# How do you find the derivative of y=x^tan(x)?

Aug 1, 2014

This is a type of problem involving logarithmic differentiation.

Whenever you're trying to differentiate a variable raised to some power also involving that variable, it's a good hint that logarithmic differentiation will help you out.

1.) $y = {x}^{\tan} x$

The first step is to take the natural log of both sides:

2.) $\ln y = \ln {x}^{\tan} x$

Using the exponents property of logarithms, we bring the exponent out in front of the log as a multiplier. This is done to make differentiating easier:

3.) $\ln y = \tan x \cdot \ln x$

Now we implicitly differentiate, taking care to use the chain rule on $\ln y$. We will also apply the product rule to the right side of the equation:

4.) $\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[\tan x\right] \cdot \ln x + \frac{d}{\mathrm{dx}} \left[\ln x\right] \cdot \tan x$

We know that the derivative of $\tan x$ is equal to ${\sec}^{2} x$, and the derivative of $\ln x$ is $\frac{1}{x}$:

5.) $\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {\sec}^{2} x \ln x + \tan \frac{x}{x}$

Multiply both sides by $y$ to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

6.) $\frac{\mathrm{dy}}{\mathrm{dx}} = y \left({\sec}^{2} x \ln x + \tan \frac{x}{x}\right)$

We know $y$ from step 1, so we will substitute:

7.) $\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\tan} x \left({\sec}^{2} x \ln x + \tan \frac{x}{x}\right)$

And there is the derivative.