Chapter 6.4, Problem 4E

To determine

To find: The derivative $\frac{dy}{dx}$ where $y={\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{1+e^{5t}}dt}$ .

Answer to Problem 4E

The derivative $\frac{dy}{dx}$ is $\sqrt{1+e^{5x}}$ where $y={\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{1+e^{5t}}dt}$ .

Explanation of Solution

Given information: The function is $y={\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{1+e^{5t}}dt}$ .

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 $\frac{dy}{dx}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{1+e^{5t}}dt}\right)\\ =\sqrt{1+e^{5x}}\end{array}$

Thus, the derivative $\frac{dy}{dx}$ is $\sqrt{1+e^{5x}}$ where $y={\displaystyle \underset{-2}{\overset{x}{\int }}\sqrt{1+e^{5t}}dt}$ .