Chapter 6.4, Problem 7E

To determine

To find: The derivative $\frac{dy}{dx}$ where $y={\displaystyle \underset{7}{\overset{x}{\int }}\frac{1+t}{1+t^2}dt}$ .

Answer to Problem 7E

The derivative $\frac{dy}{dx}$ is $\frac{1+x}{1+x^2}$ where $y={\displaystyle \underset{7}{\overset{x}{\int }}\frac{1+t}{1+t^2}dt}$ .

Explanation of Solution

Given information: The function is $y={\displaystyle \underset{7}{\overset{x}{\int }}\frac{1+t}{1+t^2}dt}$ .

Theorem used:

Fundamental Theorem :

If f is continuous on $\left[a,b\right]$ , then the function $F\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$ has a derivative at every point x in $\left[a,b\right]$ and $\frac{dF}{dx}=\frac{d}{dx}{\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}=f\left(x\right)$ .

Calculation:

By using the Fundamental Theorem, find $\frac{dy}{dx}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle \underset{7}{\overset{x}{\int }}\frac{1+t}{1+t^2}dt}\right)\\ =\frac{1+x}{1+x^2}\end{array}$

Thus, the derivative $\frac{dy}{dx}$ is $\frac{1+x}{1+x^2}$ where $y={\displaystyle \underset{7}{\overset{x}{\int }}\frac{1+t}{1+t^2}dt}$ .