)'*-sin(8x-2)=-8sin(8x-2)$