Chapter 6, Problem 39RE

To determine

To find $\frac{dy}{dx}$

Explanation of Solution

Given information:

  $y={\displaystyle {\int }_2^x\sqrt{2+{\cos }^{3}t}dt}$

Using fundamental theorem of calculus

If $\text{}\text{}\text{}\text{}y={\displaystyle {\int }_a^{x}f\left(t\right)dt}$ then $\frac{dy}{dx}=f\left(x\right)$

Now

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left({\displaystyle {\int }_2^x\sqrt{2+{\cos }^{3}t}dt}\right)\\ \frac{dy}{dx}=\sqrt{2+{\cos }^{3}x}\end{array}$

Answer:

  $\frac{dy}{dx}=\sqrt{2+{\cos }^{3}x}$.