Find the integral by first principle In my math course homework, I encountered this problem:

Find $$\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}$$ by first principle

$$\begin{split}\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_0\frac{dt}{1+e^{t^2}}-\int^{x^2}_0\frac{dt}{1+e^{t^2}}}{h}\\&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}\end{split}$$
Then I don't know how to do. Please help me!
 A: Let $x,h>0$. By the MVT there is a constant $c_{x,h}$ such that $x^2 <c_{x,h}<(x+h)^2$ and
$$\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}=\frac{2xh+h^2}{1+e^{c_{x,h}^2}}.$$
Then,
\begin{split}\lim_{h\rightarrow 0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}&=\lim_{h\rightarrow 0} \frac{2x+h}{1+e^{c_{x,h}^2}}\\
&=\frac{2x}{1+e^{x^4}}.
\end{split}
A: Compute $$\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}
$$
without using the fundamental theorem of calculus or the mean value theorem.  
If $x^2 < t < (x+h)^2$, then
$$
x^4 < t^2 < (x+h)^4
\\
1 + \exp(x^4) < 1+\exp(t^2) < 1+\exp((x+h)^4)
\\
\frac{1}{1 + \exp(x^4)} > \frac{1}{1+\exp(t^2)} > \frac{1}{1+\exp((x+h)^4)}
\\
\frac{(x+h)^2-x^2}{1 + \exp(x^4)} > \int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{(x+h)^2-x^2}{1+\exp((x+h)^4)}
\\
\frac{2hx+h^2}{h(1 + \exp(x^4))} > \frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{2hx+h^2}{h(1+\exp((x+h)^4))}
\\
\frac{2x+h}{1 + \exp(x^4)} > \frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{2x+h}{1+\exp((x+h)^4)}
$$
But
$$
\lim_{h \to 0} \frac{2x+h}{1 + \exp(x^4)} = \frac{2x}{1 + \exp(x^4)}\quad\text{and}
\quad
\lim_{h \to 0}\frac{2x+h}{1+\exp((x+h)^4)} = \frac{2x}{1 + \exp(x^4)}
$$
So our answer is squeezed between these:
$$
\lim_{h \to 0}\frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)}= \frac{2x}{1 + \exp(x^4)} .
$$
A: It might be easier to look a little more abstractly and then apply to your specific case.
Suppose $f$ is continuous and $g$ is differentiable.
Let $\phi(x) = \int_0^{g(x)} f(t) dt$. The fundamental theorem/Leibniz rule tells us that
$\phi'(x) = f(g(x)) g'(x)$.
So we want to show that $\lim_{h \to 0} |{ \phi(x+h)-\phi(x) \over h} -f(g(x))g'(x)| = 0$.
Pick $\epsilon>0$ and $\delta_1 >0$ such that if $|t-g(x)| < \delta$ then $|f(t)-f(g(x))| < \epsilon$.
Also, choose $\delta_2 \le \delta_1$ such that if $|h|< \delta_2$ then
$|{g(x+h)-g(x) \over h} - g'(x)| < \epsilon$.
\begin{eqnarray}
|{\phi(x+h)-\phi(x) \over h } - f(g(x))g'(x)  | &=& |{1 \over h}\int_{g(x)}^{g(x+h)} f(t) dt - f(g(x))g'(x)| \\&=&
|{1 \over h}\int_{g(x)}^{g(x+h)} (f(g(x))+f(t)-(g(x))) dt - f(g(x))g'(x)| \\
&\le& |{1 \over h}\int_{g(x)}^{g(x+h)} f(g(x)) dt - f(g(x))g'(x)| +|{1 \over h}\int_{g(x)}^{g(x+h)}  \epsilon dt|\\
&=& |f(g(x))| | {(g(x+h)-g(x)) \over h} - g'(x) | + \epsilon |{g(x+h)-g(x) \over h} |\\
&\le& \epsilon |f(g(x))|+ \epsilon^2
\end{eqnarray}
It follows that the limit is zero.
A: Let,
$$F(x)=\int_{0}^{x^2}\frac{dt}{1+e^{t^{2}}}$$
$$\Rightarrow F(\sqrt{x})=\int_{0}^{x}\frac{dt}{1+e^{t^{2}}}$$
Let,
$$f(x)=\frac{1}{1+e^{x^{2}}}$$
From the first principle,
$$ \frac{dF(\sqrt{x})}{dx}=f(x)$$
$$\Rightarrow \frac{F'(\sqrt{x})}{2\sqrt{x}}=\frac{1}{1+e^{x^{2}}}$$
$$\Rightarrow {F'(\sqrt{x})}=\frac{{2\sqrt{x}}}{1+e^{x^{2}}}$$
$$\Rightarrow {F'({x})}=\frac{{2{x}}}{1+e^{x^{4}}}$$
