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.