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

1 Answer
Jan 24, 2017

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

Explanation:

We will use the following:

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)#