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.