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Do you know what integration or an antiderivative is?

Try letting \(u=x^3\) and go from there.

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

True.

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

Sorry, du, so 1...

3x^2, I mean

Yes.

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

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

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

Right.

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

Are you still there?

Yeah.

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

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

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

If you had 1/13 U^13

*for x in*

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

Exactly! :)

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

Good luck then!

Thanks!