How do we find the derivative of the following function: $F(x)=\sin\Big(\int_0^x \sin\big(\int_0^y \frac{1}{1+t^4}\ dt\big)\ dy\Big)$ using FTC. Find the derivative of the following function:
$F(x)=\sin\Big(\int_0^x \sin\big(\int_0^y \frac{1}{1+t^4}\ dt\big)\ dy\Big)$
I tried applying fundamental theorem of calculus directly but the the integral $\int_0^y \frac{1}{1+t^4}\ dt$ is giving me problems, I started by deriving both sides with respect to x, and by chain rule, the derivative should be f'(g(x))*g'(x). I reached this point and couldn't complete, does anyone have the key for the solution?
 A: Well, we are trying to find:
$$\text{p}'\left(x\right):=\frac{\text{d}}{\text{d}x}\left(\sin\left(\underbrace{\int_0^x\sin\left(\int_0^y\frac{1}{1+t^4}\space\text{d}t\right)\space\text{d}y}_{:=\space\text{z}\left(x\right)}\right)\right)\tag1$$
First we use the cain rule, in order to write:
$$\text{p}'\left(x\right)=\frac{\text{d}}{\text{d}x}\left(\sin\left(\text{z}\left(x\right)\right)\right)=\cos\left(\text{z}\left(x\right)\right)\cdot\text{z}'\left(x\right)\tag2$$
Now, we must find $\text{z}'\left(x\right)$:
$$\text{z}'\left(x\right)=\frac{\text{d}}{\text{d}x}\left(\int_0^x\underbrace{\sin\left(\int_0^y\frac{1}{1+t^4}\space\text{d}t\right)}_{:=\space\text{k}\left(y\right)}\space\text{d}y\right)\tag3$$
Using FTC (by definition), we can see:
$$\text{z}'\left(x\right)=\frac{\text{d}}{\text{d}x}\left(\int_0^x\text{k}\left(y\right)\space\text{d}y\right)=\text{k}\left(x\right)\tag4$$
So, we get:
$$\text{p}'\left(x\right)=\cos\left(\text{z}\left(x\right)\right)\cdot\text{k}\left(x\right)=$$
$$\cos\left(\int_0^x\sin\left(\int_0^y\frac{1}{1+t^4}\space\text{d}t\right)\space\text{d}y\right)\cdot\sin\left(\int_0^x\frac{1}{1+t^4}\space\text{d}t\right)\tag5$$
