Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 48

Determine the derivative of the function $\displaystyle g(x) = \int^{x^2}_{\tan x} \frac{1}{\sqrt{2 + t^4}} dt$

Apply Properties of Integral


$\begin{equation} \begin{aligned} & \int^c_a f(x) dx + \int^b_c f(x) dx = \int^b_a f(x) dx, \text{ So we have} \\ \\ & g(x) = \int^1_{\tan x} \frac{1}{\sqrt{2 + t^4}} dt + \int^{x^2}_1 \frac{1}{\sqrt{2 + t^4}} dt \\ \\ & g(x) = - \int^{\tan x}_1 \frac{1}{\sqrt{2 + t^4}} dt + \int^{x^2}_1 \frac{1}{\sqrt{2 + t^4}} dt \end{aligned} \end{equation}$


Let $\displaystyle u_1 = \tan x_1 \frac{du_1}{dx} = \sec ^2 x$ and $\displaystyle u_2 = x^2, \frac{du_2}{dx} = 2x$, then


$\begin{equation} \begin{aligned} g'(x) =& - \frac{d}{dx} \left( \int^{\tan x}_1 \frac{1}{\sqrt{2 + t^4}} dt + \int^{x^2}_1 \frac{1}{\sqrt{2 + t^4}} dt \right) \frac{du}{dx} \\ \\ g'(x) =& - \frac{1}{\sqrt{2 + u^4_1}} \cdot \frac{du_1}{dx} + \frac{1}{\sqrt{2 + u^4_2}} \cdot \frac{du_2}{dx} \\ \\ g'(x) =& \left( \frac{-1}{\sqrt{2 + (\tan x)^4}} \cdot \sec ^2 x \right) + \left( \frac{1}{\sqrt{2 + (x^2)^4}} \cdot 2x \right) \\ \\ g'(x) =& \frac{- \sec^2 x}{\sqrt{2 + \tan^4 x}} + \frac{2x}{\sqrt{2 + x^8}} \end{aligned} \end{equation}$