Calculus help!

[Scrooge2]

find dy/dx if y=x^x for all x>0

I did this stuff last semester I don't remember any of it!!! WAH!!

 

[Syringer]

It's going to be x^x (ln x + d/dx (x))

So it's

x^x (ln x + 1)

 

[alphatarget1]

take the natural log of both sides

ln(y)=ln(x^x)

and then differentiate

d(ln y)= d(ln x^x)

see what you get

 

[QTArrhythmic]

Originally posted by: Syringer
It's going to be x^x (ln x + d/dx (x))

So it's

x^x (ln x + 1)

That's what I get. Take the Ln first then differentiate from there:
1.) Ln y=xLn x (then take the dy/dx)

2.) (1/y)(dy/dx)= x(1/x)+(1)Ln x

3.)(dy/dx)=y[1+Ln x]

4.) y=x^x

so, dy/dx=(x^x)(1+Ln x)

tell me if I'm wrong

 

[Gnurb]

step by step

at calc101.com