<s>[INST] 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: 
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$$\begin{align}
\left(\int_{\tan{x}}^{x^2} \frac{1}{\sqrt{2+t^4}}\,dt\right)' &= \frac{1}{\sqrt{2+(x^2)^4}}2x - \frac{1}{\sqrt{2+(\tan{x})^4}}\sec^2{x} \\
&= \frac{2x}{\sqrt{2+x^8}} - \frac{\sec^2{x}}{\sqrt{2+\tan^4{x}}} \\
\end{align}$$
Is this correct?
 [/INST] $\text{ Recall } \int_{a(x)}^{b(x)}f(t) dt=F(b(x))-F(a(x)) \text{ *note:} \text{( where } F'=f) \\ \frac{d}{dx}[F(b(x))-F(a(x))]=\frac{d}{dx}F(b(x))-\frac{d}{dx}F(a(x)) \text{ by difference rule } \\ \\ \text{ now you use chain rule and you are almost there } $
</s>