Question

defina a derivada da função:

f(x)= $( \frac{x+1}{x^2+1} )^4$

Answer 1

$f(x)=\left(\dfrac{x+1}{x^2+1} \right )^{\!4}$


Podemos enxergar a função acima como uma função composta:

$\left\{ \begin{array}{l} f(x)=[g(x)]^4\\\\ g(x)=\dfrac{x+1}{x^2+1} \end{array} \right.$


Derivaremos $f$ usando a Regra da Cadeia:

$f'(x)=\left([g(x)]^4 \right )'\\\\ f'(x)=4\,[g(x)]^{4-1}\cdot g'(x)\\\\ f'(x)=4\,[g(x)]^3\cdot g'(x)\\\\ f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \left(\dfrac{x+1}{x^2+1} \right )'$


Derivando $g$ pela Regra do Quociente:

$f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \dfrac{(x+1)'\cdot (x^2+1)-(x+1)\cdot (x^2+1)'}{(x^2+1)^2}\\\\\\ f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \dfrac{(1+0)\cdot (x^2+1)-(x+1)\cdot (2x+0)}{(x^2+1)^2}\\\\\\ f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \dfrac{x^2+1-(x+1)\cdot 2x}{(x^2+1)^2}\\\\\\ f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \dfrac{x^2+1-2x^2-2x}{(x^2+1)^2}\\\\\\ f'(x)=4\left(\dfrac{x+1}{x^2+1} \right )^{\!3}\cdot \dfrac{1-2x-x^2}{(x^2+1)^2}\\\\\\ \therefore~~\boxed{\begin{array}{c} f'(x)=\dfrac{4\,(x+1)^3\,(1-2x-x^2)}{(x^2+1)^5} \end{array}}$