# How do you find the derivative of y=xlnx?

Dec 29, 2016

Use the product rule. $y ' = \ln \left(x\right) + 1$.

#### Explanation:

You'll need the product rule for this one. The product rule is given by:

$= \left(f \left(x\right) \cdot g \left(x\right)\right) ' = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

In the case of $y = x \ln \left(x\right)$, $f \left(x\right) = x$ and $g \left(x\right) = \ln \left(x\right)$.

First we take the derivative of $f \left(x\right)$. The derivative of a single variable (no coefficient, not raised to any power) is $1$. We leave $g \left(x\right)$ alone, so the first half of the derivative is simply $1 \cdot \ln \left(x\right) = \ln \left(x\right)$.

Then we take the derivative of $g \left(x\right)$. The derivative of $\ln \left(x\right)$ is $\frac{1}{x}$. We leave $f \left(x\right)$ alone, so the second have of the derivative is $x \cdot \frac{1}{x} = 1$.

Putting it all together, we get $y ' = \ln \left(x\right) + 1$.

Hope this helps!