Integral question (Using Fundamental Theorem) Let's say I have the function
$$ x^2 = \int_{\tan(x)}^{y(x)}\frac{1}{\sqrt{2+t^2}}\,\mathrm dt  $$
can I replace the y in the upper limit of the integral with $ x^2 $ ? If not, any steps on how to proceed? How could I find $ y'(0) $ ?
 A: You can take derivatives of both sides with respect to $x$:
$$2x=\frac{y'(x)}{\sqrt{2+(y(x))^2}} - \frac{\sec^2 x}{\sqrt{2+\tan^2x}}$$
To now find $y'(0)$, we plug in $0$ for $x$ and simplify:
$$0=\frac{y'(0)}{\sqrt{2+(y(0))^2}} - \frac1{\sqrt2}$$
To solve this, we need $y(0)$, which we can get by plugging in $0$ for $x$ in the original equation:
$$ 0 = \int_0^{y(0)}\frac{1}{\sqrt{2+t^2}}\,\mathrm dt  $$
The integrand is positive for all $t$, so if $y(0)\ne 0$, the value of the definite integral would have to be non-zero as well. Thus, we have $y(0)=0$, and we can go on:
$$0=\frac{y'(0)-1}{\sqrt2}\implies y'(0)=1$$
A: By the fundamental theorem of calculus, we have
$$ F(x) := \int_{a}^{x} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
\implies F'(x) =  \frac{1}{\sqrt{2+t^2}}. $$
Splitting the original integral, we obtain
$$ x^2 = \int_{\tan(x)}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
= \int_{a}^{y(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t - \int_{a}^{\tan(x)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d}t
= F(y(x)) - F(\tan(x)). $$
Differentiating with respect to $x$ on both sides (making use of the chain rule), we get
\begin{align} 2x
&= \frac{\mathrm{d}}{\mathrm{d}x} F(y(x)) - \frac{\mathrm{d}}{\mathrm{d}x} F(\tan(x)) \\
&= F'(y(x)) y'(x) - F'(\tan(x)) \frac{\mathrm{d}}{\mathrm{d}x} \tan(x) \\
&= \frac{1}{\sqrt{2+y(x)^2}} y'(x) - \frac{1}{\sqrt{2+\tan(x)^2}} \sec(x)^2.
\end{align}
Note that this only makes sense if $y$ is differentiable.
EDIT:  To address the edited version of the original question, substitute $x=0$ into the last displayed equation in order to obtain
\begin{align} &0 = \frac{1}{\sqrt{2+y(0)^2}} y'(0) - \frac{\sec(0)^2}{\sqrt{2+\tan(0)^2}} = \frac{1}{\sqrt{2+y(0)^2}} y'(0) - \frac{1}{\sqrt{2}} \\
&\qquad\implies \frac{1}{\sqrt{2+y(0)^2}} y'(0) = \frac{1}{\sqrt{2}} \\
&\qquad\implies y'(0) = \frac{\sqrt{2+y(0)^2}}{\sqrt{2}} = \sqrt{1 + \frac{1}{2} y(0)^2}.
\end{align}
EDIT: While GTonyJacobs has already addressed this, I would feel remiss in not completing the exercise:
The original integrand is strictly positive, from which it follows that
$$ 0 = F(0) = \int_{\tan(0)}^{y(0)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d} t
= \int_{0}^{y(0)} \frac{1}{\sqrt{2+t^2}} \,\mathrm{d} t \iff y(0) = 0, $$
as the integral of a positive function over a set of non-zero measure must be positive, and every non-degenerate interval has positive measure (this can fairly easily be rephrased in terms of the the Riemann integral, but I am feeling lazy).  But we have already determined that
$$ y'(0) = \frac{\sqrt{2+y(0)^2}}{\sqrt{2}} = \sqrt{1 + \frac{1}{2} y(0)^2}, $$
and so
$$ y'(0) = \sqrt{1+\frac{1}{2}\cdot 0^2} = 1. $$
