## anonymous 5 years ago what is the derivative of x^x2

1. anonymous

is the second x raised to the 2?

2. anonymous

i think its x(2x)

3. anonymous

hmm, perhaps x^2x..

4. anonymous

This is a bit tricky and requires a little bit of thinking. Firstly, to clarify.. you wand $$\frac{d}{dx}x^{x^2}$$

5. anonymous

Err want.

6. anonymous

Right?

7. anonymous

yeah, i think that's what he means, because had it been multilication he/she would have just put the two in front..

8. anonymous

Are you going to solve this polpak? :)

9. anonymous

Crazy! How's this look? <a href=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d-7291-4b5e-be30-d944f1a0ea34/2011-04-22_1313.png "><img class="embeddedObject" src=" http://content.screencast.com/users/ffeldon/folders/Jing/media/df54790d-7291-4b5e-be30-d944f1a0ea34/2011-04-22_1313.png " width="596" height="653" border="0" /></a>

10. anonymous

Sure. The original poster seems to have gone away, but it's a fun exercise in anycase. Let $$u = x^{x^2} \implies (ln\ u) = x^2(ln\ x)$$ $\frac{d}{dx}(ln\ u) = \frac{d}{dx}x^2(ln\ x)$ $\implies \frac{d}{du}(ln\ u) * \frac{du}{dx} = x^2*\frac{d}{dx}(ln\ x) + (ln\ x)*\frac{d}{dx}(x^2)$ $\implies \frac{1}{u} * \frac{du}{dx} = \frac{x^2}{x} + 2x(ln\ x)$ $\implies \frac{du}{dx} = u(1 + 2x(ln\ x))$ $= x^{x^2}(1 + 2x(ln\ x)) = x^{x^2} + 2x^{x^2+1}(ln\ x)$ $\implies \frac{d}{dx}x^{x^2} = x^{x^2} + 2x^{x^2+1}(ln\ x)$

11. anonymous

i dont uderstand how you got to ln

12. anonymous

I took the ln(u) to get the x^2 out of the exponent.

13. anonymous

If $$u = x^{x^2}$$ then $$ln(u) = x^2ln(x)$$

14. anonymous

did you use a formula on the rest of it am kinda lost

15. anonymous

No formula. Just the chain rule and the product rule.

16. myininaya

17. anonymous

The right side of the equation on line 3 is the chain rule, the left is the product rule.

18. myininaya

you made a small error x^2/x=x not 1

19. anonymous

Ah, so I did.

20. anonymous

I had the 1 there because I was gonna factor the x from both terms, then forgot to do it ;p

21. myininaya

the equation still holds if polak takes the natural log of both sides then you use implicit differentiation to solve for y'

22. anonymous

so when trying to deferentiate x^2*lnx you use the product rule

23. myininaya

yes

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