Question

derivada de x/(x^2-1)^4 alguem me ajuda por favor....pela regra da cadeia.

Answer 1

$y=\dfrac{x}{(x^{2}-1)^{4}}$

Derivando y pela regra do quociente:

$y'=\dfrac{(x^{2}-1)^{4}\frac{d}{dx}(x)-x\frac{d}{dx}(x^{2}-1)^{4}}{[(x^{2}-1)^{4}]^{2}}\\\\\\y'=\dfrac{x^{2}-1-x\frac{d}{dx}(x^{2}-1)^{4}}{(x^{2}-1)^{8}}$

Podemos achar a derivada de (x² - 1)⁴ pela regra da cadeia

Definindo $f(x)=x^{4}$ e $g(x)=x^{2}-1$, temos que

$f(g(x))=(g(x))^{4}=(x^{2}-1)^{4}$

Derivando f(g(x)) pela regra da cadeia:

$[f(g(x))]'=f'(g(x))\cdot g'(x)\\\\\frac{d}{dx}(x^{2}-1)^{4}=4(g(x))^{3}\cdot g'(x)\\\\\frac{d}{dx}(x^{2}-1)^{4}=4(x^{2}-1)^{3}\cdot(2x-0)\\\\\frac{d}{dx}(x^{2}-1)^{4}=8x\cdot(x^{2}-1)^{3}$

Portanto:

$y'=\dfrac{(x^{2}-1)^{4}-x\frac{d}{dx}\cdot(x^{2}-1)^{4}}{(x^{2}-1)^{8}}\\\\\\y'=\dfrac{(x^{2}-1)^{4}-x\cdot8x\cdot(x^{2}-1)^{3}}{(x^{2}-1)^{8}}\\\\\\y'=\dfrac{(x^{2}-1)^{4}-8x^{2}\cdot(x^{2}-1)^{3}}{(x^{2}-1)^{8}}$

Colocando (x² - 1)³ em evidência:

$y'=\dfrac{(x^{2}-1)^{3}\cdot(x^{2}-1-8x^{2})}{(x^{2}-1)^{8}}\\\\\\y'=\dfrac{1\cdot(-7x^{2}-1)}{(x^{2}-1)^{8-3}}\\\\\\\boxed{\boxed{y'=-\dfrac{(7x^{2}+1)}{~(x^{2}-1)^{5}}}}$