5.7 Differentiation Rules

(a) Method 1: Let $y=f(x)=c$. As $x$ takes on an increment $\mathrm{\Delta }x$, the value of $y$ does not change; that is $\mathrm{\Delta }y=0$, and $$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=0.$$ Thus $$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=0.$$ Method 2:
$$\begin{array}{rl}\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{c-c}{\mathrm{\Delta }x}\\ & =0.\end{array}$$

Notice that zero divided by any number other than zero (no matter how small the number is) is zero. Because $c-c=0$ and $\mathrm{\Delta }x$ approaches zero but is never equal to zero, we have $(c-c)/\mathrm{\Delta }x=0.$
(b) Method 1: Let $y=f(x)=x$, and assume $x$ takes on an increment $\mathrm{\Delta }x$. Therefore,
$$y+\mathrm{\Delta }y=x+\mathrm{\Delta }x\Rightarrow \mathrm{\Delta }y=\mathrm{\Delta }x\Rightarrow \frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=1$$
$$\therefore \frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=1.$$
Method 2:
$$\begin{array}{rl}\underset{\mathrm{\Delta }x\to 0}{lim}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{(x+\mathrm{\Delta }x)-x}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }x}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}1=1.\end{array}$$

(a) Let $y=\varphi (x),u=f(x)$, and $v=g(x)$. Thus $y=u+v.$
Step 1: If $x$ takes on an increment $\mathrm{\Delta }x$, then $u$ and $v$ take on $\mathrm{\Delta }u$ and $\mathrm{\Delta }v$, respectively. Thus: $$y+\mathrm{\Delta }y=u+\mathrm{\Delta }u+v+\mathrm{\Delta }v$$ Step 2: Subtracting $y=u+v$ from $y+\mathrm{\Delta }y$ gives $$\mathrm{\Delta }y=\mathrm{\Delta }u+\mathrm{\Delta }v.$$ Step 3: Dividing both sides by $\mathrm{\Delta }x$, we get
$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}+\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}.$$
Step 4:
$$\begin{array}{rl}\frac{dy}{dx}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}(\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}+\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x})\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}+\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}\\ & =\frac{du}{dx}+\frac{dv}{dx}\\ {\varphi }^{\prime }(x)& =f^{\prime }(x)+g^{\prime }(x).\end{array}$$

(b) Let $y=\varphi (x)$ and $u=f(x).$ Thus $y=au.$ If $x$ takes on an increment $\mathrm{\Delta }x$, $u$ and $y$ take on increments $\mathrm{\Delta }u$ and $\mathrm{\Delta }y$, respectively.
Step 1: $y=cu$ and $$y+\mathrm{\Delta }y=c(u+\mathrm{\Delta }u)=cu+c\mathrm{\Delta }u$$ Step 2: Subtracting $y=cu$ from $y+\mathrm{\Delta }y$ gives $$\mathrm{\Delta }y=c\mathrm{\Delta }u.$$ Step 3: $$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=c\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}.$$ Step 4:
$$\begin{array}{rl}\frac{dy}{dx}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}c\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}\\ & =c\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}\\ & =c\frac{du}{dx}\\ {\varphi }^{\prime }(x)& =c{f}^{\prime }(x).\end{array}$$

(c) Let $y=\varphi (x),u=f(x)$, and $v=g(x)$. Thus $y=uv$.
Step 1: If $x$ takes on an increment $\mathrm{\Delta }x$, then $y,u$, and $v$ take on $\mathrm{\Delta }y$, $\mathrm{\Delta }u$, and $\mathrm{\Delta }v$, respectively that are related by
$$\begin{array}{rl}y+\mathrm{\Delta }y& =(u+\mathrm{\Delta }u)(v+\mathrm{\Delta }v)\\ & =uv+u\mathrm{\Delta }v+v\mathrm{\Delta }u+\mathrm{\Delta }u\mathrm{\Delta }v.\end{array}$$
Step 2: To find $\mathrm{\Delta }y$, we subtract $y=uv$ from the above expression: $$\mathrm{\Delta }y=u\mathrm{\Delta }v+v\mathrm{\Delta }u+\mathrm{\Delta }u\mathrm{\Delta }v.$$ Step 3:
$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=u\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}+v\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}+\mathrm{\Delta }u\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}.$$
Step 4:
$$\begin{array}{rl}\frac{dy}{dx}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\\ & =\underset{\mathrm{\Delta }x\to 0}{lim}\left(u\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}\right)+\underset{\mathrm{\Delta }x\to 0}{lim}\left(v\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}\right)+\underset{\mathrm{\Delta }x\to 0}{lim}\left(\mathrm{\Delta }u\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}\right)\\ & =\left(\underset{\mathrm{\Delta }x\to 0}{lim}u\right)\left(\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}\right)+\left(\underset{\mathrm{\Delta }x\to 0}{lim}v\right)\left(\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}\right)+\left(\underset{\mathrm{\Delta }x\to 0}{lim}\mathrm{\Delta }u\right)\left(\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}\right).\end{array}$$
We note that as $\mathrm{\Delta }x\to 0$, $\mathrm{\Delta }u\to 0$ and $\mathrm{\Delta }v\to 0$. Also $$\underset{\mathrm{\Delta }x\to 0}{lim}u=\underset{\mathrm{\Delta }x\to 0}{lim}u(x+\mathrm{\Delta }x)=u(x),$$ which is simply denoted by $u$. Similarly ${\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}v=v(x)}$. Thus
$$\begin{array}{rl}\frac{dy}{dx}& =u\frac{dv}{dx}+v\frac{du}{dx}+0\frac{dv}{dx}\\ & =u\frac{dv}{dx}+v\frac{du}{dx}\\ {\varphi }^{\prime }(x)& =f(x)g^{\prime }(x)+g(x)f^{\prime }(x).\end{array}$$

(d) Again $y=\varphi (x),u=f(x)$, and $v=g(x)$. Thus $$y=\frac{u}{v}(v\ne 0).$$ Step 1: Assume $y,u$, and $v$ take on $\mathrm{\Delta }y$, $\mathrm{\Delta }u$, and $\mathrm{\Delta }v$, respectively, when $x$ changes to $x+\mathrm{\Delta }x$. Thus $$y+\mathrm{\Delta }y=\frac{u+\mathrm{\Delta }u}{v+\mathrm{\Delta }v}.$$ Step 2: Computing
$\mathrm{\Delta }y$
$$\begin{array}{rl}\mathrm{\Delta }y& =\frac{u+\mathrm{\Delta }u}{v+\mathrm{\Delta }v}-\frac{u}{v}\\ & =\frac{(u+\mathrm{\Delta }u)v-u(v+\mathrm{\Delta }v)}{v(v+\mathrm{\Delta }v)}\\ & =\frac{v\mathrm{\Delta }u-v\mathrm{\Delta }v}{v(v+\mathrm{\Delta }v)}\end{array}$$
Step 3: Dividing $\mathrm{\Delta }y$ by $\mathrm{\Delta }x$
$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{v{\displaystyle \frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}}-v{\displaystyle \frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}}}{v(v+\mathrm{\Delta }v)}$$
Step 4:
$$\begin{array}{rl}\frac{dy}{dx}& =\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\\ & {=}_{}\frac{v{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}}{\displaystyle \frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}}-v{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}}{\displaystyle \frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}}}{{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}}v{\displaystyle \underset{\mathrm{\Delta }x\to 0}{lim}}(v+\mathrm{\Delta }v)}\\ & =\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v\times v}\\ {\varphi }^{\prime }(x)& =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}.\end{array}$$

(e) An increment $\mathrm{\Delta }x$ determines an increment $\mathrm{\Delta }u$, and this in turn determines an increment $\mathrm{\Delta }y$; that is,
$$\begin{array}{rl}\mathrm{\Delta }u& =g(x+\mathrm{\Delta }x)-g(x),\\ \mathrm{\Delta }y& =f(u+\mathrm{\Delta }u)-f(u).\end{array}$$
Then evidently
$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}\cdot \frac{\mathrm{\Delta }u}{\mathrm{\Delta }x},$$
Because $u=g(x)$ is a continuous function $\mathrm{\Delta }u\to 0$ as $\mathrm{\Delta }x\to 0$. Thus
$$\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }u\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}\cdot \underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x};$$
that is, $$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx},$$ or $${\varphi }^{\prime }(x)=f^{\prime }(g(x))g^{\prime }(x).$$ The proof of the case where $y=\varphi (u),u=f(v)$, and $v=g(x)$ is similar and is left for practice.

We can generalize Theorem 2 and say:

• The derivative of the sum of a finite number of functions is equal to the sum of the derivatives of the functions. That is, if $y=f_1(x)+f_2(x)+\cdots +f_n(x)$, then $$y^{\prime }=f_1^{\prime }(x)+f_2^{\prime }(x)+\cdots +f_n^{\prime }(x).$$

• The derivative of a constant times a function is equal to the constant times the derivative of the function.

• The derivative of the product of a finite number of functions is equal to the sum of the products obtained by multiplying the derivative of each factor by all the other functions. That is, if $y=f_1(x)f_2(x)\cdots f_n(x)$ then $$\begin{array}{rl}{y}^{\prime }=& [f_2(x)f_3(x)\cdots f_n(x)]\frac{d{f}_1}{dx}+[f_1(x)f_3(x)\cdots f_n(x)]\frac{d{f}_2}{dx}\\ \text{(a)}& & +\cdots +[f_1(x)f_2(x)\cdots f_{n-1}(x)]\frac{d{f}_n}{dx}.\end{array}$$

• The derivative of a fraction is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

• Notice the minus sign in the numerator in the formula of part (d). Because of this minus sign, the order here matters (unlike the product rule in part c).

• If $y=f(u)$ where $u=g(v)$ and $v=h(x)$, then it follows from the Chain Rule that $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}=f^{\prime }(g(h(x)))g^{\prime }(h(x))h^{\prime }(x).$$