Where did this derivative come from ?

Question

f(x)= x^x² 

 

I got f'(x)= x^x² (x+2xlnx) 

 

It says the answer is x^x²⁺¹ (1+2lnx) 

 

I have no clue how to get to that answer. What am i missing ? 

Answer 1

f(x) = x^{x^2} = e^{lnx^x^2} = e^{x^2lnx}

 

Let u = x²ln(x) then apply the Chain Rule:

 

     df(x)/dx = de^u/du * du/dx = e^u * du/dx = x^{x^2} * d(x²ln(x))/dx

 

Apply the Product Rule:

 

     = x^{x^2} * (2xlnx + x²/x)

     = x^{x^2} * (2xlnx + x)

     = x^{x^2} * x(2lnx + 1)

 

This is what you got, and it's correct.  If you apply the rules of exponents from algebra 1, you'll recall that x^a*x = x^a+1.  Hence x^{x^2}*x = x^{x^2 + 1}, so:

 

     = x^{x^2 + 1}*(2lnx + 1)

 

So your answer is correct; you just forgot the algebra 1 rules of exponents.