Question

Найти производную функции f(x)＝x'2/x'3+3
f（x）＝(x'3+2x)'4
f(x)＝1/（5x+1)'4
f(x)＝√3x＋11
f（x）＝√2x'4+3x-7

Answer 1

$\displaystyle f(x)=\frac{x^2}{x^3+3};\\f'(x)=\frac{2x(x^3+3)-x^2*3x^2}{(x^3+3)^2}=\frac{6x-x^4}{(x^3+3)^2};$

$\displaystyle f(x)=(x^3+2x)^4;\\f'(x)=4(x^3+2x)^3*(3x^2+2)$

$\displaystyle f(x)=\frac{1}{(5x+1)^4};\\f'(x)=\frac{-4*4}{(5x+1)^5}=\frac{-16}{(5x+1)^5}$

$\displaystyle f(x)=\sqrt{3x+11};\\f'(x)=\frac{3}{2\sqrt{3x+11}}$

$\displaystyle f(x)=\sqrt{2x^4+3x-7};\\f'(x)=\frac{8x^3+3}{2\sqrt{2x^4+3x-7}}$