Question

A regra da cadeia é o método utilizado para derivar funções compostas e a sua fórmula canônica é dy/dx=dy/du* du/dx. Dito isto, aponte a alternativa correta que apresenta o resultado da derivada da função f(x)=2/(√2x2+2.):

a) - 4x /√(2x + 2)3

b) - 4x /√(2x + 2)2

c) x /√(2x + 2)3

d) - 4 /√(2x + 2)3

e) - x /√(2x + 2)3

Answer 1

$\Large\boxed{\begin{array}{l}\sf f(x)=\dfrac{2}{\sqrt{2x^2+2}}\\\\\sf f(x)=2\cdot(2x^2+2)^{-\frac{1}{2}}\\\\\sf f'(x)=\diagup\!\!\!\!2\cdot\bigg(-\dfrac{1}{\diagup\!\!\!\!2}\bigg)\cdot(2x^2+2)^{-\frac{3}{2}}\cdot4x\\\\\sf f'(x)=-\dfrac{4x}{(2x^2+2)^{\frac{3}{2}}}\\\\\sf f'(x)=-\dfrac{4x}{(\sqrt{2x^2+2})^3}\\\\\Huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~a}}}}\end{array}}$

Answer 2

Resposta:

$\sf f(x)=\dfrac{2}{\sqrt{2x^2+2}}$

$\sf f'(x)=\dfrac{d}{dx}\bigg(\dfrac{2}{\sqrt{2x^2+2}}\bigg)$

$\sf f'(x)=2\cdot\dfrac{d}{dx}\bigg(\dfrac{1}{\sqrt{2x^2+2}}\bigg)$

Pela regra do quociente, na qual d/dx(f/g) = [d/dx(f) . g - f . d/dx(g)]/g² :

$\sf f'(x)=2\cdot\dfrac{\frac{d}{dx}(1)\cdot\sqrt{2x^2+2}-1\cdot\frac{d}{dx}\big(\sqrt{2x^2+2}\,\big)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=2\cdot\dfrac{0\cdot\sqrt{2x^2+2}-\frac{d}{dx}\big(\sqrt{2x^2+2}\,\big)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=2\cdot\dfrac{0-\frac{d}{dx}\big(\sqrt{2x^2+2}\,\big)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{d}{dx}\big(\sqrt{2x^2+2}\,\big)}{\big(\sqrt{2x^2+2}\big)^2}$

Agora chegou o momento de aplicar a regra da cadeia. Fazendo u = 2x² + 2:

$\sf f'(x)=-\,2\cdot\dfrac{\frac{d}{du}\big(\sqrt{u}\big)\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{d}{du}\big(u^{\frac{1}{2}}\big)\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2}\cdot u^{\frac{1}{2}-1}\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2}\cdot u^{-\frac{1}{2}}\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2}\cdot\frac{1}{u^{\frac{1}{2}}}\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2\sqrt{u}}\cdot\frac{d}{dx}(u)}{\big(\sqrt{2x^2+2}\big)^2}$

Substituindo de volta 2x² + 2 = u:

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2\sqrt{2x^2\:+\:2}}\cdot\frac{d}{dx}(2x^2+2)}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2\sqrt{2x^2\:+\:2}}\cdot\big[\frac{d}{dx}(2x^2)+\frac{d}{dx}(2)\big]}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2\sqrt{2x^2\:+\:2}}\cdot\big[2\cdot2x^{2-1}+0\big]}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{1}{2\sqrt{2x^2\:+\:2}}\cdot4x}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{\frac{2x}{\sqrt{2x^2\:+\:2}}}{\big(\sqrt{2x^2+2}\big)^2}$

$\sf f'(x)=-\,2\cdot\dfrac{2x}{\big(\sqrt{2x^2+2}\big)^2\big(\sqrt{2x^2+2}\big)}$

$\red{\underline{\boxed{\sf f'(x)=-\dfrac{4x}{\big(\sqrt{2x^2+2}\big)^3}}}$

Letra A