Chapter 5.5, Problem 68E

a.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(c\right)=0\left(c\text{ is a constant}\right)$

Proof:

Differentiate with respect to x .

  $\begin{array}{l}d\left(c\right)={\left(c\right)}^{\prime }\cdot dx\\ =0\cdot dx\left[{\left(c\right)}^{\prime }=0,\text{ the derivative of a constant}\right]\\ =0\end{array}$

b.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(cu\right)=cdu\left(c\text{ is a constant}\right)$

Proof:

Consider $y=cu$

Differentiate with respect to x .

  $\begin{array}{l}y=cu\\ \frac{dy}{dx}=c\frac{du}{dx}+u\frac{dc}{dx}\left[\text{chain rule}\right]\\ \frac{dy}{dx}=c\frac{du}{dx}+0\left[{\left(c\right)}^{\prime }=0,\text{ the derivative of a constant}\right]\\ dy=cdu\left[\text{multiply each side by}\left(dx\right)\right]\\ d\left(cu\right)=cdu\left[y=cu\right]\end{array}$

c.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(u+v\right)=du+dv$

Proof:

Consider $y=u+v$

Differentiate with respect to x .

  $\begin{array}{l}y=u+v\\ \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\\ dy=du+dv\left[\text{multiply each side by}\left(dx\right)\right]\\ d\left(u+v\right)=du+dv\left[y=u+v\right]\end{array}$

d.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(u\cdot v\right)=udv+vdu$

Proof:

Consider $y=\left(u\cdot v\right)$

Differentiate with respect to x .

  $\begin{array}{l}y=\left(u\cdot v\right)\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\left[\text{chain rule}\right]\\ dy=udv+vdu\left[\text{ multiply each side by}\left(dx\right)\right]\\ d\left(u\cdot v\right)=udv+vdu\left[y=\left(u\cdot v\right)\right]\end{array}$

e.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}$

Proof:

Consider $y=\frac{u}{v}$

Differentiate with respect to x .

  $\begin{array}{l}y=\frac{u}{v}\\ y=u{v}^{-1}\\ \frac{dy}{dx}=-u{v}^{-2}\frac{dv}{dx}+v^{-1}\frac{du}{dx}\left[\text{chain rule}\right]\\ dy=\frac{-u}{v^2}dv+\frac{du}{v}\\ dy=\frac{1}{v^2}\left(-udv+vdu\right)\left[\text{take}\left(\frac{1}{v^2}\right)\text{common}\right]\\ dy=\frac{vdu-udv}{v^2}\\ d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}\left[y=\frac{u}{v}\right]\end{array}$

f.

To determine

To verify the formula.

Explanation of Solution

Given information:

The given formula is $d\left(u^n\right)=n{u}^{n-1}du$

Proof:

Consider $y=u^n$

Differentiate with respect to x .

  $\begin{array}{l}y=u^n\\ \frac{dy}{dx}=n{u}^{n-1}\frac{du}{dx}\\ dy=n{u}^{n-1}du\left[\text{multiply each side by}\left(dx\right)\right]\\ d\left(u^n\right)=n{u}^{n-1}du\left[y=u^n\right]\end{array}$