# How do you differentiate y=e^x+x^10-1/x?

Aug 1, 2017

color(blue)(y'(x) = e^x + 10x^9 + 1/(x^2)

#### Explanation:

We're asked to find the derivative

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[y = {e}^{x} + {x}^{10} - \frac{1}{x}\right]$

The derivative of ${e}^{x}$ is defined as ${e}^{x}$:

$y ' \left(x\right) = {e}^{x} + \frac{d}{\mathrm{dx}} \left[{x}^{10}\right] - \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]$

Us the power rule on the ${x}^{10}$ term:

$y ' \left(x\right) = {e}^{x} + 10 {x}^{9} - \frac{d}{\mathrm{dx}} \left[\frac{1}{x}\right]$

Use the quotient rule on the $\frac{1}{x}$ term, which states

$\frac{d}{\mathrm{dx}} \left[\frac{u}{v}\right] = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

where

• $u = 1$

• $v = x$:

$y ' \left(x\right) = {e}^{x} + 10 {x}^{9} - \frac{x \frac{d}{\mathrm{dx}} \left[1\right] - 1 \frac{d}{\mathrm{dx}} \left[x\right]}{{x}^{2}}$

The derivative of $1$ is $0$ and the derivative of $x$ is $1$:

color(blue)(y'(x) = e^x + 10x^9 + 1/(x^2)