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

1. AnwarA Group Title

Do you know what integration or an antiderivative is?

2. joshuanoble2000 Group Title

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 Group Title

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

4. AnwarA Group Title

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

5. dave444 Group Title

True.

6. AnwarA Group Title

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

7. AnwarA Group Title

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

8. dave444 Group Title

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 Group Title

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 Group Title

Sorry, du, so 1...

11. joshuanoble2000 Group Title

3x^2, I mean

12. AnwarA Group Title

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 Group Title

Yes.

14. AnwarA Group Title

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 Group Title

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

16. joshuanoble2000 Group Title

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

17. AnwarA Group Title

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

18. joshuanoble2000 Group Title

Right.

19. AnwarA Group Title

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 Group Title

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

21. AnwarA Group Title

Are you still there?

22. joshuanoble2000 Group Title

Yeah.

23. AnwarA Group Title

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

24. AnwarA Group Title

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

25. AnwarA Group Title

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

26. joshuanoble2000 Group Title

27. AnwarA Group Title

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 Group Title

*for x in*

29. joshuanoble2000 Group Title

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

30. AnwarA Group Title

Exactly! :)

31. joshuanoble2000 Group Title

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 Group Title

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 Group Title

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

34. joshuanoble2000 Group Title

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 Group Title

Good luck then!

36. joshuanoble2000 Group Title

Thanks!