Question

a derivada de (x+x^-1)^3

Answer 1

Acompanhe:

$\displaystyle \mathsf{y=\bigg(x+\frac{1}{x}\bigg)^3} \\ \\ \\ \mathsf{ y' = 3 \cdot \bigg( x +\frac{1}{x}\bigg)^{3-1} \cdot \bigg( x+\frac{1}{x}\bigg)' } \\ \\ \\ \mathsf{ y' = 3 \cdot \bigg( x +\frac{1}{x}\bigg)^{2} \cdot \bigg( 1 - \frac{1}{x^2} \bigg) } \\ \\ \\ \boxed{\boxed{ \mathsf{ y' = \bigg(3-\frac{3}{x^2} \bigg) \cdot \bigg( x +\frac{1}{x}\bigg)^{2} } }}$

Obs:

$\displaystyle \mathsf{\frac{1}{x}' = x^{-1} = -1 \cdot x^{-1-1} = -1 \cdot x^{-2} = -\frac{1}{x^2} }$