Question

Qual derivada de ^4√1+2x+x^3

Answer 1

Resposta:

$\sf y=\sqrt[\sf4]{\sf1+2x+x^3}$

$\sf y'=\dfrac{d}{dx}(\sqrt[\sf4]{\sf1+2x+x^3})$

Pela regra da cadeia, dy/dx = dy/du . du/dx:

$\sf y'=\dfrac{d}{du}(\sqrt[\sf4]{\sf u})\cdot\dfrac{d}{dx}u$

$\sf y'=\dfrac{d}{du}(u^{\frac{1}{4}})\cdot\dfrac{d}{dx}u$

$\sf y'=\dfrac{1}{4}\cdot u^{\frac{1}{4}-1}\cdot\dfrac{d}{dx}u$

$\sf y'=\dfrac{1}{4}\cdot u^{-\frac{3}{4}}\cdot\dfrac{d}{dx}u$

$\sf y'=\dfrac{1}{4}\cdot \dfrac{1}{u^{\frac{3}{4}}}\cdot\frac{d}{dx}u$

$\sf y'=\dfrac{1}{4\cdot(1\,+\,2x\,+\,x^3)^{\frac{3}{4}}}\cdot\dfrac{d}{dx}(1+2x+x^3)$

$\sf y'=\dfrac{1}{4\cdot\sqrt[4]{\sf(1\,+\,2x\,+\,x^3)^3}}\cdot\dfrac{d}{dx}(1+2x+x^3)$

$\sf y'=\dfrac{1}{4\cdot\sqrt[4]{\sf(1\,+\,2x\,+\,x^3)^3}}\cdot\bigg(\dfrac{d}{dx}1+\dfrac{d}{dx}2x+\dfrac{d}{dx}x^3\bigg)$

$\sf y'=\dfrac{1}{4\cdot\sqrt[4]{\sf(1\,+\,2x\,+\,x^3)^3}}\cdot(0+2+3\cdot x^{3-1})$

$\sf y'=\dfrac{1}{4\cdot\sqrt[4]{\sf(1\,+\,2x\,+\,x^3)^3}}\cdot(3x^2+2)$

$\red{\boxed{\sf y'=\dfrac{3x^2+2}{4\cdot\sqrt[4]{\sf(1\,+\,2x\,+\,x^3)^3}}}}$