Find derivative of an integral with parameter $$F(t)=\int_{0}^{t^2}\mathbb{dx}\int_{x-t}^{x+t}\sin(x^2+y^2-t^2)\mathbb{dy}$$
find $F'(t)$
 A: Recal that
$$
\small
F'(t)=\frac{d}{dt}\int\limits_{\alpha(t)}^{\beta(t)} f(t,x)dx=
\int\limits_{\alpha(t)}^{\beta(t)} \frac{df}{dt}(t,x)dx+f(t,\beta(t))\frac{d\beta}{dt}(t)-f(t,\alpha(t))\frac{d\alpha}{dt}(t)\tag{1}
$$
In your case
$$
\small
f(t,x)= \int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy, \qquad \alpha(t)=0,\qquad \beta(t)=t^2
$$
So after substitution we get
$$
\small
F'(t)=\int\limits_{0}^{t^2}dx\left( \frac{d}{dt}\int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy\right)+\left(\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy\right)\cdot 2t-\left(\int\limits_{0-t}^{0+t}\sin(0^2+y^2-t^2)dy\right)\cdot 0
$$
Thus
$$
\small
F'(t)=\int\limits_{0}^{t^2}dx\left( \frac{d}{dt}\int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy\right)+2t\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy \tag{2}
$$
but this is not the end! Now you need to compute derivative of integral
$$
\small
\int\limits_{x-t}^{x+t}\sin(x^2+y^2-t^2)dy
$$
with respect to $t$. Though you have two parameters here you can think of $x$ as a number here. Formula $(1)$ will look like
$$
\small
G'(t)=\frac{d}{dt}\int\limits_{\gamma(t,x)}^{\delta(t,x)} g(t,x,y)dy
=\int\limits_{\gamma(t,x)}^{\delta(t,x)} \frac{dg}{dt}(t,x,y)dy+g(t,x,\delta(t,x))\frac{d\delta}{dt}(t,x)-g(t,x,\gamma(t,x))\frac{d\gamma}{dt}(t,x)\tag{3}
$$
In your case
$$
\small
g(t,x,y)=\sin(x^2+y^2-t^2)\qquad\gamma(t,x)=x-t\qquad\delta(t,x)=x+t
$$
So after substitution we get
$$
\small
G'(t)=\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)(-2t)dy+\sin(x^2+(x+t)^2-t^2)\cdot 1-\sin(x^2+(x-t)^2-t^2)\cdot (-1)
$$
$$
\small
=-2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+\sin(2x^2+2xt)+\sin(2x^2-2xt)
$$
$$
\small
=-2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+2\sin(2x^2)\cos(2xt)
$$
This finall result is
$$
\small
F'(t)=\int\limits_{0}^{t^2}dx\left( -2t\int\limits_{x-t}^{x+t}\cos(x^2+y^2-t^2)dy+2\sin(2x^2)\cos(2xt)\right)+2t\int\limits_{t^2-t}^{t^2+t}\sin(t^4+y^2-t^2)dy
$$
