Question

Помогите решить задачу на производных 1. f(x)=x^5+tg8x f(x)=6x^3-cos5x f(x)= -6√5+x

Answer 1

$(x^5)'+(tg(8x))'=5x^4+\frac{1}{cos^2(8x)}*(8x)'=5x^4+\frac{8}{cos^2(8x)}$

$f'(x)=(6x^3-cos(5x))'=6(x^3)'-(cos(5x))'=6*3x^2-(-sin(5x)*(5x)')=18x^2+5sin(5x)$

$f'(x)=(-6\sqrt{5+x})'=-6(\sqrt{5+x})'=-6((5+x)^{\frac{1}{2} })'=-6*\frac{1}{2} (5+x)^{-\frac{1}{2}} *(5+x)'=-3(5+x)^{-\frac{1}{2}} *1=-\frac{3}{\sqrt{5+x} }$