## Calculus: Early Transcendentals 8th Edition

$y'=5(x+\dfrac{1}{x})^{4}(1-\dfrac{1}{x^{2}})$
$y=(x+\dfrac{1}{x})^{5}$ Let's rewrite the function like this: $y=(x+x^{-1})^{5}$ Now, differentiate using the chain rule: $y'=5(x+x^{-1})^{4}(x+x^{-1})'=5(x+\dfrac{1}{x})^{4}(1-x^{-2})$ $y'=5(x+\dfrac{1}{x})^{4}(1-\dfrac{1}{x^{2}})$