Chapter 4.5, Problem 68E

a.

To determine

Verify the formula $d(c)=0$ ( $c$ a constant).

Explanation of Solution

Given: $d(c)=0$ ( $c$ a constant).

Concept Used:

Derivative:- of a function $f(x)$

  $f\text{'}(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Also, $\frac{df}{dx}=f\text{'}(x)$ or $df=f\text{'}(x)dx$ .

Calculation:

Take $f(x)=c$ , ( $c$ a constant) then

  $\begin{array}{l}f(x)=c\\ f(x+h)=c\\ \frac{d}{dx}f(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}\\ df=\left({\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}\right)dx\\ d(c)=\left({\lim }_{h\to 0}\frac{c-c}{h}\right)dx=\left(0\right)dx\\ d(c)=0\end{array}$

b.

To determine

Verify the formula $d(cu)=cdu$ ( $c$ a constant).

Explanation of Solution

Given: $d(cu)=cdu$ ( $c$ a constant).

Concept Used:

Derivative:- of a function $f(x)$

  $f\text{'}(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Also, $\frac{df}{dx}=f\text{'}(x)$ or $df=f\text{'}(x)dx$ .

Calculation:

Take $f(u)=cu$ , ( $c$ a constant) then

  $\begin{array}{l}f(u)=cu\\ f(u+h)=c\left(u+h\right)\\ \frac{d}{du}f(u)={\lim }_{h\to 0}\frac{f(u+h)-f(u)}{h}\\ df=\left({\lim }_{h\to 0}\frac{f(u+h)-f(u)}{h}\right)du\\ d(cu)=\left({\lim }_{h\to 0}\frac{c\left(u+h\right)-cu}{h}\right)du\\ d(cu)=\left({\lim }_{h\to 0}\frac{cu+ch-cu}{h}\right)du\\ d(cu)=\left({\lim }_{h\to 0}\frac{ch}{h}\right)du\\ d(cu)=cdu\end{array}$

c.

To determine

Verify the formula $d(u+v)=du+dv$ .

Explanation of Solution

Given: $d(u+v)=du+dv$ .

Calculation:

Take $u$ and $v$ as dependent variable and depends on independent variable $x$ .

  $\begin{array}{l}\frac{d}{dx}(u+v)=\frac{d}{dx}\left(u\right)+\frac{d}{dx}\left(v\right)\\ =\frac{du}{dx}+\frac{dv}{dx}\\ =\frac{du+dv}{dx}\\ \\ \frac{d}{dx}(u+v)=\frac{du+dv}{dx}\\ d(u+v)=du+dv\end{array}$

d.

To determine

Verify the formula $d(u·v)=udv+vdu$ .

Explanation of Solution

Given: $d(u·v)=udv+vdu$ .

Calculation:

Take $u$ and $v$ as dependent variable and depends on independent variable $x$ .

  $\begin{array}{l}\frac{d}{dx}(u·v)=u\frac{d}{dx}\left(v\right)+v\frac{d}{dx}\left(u\right)\\ =u\frac{dv}{dx}+v\frac{du}{dx}\\ =\frac{udv+vdu}{dx}\\ \\ \frac{d}{dx}(u·v)=\frac{udv+vdu}{dx}\\ d(u·v)=udv+vdu\end{array}$

e.

To determine

Verify the formula $d\left(\frac{u}{v}\right)=\frac{\text{v }du-udv}{v^2}$ .

Explanation of Solution

Given: $d\left(\frac{u}{v}\right)=\frac{\text{v }du-udv}{v^2}$ .

Calculation:

Take $u$ and $v$ as dependent variable and depends on independent variable $x$ .

  $\begin{array}{l}\text{Note: }\frac{u}{v}=u{v}^{-1}\\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{d}{dx}\left(u{v}^{-1}\right)\\ =v^{-1}\frac{d}{dx}\left(u\right)+u\frac{d}{dx}\left(v^{-1}\right)\\ =v^{-1}\frac{du}{dx}+u\left(-v^{-2}\right)\frac{dv}{dx}\\ =\frac{1}{v}\frac{du}{dx}-\frac{u}{v^2}\frac{dv}{dx}\\ =\frac{1}{vdx}\left(du-\frac{u}{v}dv\right)\\ =\frac{1}{vdx}\left(\frac{vdu-udv}{v}\right)\\ =\frac{1}{dx}\left(\frac{vdu-udv}{v^2}\right)\\ \\ \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{1}{dx}\left(\frac{vdu-udv}{v^2}\right)\\ d\left(\frac{u}{v}\right)=\frac{vdu-udv}{v^2}\end{array}$

f.

To determine

Verify the formula $d(u^n)=n{u}^{n-1}du$

Explanation of Solution

Given: $d(u^n)=n{u}^{n-1}du$

Concept Used:

Derivative:- of a function $f(x)$

  $f\text{'}(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Also, $\frac{df}{dx}=f\text{'}(x)$ or $df=f\text{'}(x)dx$ .

Calculation:

Take $f(u)=u^n$ , then

  $\begin{array}{l}f(u)=u^n\\ f(u+h)={\left(u+h\right)}^n\\ \frac{d}{du}f(u)={\lim }_{h\to 0}\frac{f(u+h)-f(u)}{h}\\ df=\left({\lim }_{h\to 0}\frac{f(u+h)-f(u)}{h}\right)du\\ d(u^n)=\left({\lim }_{h\to 0}\frac{{\left(u+h\right)}^n-u^n}{h}\right)du\\ d(u^n)=\left({\lim }_{h\to 0}\frac{u^n+c{}_1{u}^{n-1}h+c{}_2{u}^{n-2}{h}^2+\mathrm{.....}-u^n}{h}\right)du\\ d(u^n)=\left({\lim }_{h\to 0}\frac{h\left(c{}_1{u}^{n-1}+c{}_2{u}^{n-2}h+\mathrm{.....}\right)}{h}\right)du\\ d(u^n)=\left({\lim }_{h\to 0}\left(c{}_1{u}^{n-1}+c{}_2{u}^{n-2}h+\mathrm{.....}\right)\right)du\\ d(u^n)=\left(\left(\frac{n!}{\left(n-1\right)!1!}\right)u^{n-1}+0+\mathrm{0.....}\right)du\\ d(u^n)=\left(n{u}^{n-1}+0+\mathrm{0.....}\right)du\\ d(u^n)=n{u}^{n-1}du\end{array}$