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

One version of the fundamental theorem of calculus is $$\frac{\mathrm{d}}{\mathrm{d}x}\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t=f(x).$$

In our case, $F$ is a composite $H\circ g$ of the two functions $$\begin{array}{rl}H(x)& =\underset{0}{\overset{x}{\int }}\frac{4-t}{1+{\cos }^2(t)}\mathrm{d}t,\\ g(x)& =x^2.\end{array}$$ We wish to apply the chain rule. The derivatives are $$\begin{array}{rl}{H}^{\prime }(x)& =\frac{4-x}{1+{\cos }^2(x)}\\ g^{\prime }(x)& =2x,\end{array}$$ so we have $$F^{\prime }(x)=H^{\prime }(g(x))\cdot g^{\prime }(x)=\frac{4-x^2}{1+{\cos }^2(x^2)}\cdot 2x.$$