Derivative of an integral, with limits in terms of x.

**Timbuk2**

OK, this is confusing me. I'm sure it's easy - I'm just not sure how to start it.

If [latex]f(x) = \int_x^{x^2}t^2dt[/latex] , what is f'(x)?

I thought f'(x) = [latex]x^4 - x^2[/latex] by the second fundamental theorem of calculus, but this doesn't seem to be the answer. Also, I don't think f'(x) = [latex]x^2[/latex] either.

How would I start this off? It's the way the function is defined in terms of the limits that is confusing me. I'm probably missing something very very basic.

Grudaire

It's easy to integrate this function, do that and you'll get a function in terms of x, this can be differentiated.

My first idea was the FTC, but I think because the limits are complicated it isn't as straightforward

LeixlipRed

Hint: Chain Rule! Look at the initial definition of the FTC and what it's limits are.

LeixlipRed

Second hint You'll need more than one integral....

dowtcha

At the risk of exposing my lack of knowledge here, is it not a matter of integrating t**2
(t**3/3) and applying limits of x**2 and x, ie. (x**6-x**3)/3, and then differentiating
wrt x, (6x**5-3x**2)/3

Have I missed the plot here??

LeixlipRed

Yes you can do it that way, work it out explicitly. But the FTC tells us that that the derivative of an integral of a function is the function itself, so there's a shortcut here to be had. Keep an eye on the thread and you'll see what I mean

**Timbuk2**

Thanks for the answers

At first I read through this thread and thought "I still don't really know what to do". Then I read it again and it clicked! I did this first:

[latex]f(x) = \int_x^{x^2}t^2dt[/latex]
[latex]f(x) = [\frac{1}{3}t^3]_x^{x^2}[/latex]
[latex]f(x) = \frac{1}{3}x^6 - \frac{1}{3}x^3[/latex]
[latex]$thus$ f'(x) = 2x^5 - x^2[/latex]

Which of course is exactly identical to dowtcha's method! I went back and had a look at FTC.

If [latex]g(t)=t^2[/latex] then I can write the antiderivative [latex]G(t)=\frac{t^3}{3}[/latex].

So, [latex]f(x) = \int_x^{x^2}g(t)dt[/latex].
Therefore I can write the derivative as
[latex]f'(x)=G'(x^2) - G'(x)[/latex]
[latex]f'(x)=\frac{d}{dx}(\frac{x^6}{3}) - \frac{d}{dx}(\frac{x^3}{3})[/latex]
[latex]f'(x) = 2x^5 - x^2[/latex]

So I got the same answer both methods. LeixlipRed, is the 'shortcut' you refer to what I just did? I have a better understanding of FTC now, it always confused me having the t's and the x's mixed

Thanks again for the help. Hopefully what I've just done is correct!

MathsManiac

I think that LeixlipRed was suggesting you can do it without ever writing G(x) explicitly.

Since G'(x) is your original g(x), it follows that G'(x^2) is g(x^2).[d/dx(x^2)] = 2x.g(x^2).

So the whole thing can be done without any explicit integration and only one minor piece of differentiation.

LeixlipRed

The FTC tells us that if [latex]F(x) = \int_a^{x}f(t)dt[/latex] then[latex]F'(x) = f(x)[/latex]. So we can write [latex]f(x) = \int_x^{x^2}t^2dt[/latex] as [latex]f(x) = \int_x^{a}t^2dt + \int_a^{x^2}t^2dt[/latex] and then apply the FTC directly noting that the second integral requires an application of the Chain Rule.

We deal with that by saying that if [latex]F(x) = \int_a^{g(x)}f(t)dt[/latex] then [latex]F'(x)=f(g(x))g'(x)[/latex]

LeixlipRed

Should be [latex]f(x) = \int_x^{a}t^2dt + \int_a^{x^2}t^2dt[/latex] there, editing it made the plus disappear and I can't remember how you undo that.

LeixlipRed

And it's carried over, ah I give up, you get what I mean, bed time!