Compute define integral written in alternative version: $\frac{d}{dt} \int_0^{t^2} \frac{dx}{1+x^2}$ $$\dfrac{d}{dt} \displaystyle\int_0^{t^2} \frac{dx}{1+x^2}$$
I don't understand what to do with the $t^{2}$ term in the integrals upper limit. As far as I know, if it is $t$ in the upper limit then 
$$\dfrac{d}{dt} \displaystyle\int_0^{t} \frac{dx}{1+x^2} = \dfrac{1}{1+t^2}$$
 A: Well, it seems to me you could just apply the chain rule. That is you can get, setting $t^2=u$, $$\dfrac{\mathrm{d}u}{\mathrm{d}t} \times \dfrac{\mathrm{d}}{\mathrm{d}u}\left(\int_{0}^{u} \dfrac{\mathrm{d}t}{t^2+1}\right)$$
Now you can apply the Fundamental theorem of calculus to get $$\dfrac{\mathrm{d}u}{\mathrm{d}t}\times \dfrac{1}{1+u^2}=\dfrac{2t}{1+t^4}$$
As $u=t^2$. 
A: Observing that $\frac{1}{1+x^2}=(\arctan(x))'$
$$
\dfrac{d}{dt} \displaystyle\int_0^{t^2} \frac{1}{1+x^2}\,dx = \dfrac{d}{dt} \displaystyle\arctan(t^2) = \frac{2t}{1+t^4}
$$
A: We have $\dfrac{d}{dt} \displaystyle\int_0^{t^2} \frac{1}{1+x^2}\,dx=\frac{1}{1+t^{4}}\cdot2t$ using the fundamental theorem of calculus. 
A: Probably the issue is that you don't recognize the function $t \mapsto \int_0^{t^2} f(x)dx$ as a composition of functions, and so you don't see how to apply the chain rule. It is a good idea in the first examples you come across to be very explicit in your use of the chain rule.
Specifically, let's call this function (the one taking $t \mapsto \int_0^{t^2} f(x)dx$) by $h$. Call the function $t \mapsto t^2$ by $f$, and the function $t \mapsto \int_0^t f(x)dx$ by $g$. You know how to compute the derivative of $g$ (you wrote it yourself!), and also of $f$, which is a polynomial. Hence, you know the derivative of $g \circ f$ by the chain rule. But $g \circ f$ is precisely $h$. Therefore,
\begin{align*}
h'(t)&=(g \circ f)'(t)\\
&=g'(f(t))\cdot f'(t) \\
&=\frac{1}{1+(t^2)^2}\cdot 2t\\
&=\frac{2t}{1+t^4}.
\end{align*}
