Chapter 5.4, Problem 13E

To determine

To find: use part 1 of the fundamental theorem of calculus to find the derivative of the function.

Answer to Problem 13E

The derivative of the function is $\text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=-\left(\frac{1}{{\text{x}}^2}\right)·\text{arctan}\left(\frac{1}{\text{x}}\right)$

Explanation of Solution

Given information: The function is $\text{h(x)=}{\displaystyle {\int }_2^{\frac{1}{\text{x}}}\text{arctan t dt}}$

Let’s remind of the fundamental theorem of calculus part 1:

The fundamental theorem of calculus part 1: If f is continuous on $[\text{a,b}]$ then the function of g defined by

  $\text{g(x)=}{\displaystyle {\int }_{\text{a}}^{\text{x}}\text{f(t)}}\text{ dt}$ where $\text{a}\le \text{x}\le \text{b}$ is continuous on $[\text{a,b}]$ and differentiable on $(\text{a,b})$ and $\text{<msup><mo>g</mo><mo>′</mo></msup><mo>(x)=f(x)</mo>}$ .

we have $\text{h(x)=}{\displaystyle {\int }_2^{\frac{1}{\text{x}}}\text{arctan t dt}}$ .

Since the upper limit of integration is not $\text{x}$ , we apply the chain rule.

Let $\text{u}=\frac{1}{\text{x}}$ , $\text{<msup><mo>u</mo><mo>′</mo></msup>}=\frac{1}{{\text{x}}^2}$

Consider the new function

  $\text{f(u)=}{\displaystyle {\int }_2^{\text{u}}\arctan }\text{ t dt}$

Differentiation with respect to u,

  $\begin{array}{l}\frac{\text{df(u)}}{\text{du}}=\frac{\text{d}}{\text{du}}\left[{\displaystyle {\int }_2^{\text{u}}\text{arctan t dt}}\right]\\ \text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{u}\right)=\text{arc tan u}\\ \text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{u}\right)=\text{u(x)}\\ \text{u(x)}=\text{f}\left(\frac{1}{\text{x}}\right).\\ \text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)={\left[\text{f}\left(\frac{1}{\text{x}}\right)\right]}^{\prime }\\ \text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=\text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\frac{1}{\text{x}}\right){\left(\frac{1}{\text{x}}\right)}^{\prime }\\ \text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=\text{arctan}\left(\frac{1}{\text{x}}\right)·{\left(\frac{1}{\text{x}}\right)}^{\prime }\\ \text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=\text{arctan}\left(\frac{1}{\text{x}}\right)·-\left(\frac{1}{{\text{x}}^2}\right)\\ \text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=-\left(\frac{1}{{\text{x}}^2}\right)·\text{arctan}\left(\frac{1}{\text{x}}\right)\end{array}$

Thus $\text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=-\left(\frac{1}{{\text{x}}^2}\right)·\text{arctan}\left(\frac{1}{\text{x}}\right)$

Hence, The derivative of the function is $\text{<msup><mo>u</mo><mo>′</mo></msup>}\left(\text{x}\right)=-\left(\frac{1}{{\text{x}}^2}\right)·\text{arctan}\left(\frac{1}{\text{x}}\right)$