Chapter 5.4, Problem 11E

To determine

To find: The value of $dy/dx$ .

Answer to Problem 11E

The answer is: $\frac{dy}{dx}=\frac{\sqrt{1+25x^2}}{x}$

Explanation of Solution

Given information:

The expression is: $y={{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du$

Calculation:

The given expression is: $y={{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du$

The calculus fundamental theorem, part 1:

If $f$ is continuous on $[a,b]$ and

  $F(x)={{\displaystyle \int }}_a^{x}f(t)dt$

The derivative of $F(x)$ is:

  $\begin{array}{c}{F}^{\prime }(x)=\frac{d}{dx}\left[{{\displaystyle \int }}_a^{x}f(t)dt\right]\\ =f(x)\end{array}$

It follows that,

  $y={{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du$

Differentiate with relation to $x$ :

  $\frac{dy}{dx}=\frac{d}{dx}\left[{{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du\right]$

Because the highest limit is $x^2$ rather than $x$ , it is impossible to simply use Part 1 of the calculus fundamental theorem. Instead, utilize the chain rule as demonstrated below:

  $\frac{dy}{dx}=\frac{d}{d(5x)}\left[{{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du\right]\cdot \frac{d(5x)}{dx}$

Note that the variable differentiate against (i.e. $x^2,$ ) has the same upper limit that it does now

  $\begin{array}{c}\frac{dy}{dx}=\frac{\sqrt{1+{(5x)}^2}}{5x}\\ \frac{dy}{dx}=\frac{\sqrt{1+25x^2}}{x}\end{array}$

Therefore the required value of the given expression $y={{\displaystyle \int }}_2^{5x}\frac{\sqrt{1+u^2}}{u}du$ is

  $\frac{dy}{dx}=\frac{\sqrt{1+25x^2}}{x}$ .