Надо найти производную.

Решите задачу:

$f(x)= \frac{7x^3-21x^8}{x^5+x}$

$f'(x)= (\frac{7x^3-21x^8}{x^5+x} )'= \frac{(7x^3-21x^8)'*(x^5+x)-(7x^3-21x^8)*(x^5+x)'}{(x^5+x)^2}=$ $= \frac{(21x^2-168x^7)*(x^5+x)-(7x^3-21x^8)*(5x^4+1)}{(x^5+x)^2}=$ $=\frac{(21x^7+21x^3-168x^{12}-168x^8-(35x^7+7x^3-105x^{12}-21x^8)}{(x^5+x)^2}=$ $=\frac{21x^7+21x^3-168x^{12}-168x^8-35x^7-7x^3+105x^{12}+21x^8}{(x^5+x)^2}=$ $=\frac{-14x^7+14x^3-63x^{12}-147x^8}{(x(x^4+1))^2} =\frac{-14x^7+14x^3-63x^{12}-147x^8}{x^2(x^4+1)^2} =$ $=\frac{x^2(-14x^5+14x-63x^{10}-147x^6)}{x^2(x^4+1)^2} =\frac{-14x^5+14x-63x^{10}-147x^6}{(x^4+1)^2}$