Question

alguem me ajuda essa ? ficarei grato eternamente
Usando a regra da cadeia determine as derivadas das funções a seguir:

Answer 1

Explicação passo-a-passo:

a)

$\sf f(x)=ln~(x^2+3)$

$\sf y=ln~(x^2+3)$

Seja $\sf u=x^2+3$

$\sf y=ln~u$

Regra da cadeia:

$\sf \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$

Temos:

• $\sf y=ln~u$

$\sf \dfrac{dy}{du}=\dfrac{1}{u}$

• $\sf u=x^2+3$

$\sf \dfrac{du}{dx}=(x^2+3)'$

$\sf \dfrac{du}{dx}=2x$

Assim:

$\sf \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$

$\sf \dfrac{dy}{dx}=\dfrac{1}{u}\cdot 2x$

$\sf \dfrac{dy}{dx}=\dfrac{1}{x^2+3}\cdot 2x$

$\sf \red{\dfrac{dy}{dx}=\dfrac{2x}{x^2+3}}$

b)

$\sf y=\Big(\dfrac{x+1}{x^2+1}\Big)^4$

Seja $\sf u=\dfrac{x+1}{x^2+1}$

$\sf y=u^4$

Regra da cadeia:

$\sf \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$

Temos:

• $\sf y=u^4$

$\sf y=4\cdot u^3$

• $\sf u=\dfrac{x+1}{x^2+1}$

Regra do quociente:

$\sf \Big(\dfrac{f}{g}\Big)'=\dfrac{f'\cdot g-f\cdot g'}{g^2}$

$\sf u=\dfrac{x+1}{x^2+1}$

$\sf \dfrac{du}{dx}=\dfrac{(x+1)'\cdot(x^2+1)-(x+1)\cdot(x^2+1)'}{(x^2+1)^2}$

$\sf \dfrac{du}{dx}=\dfrac{1\cdot(x^2+1)-(x+1)\cdot2x}{(x^2+1)^2}$

$\sf \dfrac{du}{dx}=\dfrac{x^2+1-2x^2-2x}{(x^2+1)^2}$

$\sf \dfrac{du}{dx}=\dfrac{1-2x-x^2}{(x^2+1)^2}$

Assim:

$\sf \dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$

$\sf \dfrac{dy}{dx}=4\cdot u^3\cdot\dfrac{1-2x-x^2}{(x^2+1)^2}$

$\sf \dfrac{dy}{dx}=4\cdot\Big(\dfrac{x+1}{x^2+1}\Big)^3\cdot\dfrac{1-2x-x^2}{(x^2+1)^2}$

$\sf \red{\dfrac{dy}{dx}=\dfrac{4\cdot(x+1)^3\cdot(1-2x-x^2)}{(x^2+1)^5}}$