Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 18

Find $y'$ of $\displaystyle y = \left( x + \frac{1}{x^2} \right)^{\sqrt{7}}$


$\begin{equation} \begin{aligned} y' &= \frac{d}{dx} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}}\\ \\ y' &= \sqrt{7} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}-1} \frac{d}{dx} \left( x + \frac{1}{x^2} \right)\\ \\ y' &= \sqrt{7} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}-1} \left[ 1 + \frac{x^2 \frac{d}{dx}(1) - (1) \frac{d}{dx}(x^2) }{(x^2)^2} \right]\\ \\ y' &= \sqrt{7} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}-1} \left[ 1 + \frac{(x^2)(0)-(1)(2x)}{x^4} \right]\\ \\ y' &= \sqrt{7} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}-1} \left[ 1 + \left( \frac{-2x}{x^4} \right) \right]\\ \\ y' &= \sqrt{7} \left( x + \frac{1}{x^2} \right)^{\sqrt{7}-1} \left( 1 - \frac{2}{x^3} \right) \end{aligned} \end{equation}$