Question

What is the derivative of `f(x)=e^(x^2lnx)`?

Answer 1

`f'(x)=xe^(x^2ln(x))(2ln(x)+1)`

Explanation:

We will use the following:

• The chain rule.

• The product rule.

• `d/dx e^x = e^x`

• `d/dx x^n = nx^(n-1)`

• `d/dx ln(x) = 1/x`

With those:

`f'(x) = d/dxe^(x^2ln(x))`

`= e^(x^2ln(x))(d/dxx^2ln(x))`

`=e^(x^2ln(x))(x^2(d/dxln(x))+ln(x)(d/dxx^2))`

`=e^(x^2ln(x))(x^2(1/x)+ln(x)(2x))`

`=e^(x^2ln(x))(x+2xln(x))`

`=xe^(x^2ln(x))(2ln(x)+1)`