Find the gradient of the curve y = x^2(ln(x)) at x = e

We'll need to use the product rule.

Let's take u = x^2 -> du/dx = 2x, and v = ln(x) -> dv/dx = 1/x

Then dy/dx = x^2(1/x) + 2xln(x) = x + 2xln(x)

Substituting our x value gives (dy/dx)|(x = e) = e + 2eln(e) = 3e