Answer
\begin{aligned}
g'(x) &= \frac{-\sec^2(x)}{\sqrt{2+\tan^4(x)}}+ \frac{2x}{\sqrt{2+x^8}}\end{aligned}
Work Step by Step
Given
$$g(x)=\int_{\tan x}^{x^2} \frac{1}{\sqrt{2+t^4}} d t$$
Then
\begin{aligned}
g'(x)&=\frac{d}{dx}\int_{\tan x}^{x^2} \frac{1}{\sqrt{2+t^4}} d t\\
&= \frac{d}{dx}\int_{\tan x}^{a} \frac{1}{\sqrt{2+t^4}} d t+\frac{d}{dx}\int_{a}^{x^2} \frac{1}{\sqrt{2+t^4}} d t\\
&= -\int_{a}^{\tan x} \frac{1}{\sqrt{2+t^4}} d t+\frac{d}{dx}\int_{a}^{x^2} \frac{1}{\sqrt{2+t^4}} d t\\
&= \frac{-1}{\sqrt{2+\tan^4(x)}}\frac{d}{dx}(\tan (x)) + \frac{1}{\sqrt{2+x^8}}\frac{d}{dx}(x^2)\\
&= \frac{-\sec^2(x)}{\sqrt{2+\tan^4(x)}}+ \frac{2x}{\sqrt{2+x^8}}\end{aligned}
