# Derivative with natural log

1. Jan 11, 2005

### Shay10825

Hi everyone!

(d/dx) [xlnx]

Why is the answer 1+lnx and not 1??

~Thanks

2. Jan 11, 2005

### Curious3141

Do you know the product rule ?

3. Jan 11, 2005

### recon

$$\frac{d\ln x}{dx} = \frac{1}{x}$$, in case you didn't know.

4. Jan 11, 2005

$$f'(x) = x * (1/x) + ln x * 1 = 1 + ln x$$
$$d(uv) = vdu + udv$$

5. Jan 11, 2005

### Shay10825

Wow I'm really stupid!!! I don't know how I did not see that. Thanks

6. Jan 11, 2005

### Shay10825

At first I did:
x(1/x) which gave me 1 but why does this not work:

xlnx
ln (x^x)
[x(x^x-1)]/(x^x) ?????

I know the second way is not how you would usually do it but why does it not work??

7. Jan 11, 2005

### recon

What you've just written is very confusing, do you mind trying to make it a bit more clear?

8. Jan 11, 2005

### Shay10825

9. Jan 11, 2005

### Curious3141

$$\frac{d}{dx}(x^n) = nx^{n-1}$$ only works when n is a constant. Can you see your error now ?

10. Jan 11, 2005

the derivative of x^x is not $$x( x^x - 1)$$. It's $$x^x((ln(x) + 1))$$

Last edited: Jan 11, 2005
11. Jan 11, 2005

### Shay10825

Yeah I see my error. Thanks.

12. Jan 11, 2005

### Shay10825

Is there a rule for this or something?

13. Jan 11, 2005

$$x^x(ln(x) +1 )$$

$$f(x) = a^x$$ then $$f'(x) = a^x(ln(a))$$

14. Jan 11, 2005

### Shay10825

Thanks

15. Jan 11, 2005

### Curious3141

You don't need a rule.

$$x^x = e^{x\ln x}$$. Can you see how to differentiate it now ?

The other less direct but "easier to see" way is to use implicit differentiation.

I once "just used" the derivative of $x^x$ in exam and got docked a couple of points spoiling an otherwise perfect score. The teacher refused to believe I just did it in my head. :tongue2:

16. Jan 11, 2005

### Shay10825

I'm sorry but you just lost me. How did you get $$x^x = e^{x\ln x}$$?

17. Jan 11, 2005

### Curious3141

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

That's actually the easy part.

18. Jan 11, 2005

### Shay10825

Do you have to memorize this or something?

19. Jan 11, 2005

### Curious3141

No, isn't it obvious ? I'm just using $a = e^{\ln a}$ and ${(a^b)}^c = a^{bc}$

20. Jan 11, 2005

$$x^x = (e^{\ln x})^x = e^{x\ln x}$$ there is also a rule which states this. But as Curious said, its pretty obvious.