Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 01/Solution 1

One version of the fundamental theorem of calculus is

$${\frac {\mathrm {d} }{\mathrm {d} x}}\int _{a}^{x}f(t)\mathrm {d} t=f(x).$$

In our case, $F$ is a composite $H\circ g$ of the two functions

$${\begin{aligned}H(x)&=\int _{0}^{x}{\frac {4-t}{1+\cos ^{2}(t)}}\mathrm {d} t,\\g(x)&=x^{2}.\end{aligned}}$$

We wish to apply the chain rule. The derivatives are

$${\begin{aligned}H'(x)&={\frac {4-x}{1+\cos ^{2}(x)}}\\g'(x)&=2x,\end{aligned}}$$

so we have

$$F'(x)=H'(g(x))\cdot g'(x)={\frac {4-x^{2}}{1+\cos ^{2}(x^{2})}}\cdot 2x.$$