# Dy/dx exponential function

1. Dec 1, 2011

### iRaid

1. The problem statement, all variables and given/known data
Find dy/dx.
$y=x^{lnx}$

2. Relevant equations

3. 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 :?

2. Dec 1, 2011

### Staff: Mentor

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

3. Dec 1, 2011

### McAfee

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

4. Dec 1, 2011

### iRaid

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

So then I get:

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

5. Dec 1, 2011

### iRaid

Too bad the chain rule doesn't work here.

6. Dec 1, 2011

### dextercioby

Looks ok now.

7. Dec 1, 2011

### McAfee

anything is possible

8. Dec 1, 2011

### iRaid

Not the answer the book gives..

9. Dec 1, 2011

### McAfee

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

10. Dec 1, 2011

### dextercioby

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

11. Dec 2, 2011

### HallsofIvy

Staff Emeritus
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".

12. Dec 2, 2011

### Curious3141

My preferred way to do this:

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

Differentiate with Chain Rule,

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