How do we differentiate an integral with a different argument Let's say we have:
$$y = \int_0^x e^{t^2} dt$$
and we want $dy/dx$.
Do we consider $t$ as being $x$?
 A: HINT: Let $$y = \int^x_a f(t) dt$$
if we know the anti-derivative of $f$ is $F$, then by fundamental theorem of calculus
$$
y = F(x) - F(a)
$$
Hence
$$
\frac{dy}{dx} = \frac{d}{dx}F(x) - \frac{d}{dx}F(a) = \frac{d}{dx}F(x) = f(x).
$$
A: Shortly yes. Longer derivation :
\begin{align}
y'(x):=\lim_{h\to 0}\frac{y(x+h)-y(x)}h&=\lim_{h\to 0}\frac 1h\left[\int_0^{x+h} e^{t^2} dt-\int_0^{x} e^{t^2} dt\right]\\
&=\lim_{h\to 0}\frac 1h\int_x^{x+h} e^{t^2} dt\\
&=e^{x^2}\\
\end{align}
A: $\dfrac{dy}{dx} = e^{x^2}$ by a direct use of FTC. Note that $t$ is a dummy variable.
A: A related problem.
$$ y(x+h)-y(x) = \int_0^{x+h} e^{t^2} dt - \int_0^{x} e^{t^2} dt $$
$$ =  \int_0^{x} e^{t^2} dt + \int_{x}^{x+h} e^{t^2} dt - \int_0^{x} e^{t^2} dt $$
$$ \implies \frac{y(x+h)-y(x)}{h}= \frac{1}{h}\int_{x}^{x+h} e^{t^2} dt $$
Take the limit of the above as $h\to 0$ and see what you get.
Added: For the sake of completion. To see that the limit is $e^{x^2}$, see this
$$ \Big| \frac{y(x+h)-y(x)}{h}-e^{x^2}  \Big|= \Big|\frac{1}{h}\int_{x}^{x+h} e^{t^2} dt-\frac{1}{h}\int_{x}^{x+h} e^{x^2} dt   \Big| $$
$$ =\Big| \frac{1}{h}\int_{x}^{x+h} \left(e^{t^2}-e^{x^2}\right) dt  \Big|\leq \frac{1}{h}\int_{x}^{x+h}\Big| e^{t^2}-e^{x^2} \Big|dt $$
$$ < \epsilon \frac{1}{h}\int_{x}^{x+h} dt = \epsilon. $$
Note that we used the continuity of the function on the interval $[x,x+h]$, that is
$$ |h|<\delta \implies \Big| e^{t^2}-e^{x^2} \Big|< \epsilon . $$
A: In general we have by chain rule
$$\frac{d}{dx}\left(  \int_{v(x)}^{u(x)} f(t) dt\right)=f\left(u(x)\right)\times u'(x)-f\left(v(x)\right)\times v'(x)$$
