Question

Qual a derivada da raiz quadrada de 4x^5-x

Answer 1

$\frac{d}{dx}\left(\sqrt{4x^{5-x}}\right)$

$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(4x^{5-x}\right)$

$=\frac{1}{2\sqrt{u}}\cdot \:4x^{5-x}\left(-\ln \left(x\right)+\frac{5-x}{x}\right)$

$=\frac{1}{2\sqrt{4x^{5-x}}}\cdot \:4x^{5-x}\left(-\ln \left(x\right)+\frac{5-x}{x}\right)$

$=\frac{x^{-x+4}\left(-x\ln \left(x\right)+5-x\right)}{\sqrt{x^{5-x}}}$

então...

$\frac{d}{dx}\left(\sqrt{4x^{5-x}}\right)=\frac{x^{-x+4}\left(-x\ln \left(x\right)+5-x\right)}{\sqrt{x^{5-x}}}$

espero ter ajudado