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AnwarABest ResponseYou've already chosen the best response.0
Do you know what integration or an antiderivative is?
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
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.
 2 years ago

dave444Best ResponseYou've already chosen the best response.1
Try letting \(u=x^3\) and go from there.
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
I see, we can do that. Actually try putting \(u=x^3+2\), that would work better.
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
You will have then something like \(u'u^{12}\). Now try to relate this to the product rule, can you do that?
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
Oh sorry, I mean relate it ot the chain rule.
 2 years ago

dave444Best ResponseYou've already chosen the best response.1
Or note that if \(u=x^3+2\), then \( du = 3x^2\;dx\) and so \[\int 3x^2(x^3+2)^{12}\;dx = \int u^{12} du = \frac{u^{13}}{13}+C = \frac{(x^3+2)^{13}}{13} +C \]
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
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?
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
Sorry, du, so 1...
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
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?
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
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?
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
f'(x) = f'(g(x)) * g'(x)
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
(assuming the original functions were stated f(g(x))
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
True, you mean \([f(g(x))]'\), or \(\frac{d}{dx}[f(g(x))]\)
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
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 \(f(g(x))\)?
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
Simply you have \(f'=u^{12}\), so \(f\)=..?!
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
All you need here is to apply the power rule. You have f'(u)=u^12, so f(u)=..?
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
The symbols I just used are not really accurate, but I used them for simplification.
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
You can see that \(\large f(u)=\frac{u^{13}}{13}\)?
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
If you had 1/13 U^13
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
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 \(u=x^3+2\).
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
Oh my, I feel silly now. It should be 1/13(x^3+2)^13?
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
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.
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
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!
 2 years ago

AnwarABest ResponseYou've already chosen the best response.0
In our case f(x) is obviously x^3+2.
 2 years ago

joshuanoble2000Best ResponseYou've already chosen the best response.1
No, you're thoroughly helpful and when I'm done with my homework I'll more carefully look over what you've written.
 2 years ago
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