## joshuanoble2000 3 years ago Find a function whose derivative is 3x^2(x^3 + 2)^12

1. AnwarA

Do you know what integration or an antiderivative is?

2. joshuanoble2000

No. I think we're meant to be doing this by figuring one out that would work by simply working backwards...we're working on the chain rule now.

3. dave444

Try letting $$u=x^3$$ and go from there.

4. AnwarA

I see, we can do that. Actually try putting $$u=x^3+2$$, that would work better.

5. dave444

True.

6. AnwarA

You will have then something like $$u'u^{12}$$. Now try to relate this to the product rule, can you do that?

7. AnwarA

Oh sorry, I mean relate it ot the chain rule.

8. dave444

Or note that if $$u=x^3+2$$, then $$du = 3x^2\;dx$$ and so $\int 3x^2(x^3+2)^{12}\;dx = \int u^{12} du = \frac{u^{13}}{13}+C = \frac{(x^3+2)^{13}}{13} +C$

9. joshuanoble2000

Sorry, I'm going to have to go through this rather painstakingly, I fear. U is x^3 + 2. So I say dy/du is (dy/(u^3 + 2) * (u^3 +2) / dx) to begin with?

10. joshuanoble2000

Sorry, du, so 1...

11. joshuanoble2000

3x^2, I mean

12. AnwarA

I can't read what you wrote well. But what I did is that I just found the derivative of $$u$$ w.r.t $$x$$. $$u=x^3 + 2 \implies \frac{du}{dx}=3x^2$$. So the expression we have can then be written as $$\large u'u^{12}$$. So far so good?

13. joshuanoble2000

Yes.

14. AnwarA

Good. Now you know from your study of derivatives, a method for finding derivatives of composite functions called the Chain rule. Can you state it for me?

15. joshuanoble2000

f'(x) = f'(g(x)) * g'(x)

16. joshuanoble2000

(assuming the original functions were stated f(g(x))

17. AnwarA

True, you mean $$[f(g(x))]'$$, or $$\frac{d}{dx}[f(g(x))]$$

18. joshuanoble2000

Right.

19. AnwarA

Now we want to relate this formula you wrote with the expression we have. Consider $$g'(x)=u' \implies g(x)=u,$$ and $$f'(g(x))=u^{12}=g^{12}.$$ Can you then find $$f(g(x))$$?

20. AnwarA

Simply you have $$f'=u^{12}$$, so $$f$$=..?!

21. AnwarA

Are you still there?

22. joshuanoble2000

Yeah.

23. AnwarA

All you need here is to apply the power rule. You have f'(u)=u^12, so f(u)=..?

24. AnwarA

The symbols I just used are not really accurate, but I used them for simplification.

25. AnwarA

You can see that $$\large f(u)=\frac{u^{13}}{13}$$?

26. joshuanoble2000

27. AnwarA

Very good. So we're saying u^13/13 is a function that has a derivative $$u'u^{12}$$. Now just resubstitute for $$x$$ is $$\large \frac{u^{13}}{13}$$, and this should be your final answer. Don't forget that $$u=x^3+2$$.

28. AnwarA

*for x in*

29. joshuanoble2000

Oh my, I feel silly now. It should be 1/13(x^3+2)^13?

30. AnwarA

Exactly! :)

31. joshuanoble2000

This whole trouble was because the 3x^2 was moved to the front of the equation...quite dense on my part. Thanks so much for being patient.

32. AnwarA

So we can write a general rule from what we just did, that states: An anti derivative of $$f'(x)[f(x)]^n$$ is $$\large \frac{[f(x)]^{n+1}}{n+1}$$. I hope this doesn't make it any more complicated. You're welcome!

33. AnwarA

In our case f(x) is obviously x^3+2.

34. joshuanoble2000

No, you're thoroughly helpful and when I'm done with my homework I'll more carefully look over what you've written.

35. AnwarA

Good luck then!

36. joshuanoble2000

Thanks!