Derivative of x ln x

Differentiate again: \frac{d^2}{dx^2}\left[x \ln\left(x\right)\right] = \frac{1}{x}.

Plug into the derivative: f'(1.8) = 1.58779. That is the slope of the tangent line there; drag the graph to see it.

\frac{d}{dx}\left[x \ln\left(x\right)\right] = \ln\left(x\right) + 1

A standard product-rule exercise: (x)′·ln x + x·(ln x)′ = ln x + 1. The minimum of x ln x sits where ln x = −1, at x = 1/e.

Logarithm rule
$\frac{d}{dx}\left[\ln\left(x\right)\right] = \frac{1}{x}$
Product rule
With u = $x$ and v = $\ln\left(x\right)$ : (uv)′ = u′v + uv′.
$\frac{d}{dx}\left[x \ln\left(x\right)\right] = \left(1\right)\left(\ln\left(x\right)\right) + \left(x\right)\left(\frac{1}{x}\right)$

Drag the graph · at x = 1.96 the slope is 1.6729, the height of the dashed curve.

Questions students ask

What is the derivative of x ln x?

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

What is the second derivative of x ln x?

Differentiate again: $\frac{d^2}{dx^2}\left[x \ln\left(x\right)\right] = \frac{1}{x}$ .

What is the slope of x ln x at x = 1.8?

Plug into the derivative: $f'(1.8) = 1.58779$ . That is the slope of the tangent line there; drag the graph to see it.

How do you find the derivative of x ln x?

A standard product-rule exercise: (x)′·ln x + x·(ln x)′ = ln x + 1. The minimum of x ln x sits where ln x = −1, at x = 1/e.

What is the integral of x ln x?

See the integral of x ln x page for the antiderivative with steps.