# What is the derivative of ((x^2 + 1)/x)^5?

Mar 29, 2017

$\frac{d}{\mathrm{dx}} {\left(\frac{{x}^{2} + 1}{x}\right)}^{5} = \frac{5 {\left({x}^{2} + 1\right)}^{4} \left({x}^{2} - 1\right)}{x} ^ 6$

#### Explanation:

Using the chain rule:

$\frac{d}{\mathrm{dx}} {\left(\frac{{x}^{2} + 1}{x}\right)}^{5} = 5 {\left(\frac{{x}^{2} + 1}{x}\right)}^{4} \cdot \frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} + 1}{x}\right)$

We can now calculate:

$\frac{d}{\mathrm{dx}} \left(\frac{{x}^{2} + 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(x + \frac{1}{x}\right) = 1 - \frac{1}{x} ^ 2 = \frac{{x}^{2} - 1}{x} ^ 2$

and conclude that:

$\frac{d}{\mathrm{dx}} {\left(\frac{{x}^{2} + 1}{x}\right)}^{5} = 5 {\left(\frac{{x}^{2} + 1}{x}\right)}^{4} \cdot \left(\frac{{x}^{2} - 1}{x} ^ 2\right) = \frac{5 {\left({x}^{2} + 1\right)}^{4} \left({x}^{2} - 1\right)}{x} ^ 6$