Combining FTC$-1$ with the Chain Rule, we have

Corollary. If $f$ is a continuous function and $u$ and $v$ are differentiable functions, then the function $g$ given by

$g(x)={\displaystyle \underset{v(x)}{\overset{u(x)}{}}}f(t)dt$

is differentiable and $g^{\text{'}}(x)=f(u(x))u^{\text{'}}(x)-f(v(x))v^{\text{'}}(x).$

In particular, if $g(x)={\displaystyle \underset{a}{\overset{u(x)}{}}}f(t)dt,$ then $g^{\text{'}}(x)=f(u(x))u^{\text{'}}(x).$

Example $5.$ Find $g^{\text{'}}(x)$ by using FTC$-1.$

$($i$)g(x)={\displaystyle \underset{0}{\overset{x}{}}}\sqrt[\phantom{2}]{1+t^2}dt.$

$($ii$)g(x)={\displaystyle \underset{1}{\overset{x}{}}}(e^{t^2}+si{n}^{2}t+ln(t^2))dt.$

$($iii$)g(x)={\displaystyle \underset{0}{\overset{x^3}{}}}\sqrt[\phantom{2}]{1+t^2}dt.$