# What is the derivative of  y= xlnx?

May 16, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \ln x$

#### Explanation:

We have:

$y = x \ln x$

We can apply the product rule to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(x\right) \left(\frac{d}{\mathrm{dx}} \ln x\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left(\ln x\right)$

Noting a standard calculus result:

$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$

We get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \cdot \frac{1}{x} + 1 \cdot \ln x$

$\text{ } = 1 + \ln x$

Corollary
We have just shown that:

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

If we now integrate both sides, then we get:

$\setminus \setminus \setminus \setminus \setminus x \ln x = \int \setminus \left(1 + \ln x\right) \setminus \mathrm{dx}$

$\therefore x \ln x = x - c + \int \setminus \ln x \setminus \mathrm{dx}$

Hence:

$\int \setminus \ln x \setminus \mathrm{dx} = x \ln x - x + c$