Question

Помогите решить производную f(x)=(thx)(3-2x) f(x)=cos(4x^5+3x)

Answer 1

$f(x)=Ctg(3-2x)\\\\f'(x) = [Ctg(3-2x)]'=- \frac{1}{Sin ^{2}(3-2x) }*(3-2x)'= \frac{2}{Sin ^{2}(3-2x) }$

$f(x)=Cos(4 x^{5} +3x)\\\\f'(x)=[Cos(4 x^{5} +3x)]'=-Sin(4 x^{5}+3x)*(4 x^{5} +3x)'=$$=-Sin(4 x^{5}+3x)*(20 x^{4}+3)$