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Argh I had a picture for this but don't have enough reputation for anything on here. I also don't understand how to insert the math notation in here.

If $f(x) = \int_0^{g(x)}(1+t^3)^{-\frac12} \mathrm{d}t$ where $g(x) = \int_0^{\cos x}(1+\sin (t^2))\mathrm{d}t$

I have to find $f'(\frac{\pi}{2}).$

I know it has something to do with substitution and I've tried integrating by parts and things like that too but its not working out. I think there is something about the way that the questions been asked thats not helping anyway that is the first question.

There'll be more in the future I'm sure.

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    okay thanks for the advice. I'll play guesswork then. thanks henning for adding that in for me2012-02-24

3 Answers 3

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Note that $g(x) = \tilde{g}(y) = \int_0^y(1+\sin (t^2)) dt$, where $y=\cos(x)$. Moreover $f(x)=\tilde{f}(z) =\int_0^z(1+t^3)^{-1/2} dt$, where $z=\tilde{g}(y)$. So $f(x) = \tilde{f}(\tilde{g}(\cos(x))) = (\mathrm{notation}) = \tilde{f}(z(y(x)))$.

f' = \frac{df}{dx} = \frac{d\tilde{f}(z(y(x)))}{dx} = \frac{d\tilde{f}}{dz} \cdot \frac{dz}{dy} \cdot \frac{dy}{dx} = (1 + z^3)^{-1/2} \cdot (1 + \sin(y^2)) \cdot (-\sin(x)) = (1 + 0^3)^{-1/2} \cdot (1+0) \cdot (-1) = -1 (note, that $z=g(x)=\tilde{g}(\cos(\pi/2))$ is an integral from $0$ to $0$, so $z=g(x)=0$).

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    @TessaDangerBamkin: I rewrited that, is it clearer? I only use the chain rule and the fact that $d(\int_a^x f(t)dt)/dx = f(x)$.2012-02-24
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To make the work clear, put

$h(x) =\int_0^{x}(1+t^3)^{-1/2} dt$

$m(x) = \int_0^{ x}(1+\sin (t^2)) dt$

$f(x) = \int_0^{g(x)}(1+t^3)^{-1/2} dt$

$g(x) = \int_0^{\cos x}(1+\sin (t^2)) dt$

Then, we have that

$F(x) = h \circ m \circ \cos(x)$

Because of the chain rule we can solve this as follows:

$F'(x) = h' \circ m \circ \cos(x) \cdot (m \circ \cos (x))'$

$F'(x) = h' \circ m \circ \cos(x) \cdot (m' \circ \cos (x)) \cdot (\cos (x))'$

$F'(x) = h' \circ m \circ \cos(x) \cdot (m' \circ \cos (x)) \cdot (-\sin(x))$

So we now calculate our derivatives:

$h'(x) = (1+x^3)^{-1/2}$

$m'(x) =1+\sin(x^2)$

Now we're ready to plug in $\pi/2$. We have

$-\sin (\pi/2)=-1$

$\cos (\pi/2) = 0 $

so that

$m'(0) = 1+\sin (0) = 1$

So for the last one we have

$h' \circ m(0) = h'(0) = 1$

Finally, we get

$F'(\pi/2) = -1$

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    okay think i get it now thanks :). I just need to work on understanding chain rule as a composite functio I guess and not in the leibniz notation that I was first taught.2012-02-24
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Write $H(t) $ for the first integrand.

If $f(x) = \int_0^{g(x)} H(t) \,dt$, then f'(x) is defined as the limit of $(f(x+h)-f(x))/h$, for $h\to 0$, so you have to compute $\lim_{h\to 0} \frac1h \int_{g(x)}^{g(x+h)} H(t)\,dt$ The integral is "difference between bounds" times "average value"; for continuous $H$ and small $h$, the average value is close to $H(g(x))$. So the integral is close to $(g(x+h)-g(x)) \cdot H(g(x))$. The whole limit then comes out to \lim_{h\to 0} \frac1h (g(x+h)-g(x)) \cdot H(g(x))= g'(x)\cdot H(g(x)). By the same reasoning, g'(x) = (\cos x)'\cdot (1+\sin(\cos(x)^2)).

Now plug in $x=\pi/2$, to get g'(\pi/2) = (-1) \cdot (1+\sin(0)) = -1.

Multiply this by $H(g(\pi/2))$. $\cos(\pi/2)=0$, so $g(\pi/2) = 0$, and $H(g(\pi/2)) = H(0) = 1$.

So the result, if I have not miscalculated, is $-1$.