Chapter 5.4, Problem 12E

To determine

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

Answer to Problem 12E

The derivative of the function is $\text{<msup><mo>G</mo><mo>′</mo></msup><mo>(x)=-cos </mo>}\sqrt{\text{x}}$

Explanation of Solution

Given information: The function is $\text{G(x)=-}{\displaystyle {\int }_1^{\text{x}}\text{cos}\sqrt{\text{t}}}\text{ dt}$

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 $\text{<msup><mo>g</mo><mo>′</mo></msup><mo>(x)=f(x)</mo>}$ .

First, we will use the properties of definite integral to match the form in fundamental theorem of calculus part 1.

So, ${\displaystyle {\int }_{\text{x}}^1\text{cos}\sqrt{\text{t}}}\text{ dt=-}{\displaystyle {\int }_1^{\text{x}}\text{cos}\sqrt{\text{t}}}\text{ dt}$ where $\text{G(x)=-G(x)}$

Hence, $\text{G(x)=-}{\displaystyle {\int }_1^{\text{x}}\text{cos}\sqrt{\text{t}}}\text{ dt}$

Now given, $\text{G(x)=-}{\displaystyle {\int }_1^{\text{x}}\text{cos}\sqrt{\text{t}}}\text{ dt}$

Differentiation with respect to x,

  $\Rightarrow \frac{\text{d G(x)}}{\text{dx}}=\frac{\text{d}}{\text{dx}}[-{\displaystyle {\int }_1^{\text{x}}\text{cos}\sqrt{\text{t}}}\text{ dt]}$

  $\Rightarrow \text{<msup><mo>G</mo><mo>′</mo></msup><mo>(x)=-cos </mo>}\sqrt{\text{x}}$[where $\text{<msup><mo>G</mo><mo>′</mo></msup><mo>(x)=F(x)</mo>}$ ]

Thus, $\text{<msup><mo>G</mo><mo>′</mo></msup><mo>(x)=-cos </mo>}\sqrt{\text{x}}$

Hence, The derivative of the function is $\text{<msup><mo>G</mo><mo>′</mo></msup><mo>(x)=-cos </mo>}\sqrt{\text{x}}$