Chapter 6, Problem 41RE

To determine

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

Answer to Problem 41RE

  $\frac{dy}{dx}=-\frac{6}{3+x^4}$

Explanation of Solution

Given information:

  $y={\displaystyle {\int }_x^1\frac{6}{3+t^4}dt}$

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 it can be written as

  $\begin{array}{l}y={\displaystyle {\int }_x^1\frac{6}{3+t^4}dt}\\ y=-{\displaystyle {\int }_1^x\frac{6}{3+t^4}dt}\end{array}$

The derivative of the y is

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(-{\displaystyle {\int }_1^x\frac{6}{3+t^4}dt}\right)\\ \frac{dy}{dx}=-\frac{d}{dx}\left({\displaystyle {\int }_1^x\frac{6}{3+t^4}dt}\right)\\ \frac{dy}{dx}=-\frac{6}{3+x^4}\end{array}$.