Find the derivative of the function $F(x) = \int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt$. $$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+t^4}}2x - \frac{1}{\sqrt{2+t^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+t^4}} - \frac{\sec^2{x}}{\sqrt{2+t^4}} \\
&= \frac{2x-\sec^2{x}}{\sqrt{2+t^4}} \\
\end{align}$$
Update: 
Is this looking better??
$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?

No. $t$ is a mute variable. The answer must have $x$ as the only variable. Yor computations are similar to
$$
\int_1^2t\,dt=2\cdot t-1\cdot t=t,
$$
which is obviously wrong.
