Question

Qual a derivada da função: f(x) = 8x/[(4-x^2)^2]

Answer 1

Olá, siga a explicação:

Diferenciando pela Regra do Quociente:

$f(x)= \dfrac{8x}{[(4-x^2)^2]}$

Sendo:

$\dfrac{d}{dx} \left [ \dfrac{f(x)}{g(x)} \right ]= \dfrac{g(x)\dfrac{d}{dx} [f(x)] - f(x) \dfrac{d}{dx} [g(x)] }{g(x)^2} .$

$\dfrac{24x^2+32}{(-x^2+4)^3}$

• Att. MatiasHP

Answer 2

Queremos determinar a derivada da função:

$\begin{array}{l}\\\sf f(x)=\dfrac{8x}{(4-x^2)^2}\end{array}$

Assim...

$\begin{array}{l}\sf f'(x)=\bigg[\dfrac{8x}{(4-x^2)^2}\bigg]'\\\\\end{array}$

Pela regra do quociente:

• [f(x)/g(x)]' = [f'(x) * g(x) – g'(x) * f(x)]/(g(x))²

$\begin{array}{l}\\\sf f'(x)=\dfrac{[8x]'\cdot(4-4x^2)^2-[(4-x^2)^2]'\cdot8x}{[(4-x^2)^2]^2}\\\\\sf f'(x)=\dfrac{[8x]'\cdot(4-4x^2)^2-[(4-x^2)^2]'\cdot8x}{(4-x^2)^4}\\\\\end{array}$

A derivada de uma variável se torna uma constante

• [ax]' = a

$\begin{array}{l}\\\sf f'(x)=\dfrac{8\cdot(4-4x^2)^2-[(4-x^2)^2]'\cdot8x}{(4-x^2)^4}\\\\\end{array}$

Para derivar (4 - x²)² vamos aplicar a regra:

• f(g) = [f]' * [g]'

$\begin{array}{l}\\\sf f'(x)=\dfrac{8\cdot(4-4x^2)^2-[(4-x^2)^2]'\cdot[4-x^2]'\cdot8x}{(4-x^2)^4}\\\\\end{array}$

Pelas propriedades:

• [axⁿ]' = n * axⁿ⁻¹
• [a]' = 0

$\begin{array}{l}\\\sf f'(x)=\dfrac{8\cdot(4-x^2)^2-2\cdot(4-x^2)^{2-1}\cdot(0-2\cdot x^{2-1})\cdot8x}{(4-x^2)^4}\\\\\sf f'(x)=\dfrac{8\cdot(4-x^2)^2-2(4-x^2)\cdot(-2x)\cdot8x}{(4-x^2)^4}\\\\\sf f'(x)=\dfrac{8\cdot(4-x^2)^2+32x^2(4-x^2)}{(4-x^2)^4}\end{array}$

Por fator comum em evidência:

$\begin{array}{l}\sf f'(x)=\dfrac{(8\cdot(4-x^2)+32x^2)\cdot(4-x^2)}{(4-x^2)^4}\\\\\sf f'(x)=\dfrac{8\cdot(4-x^2)+32x^2}{(4-x^2)^3}\\\\\sf f'(x)=\dfrac{8\cdot(4)+8\cdot(-x^2)+32x^2}{(4-x^2)^3}\\\\\sf f'(x)=\dfrac{32-8x^2+32x^2}{(4-x^2)^3}\\\\\!\boxed{\sf f'(x)=\dfrac{32+24x^2}{(4-x^2)^3}}\end{array}$

$~~$

Att. Nasgovaskov

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