Question

Dada a função 

$f(x)=4x^4- \frac{1}{4x^4}$

Determine f'(x)

Answer 1

f = u - v
f' = u' - v'

$u=4x^4$
$u'=16x^3$
$v=\frac{1}{4}.\frac{1}{x^4}=\frac{1}{4} \ . \ x^{-4}\\ \\ v'=\frac{1}{4}.-4x^{-5}=-x^{-5}$

$f'(x)=16x^3-(-x^{-5})=16x^3+x^{-5}=\boxed{16x^3+\frac{1}{x^5}}$