Problem in the derivate of x^x

[soopo]

Homework Statement

Why is the derivate of x^x
x^x(1 + lnx)?

The Attempt at a Solution

I know the derivate can be explained by calculus.
However, I am not exactly sure how.

 

[Ofey]

One way to approach the problem is from the equivalence

e^(lnx)=x

Which gives that

e^(lnx^x)=x^x

Or

e^(xlnx)=x^x

This you should be able to derivate using the set of rules you already know.

 

[HallsofIvy]

Or, let y= x^x. Then ln(y)= x ln (x). Differentiate both sides with respect to x (implicit differentiation) and solve for dy/dx.

 

[soopo]

One way to approach the problem is from the equivalence

e^(lnx)=x

Which gives that

e^(lnx^x)=x^x

Or

e^(xlnx)=x^x

This you should be able to derivate using the set of rules you already know.

Thank you!

So we get the solution by differenting
y = e^{g(x)},
where g(x) = xlnx.

y' = g'(x) * e^{g(x)},
where g'(x) = lnx + 1
y' = (lnx + 1) * e^{g(x)}
y' = (lnx + 1) x^{x}

 

[soopo]

Or, let y= x^x. Then ln(y)= x ln (x). Differentiate both sides with respect to x (implicit differentiation) and solve for dy/dx.

Thank you HallofIvy!
Your answer is excellent.