Derivatives of Functions Defined by Integrals

Example

Consider $g(x)={\int }_{\sin (x)}^{\cos (x)}{\tan }^{-1}(t)dt$ Find $g^{\prime }(x)$

Soln

1. Note that $f(x)=ta{n}^{-1}(x)$
2. Note that $f(x)$ is Continuous on $\mathbb{R}$
3. This means that $f(x)$ is also Continuous on $[\sin (x),\cos (x)]$
4. Define $F(x)={\int }_c^{x}f(x)dt$ where $c$ is some constant
5. Then, consider $g^{\prime }(x)=\frac{d}{dx}({\int }_{\sin (x)}^{\cos (x)}f(x)dt)$
6. $=\frac{d}{dx}({\int }_{\sin (x)}^c)f(t)dt+{\int }_c^{\cos (x)}f(t)dt)$ by Integral union property
7. $=\frac{d}{dx}(-{\int }_c^{\sin (x)})f(t)dt+{\int }_c^{\cos (x)}f(t)dx)$ by Integral property
8. $=\frac{d}{dx}(F(sin(x))+F(cos(x)))$ by defn of $F(x)$
9. $=F^{\prime }(sin(x))cos(x)-F^{\prime }(cos(x))sin(x)$ by Chain Rule
10. By Fundamental Theorem of Calculus Part 2 $F^{\prime }(x)=f(x)$
11. Thus, $=f(sin(x))cos(x)-f(cos(x))sin(x)$
12. $=-cos(x)ta{n}^{-1}(sin(x))-sin(x)ta{n}^{-1}(cos(x))$ by defn of $f(x)$ Done