Question

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

Answer 1

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

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