How to find $f'(\frac{\pi}{2})$ knowing $f(x) = \int_0^{g(x)}(1+t^3)^{-\frac12} \mathrm{d}t$? 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. 
 A: 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 $z=g(x)=0$).
A: 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
$$F'(\pi/2) =  -1$$
A: 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 $-1$. 
