Question

DERIVADA DA FUNÇÃO f(x) = (x^2-x+4)^5/4

Answer 1

$df/dx= \frac{5}{4}(x^2-x+40)^[tex] \frac{1}{4}$(2x-1)[/tex]

Answer 2

$f(x) \ = \ (x^2 \ - \ x \ + \ 4)^{\frac{5}{4}}$

Utilizando a regra da cadeia:

$f'(x) \ = \ \frac{5}{4} \ . \ (x^2 \ - \ x \ + \ 4)^{\frac{5}{4} \ - \ 1} \ . \ (2 \ . \ x^{2 \ - \ 1} \ - \ 1 \ . \ x^{1 \ - \ 1} \ + \ 0)$

$f'(x) \ = \ \frac{5}{4} \ . \ (x^2 \ - \ x \ + \ 4)^{\frac{5}{4} \ - \ 1} \ . \ (2 \ . \ x^{1} \ - \ 1 \ . \ x^{0})$

$f'(x) \ = \ \frac{5}{4} \ . \ (x^2 \ - \ x \ + \ 4)^{\frac{5 \ - \ 4}{4}} \ . \ (2x \ - \ 1)$

$\boxed{\bold{f'(x) \ = \ \frac{5}{4} \ . \ (x^2 \ - \ x \ + \ 4)^{\frac{1}{4}} \ . \ (2x \ - \ 1)}}$