How Do You Differentiate y = x^(ln x)?

[iRaid]

Homework Statement

Find dy/dx.
y=x^{lnx}

Homework Equations

The Attempt at a Solution

ln y= ln x^{ln x}
ln y= (ln x)(ln x)

Taking the derivative now:

\frac{1}{y}y'= (\frac{1}{x})(\frac{1}{x})
\frac{1}{y}y'= (\frac{1}{x})^2

Multiply by y:

y'= (\frac{x^{lnx}}{x^2})

But it's not the right answer :?

 

[Mark44]

Homework Statement

Find dy/dx.
y=x^{lnx}

Homework Equations

The Attempt at a Solution

ln y= ln x^{ln x}
ln y= (ln x)(ln x)

Taking the derivative now:

\frac{1}{y}y'= (\frac{1}{x})(\frac{1}{x})

Are you trying to say that d/dx(fg) = f'g'?

\frac{1}{y}y'= (\frac{1}{x})^2

Multiply by y:

y'= (\frac{x^{lnx}}{x^2})

But it's not the right answer :?

 

[McAfee]

I'll give you a hint for the way I would solve this problem. Use the chain rule!

 

[iRaid]

Are you trying to say that d/dx(fg) = f'g'?

Wow forget the basics and you completely screw up lmao..

So then I get:

y'=(\frac{lnx}{x}+\frac{lnx}{x})(y)
y'=(\frac{2lnx}{x})(x^{lnx})
y'=\frac{2x^{lnx}lnx}{x}

 

[iRaid]

I'll give you a hint for the way I would solve this problem. Use the chain rule!

Too bad the chain rule doesn't work here.

 

[dextercioby]

Looks ok now.

 

[McAfee]

Too bad the chain rule doesn't work here.

anything is possible

 

[iRaid]

anything is possible

Not the answer the book gives..

 

[McAfee]

Not the answer the book gives..

Yeah. I know. It is a different way of solving the problem, but the answer is still right.

 

[dextercioby]

Needn't have copied the fine (and smart) calculation. The answer is nonetheless correct.

 

[HallsofIvy]

Just to add: there are two serious mistakes one could make in differentiating
f(x)^{g(x)}:

1) Treat g(x) as if it were a constant- g(x)f(x)^{g(x)- 1}
2) Treat f(x) as if it were a constant- f(x)^{g(x)}ln(g(x))

The peculiar thing is that the correct derivative is the sum of those two "mistakes".

 

[Curious3141]

Homework Statement

Find dy/dx.
y=x^{lnx}

My preferred way to do this:

y={(e^{\ln x})}^{\ln x} = e^{(\ln x)^2}

Differentiate with Chain Rule,

y' = e^{(\ln x)^2}(\frac{2\ln x}{x}) = \frac{2x^{\ln x}}{x}