5.7 Differentiation Rules

In this section, we learn some formulas that enable us to differentiate many functions without having to use the definition of a derivative.

Theorem 1. (a) If $f$ is a constant function, $f(x)=c$ where $c$ is a constant, then $f'(x)=0$.
(b) If $f$ is the identity function, $f(x)=x$, then $f'(x)=1$.

This theorem states:

The derivative of a constant is zero.
The derivative of a variable with respect to itself is unity.

Show the proofs

(a) Method 1: Let $y=f(x)=c$. As $x$ takes on an increment $\Delta x$, the value of $y$ does not change; that is $\Delta y=0$, and \[\frac{\Delta y}{\Delta x}=0.\] Thus \[\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=0.\] Method 2:
\[\begin{align} \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & =\lim_{\Delta x\to0}\frac{c-c}{\Delta x}\\ & =0.\end{align}\]

Notice that zero divided by any number other than zero (no matter how small the number is) is zero. Because $c-c=0$ and $\Delta x$ approaches zero but is never equal to zero, we have $(c-c)/\Delta x=0.$
(b) Method 1: Let $y=f(x)=x$, and assume $x$ takes on an increment $\Delta x$. Therefore,
\[y+\Delta y=x+\Delta x\Rightarrow\Delta y=\Delta x\Rightarrow\frac{\Delta y}{\Delta x}=1\]
\[\therefore\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=1.\]
Method 2:
\[\begin{align} \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & =\lim_{\Delta x\to0}\frac{(x+\Delta x)-x}{\Delta x}\\ & =\lim_{\Delta x\to0}\frac{\Delta x}{\Delta x}\\ & =\lim_{\Delta x\to0}1=1.\end{align}\]

Theorem 2. Suppose $f(x)$ and $g(x)$ have derivatives $f'(x)$ and $g'(x)$ for the values of $x$ considered. The following differentiation rules are valid.

The Sum Rule: If $\phi(x)=f(x)+g(x)$, then $\phi(x)$ has a derivative \[\phi'(x)=f'(x)+g'(x).\]
The Constant Multiple Rule: If $\phi(x)=af(x)$, where $a$ is a constant, then $\phi(x)$ has a derivative \[\phi'(x)=af'(x).\]
The Product Rule: If $\phi(x)=f(x)g(x),$ then $\phi(x)$ has a derivative \[\phi'(x)=f'(x)g(x)+f(x)g'(x).\]
The Quotient Rule: If $\phi(x)=\frac{f(x)}{g(x)}$ and $f'(x)\neq0$, then $\phi(x)$ has a derivative \[\phi'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}.\]
The Chain Rule: If $\phi(x)=f(g(x))$, then $\phi(x)$ has a derivative
\[\phi'(x)=f'(g(x))g'(x).\]
If $y=f(u)$ where $u=g(x)$, then we can write the above equation as
\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(u)g'(x).\]

Show the proofs

(a) Let $y=\phi(x),u=f(x)$, and $v=g(x)$. Thus $y=u+v.$
Step 1: If $x$ takes on an increment $\Delta x$, then $u$ and $v$ take on $\Delta u$ and $\Delta v$, respectively. Thus: \[y+\Delta y=u+\Delta u+v+\Delta v\] Step 2: Subtracting $y=u+v$ from $y+\Delta y$ gives \[\Delta y=\Delta u+\Delta v.\] Step 3: Dividing both sides by $\Delta x$, we get
\[\frac{\Delta y}{\Delta x}=\frac{\Delta u}{\Delta x}+\frac{\Delta v}{\Delta x}.\]
Step 4:
\[\begin{align} \frac{dy}{dx} & =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\ & =\lim_{\Delta x\to0}\left(\frac{\Delta u}{\Delta x}+\frac{\Delta v}{\Delta x}\right)\\ & =\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}+\lim_{\Delta x\to0}\frac{\Delta v}{\Delta x}\\ & =\frac{du}{dx}+\frac{dv}{dx}\\ \phi'(x) & =f'(x)+g'(x).\end{align}\]

(b) Let $y=\phi(x)$ and $u=f(x).$ Thus $y=au.$ If $x$ takes on an increment $\Delta x$, $u$ and $y$ take on increments $\Delta u$ and $\Delta y$, respectively.
Step 1: $y=cu$ and \[y+\Delta y=c(u+\Delta u)=cu+c\Delta u\] Step 2: Subtracting $y=cu$ from $y+\Delta y$ gives \[\Delta y=c\Delta u.\] Step 3: \[\frac{\Delta y}{\Delta x}=c\frac{\Delta u}{\Delta x}.\] Step 4:
\[\begin{align} \frac{dy}{dx} & =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\ & =\lim_{\Delta x\to0}c\frac{\Delta u}{\Delta x}\\ & =c\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}\\ & =c\frac{du}{dx}\\ \phi'(x) & =cf'(x).\end{align}\]

(c) Let $y=\phi(x),u=f(x)$, and $v=g(x)$. Thus $y=uv$.
Step 1: If $x$ takes on an increment $\Delta x$, then $y,u$, and $v$ take on $\Delta y$, $\Delta u$, and $\Delta v$, respectively that are related by
\[\begin{align} y+\Delta y & =(u+\Delta u)(v+\Delta v)\\ & =uv+u\Delta v+v\Delta u+\Delta u\Delta v.\end{align}\]
Step 2: To find $\Delta y$, we subtract $y=uv$ from the above expression: \[\Delta y=u\Delta v+v\Delta u+\Delta u\Delta v.\] Step 3:
\[\frac{\Delta y}{\Delta x}=u\frac{\Delta v}{\Delta x}+v\frac{\Delta u}{\Delta x}+\Delta u\frac{\Delta v}{\Delta x}.\]
Step 4:
\[\begin{align} \frac{dy}{dx} & =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\ & =\lim_{\Delta x\to0}\left(u\frac{\Delta v}{\Delta x}\right)+\lim_{\Delta x\to0}\left(v\frac{\Delta u}{\Delta x}\right)+\lim_{\Delta x\to0}\left(\Delta u\frac{\Delta v}{\Delta x}\right)\\ & =\left(\lim_{\Delta x\to0}u\right)\left(\lim_{\Delta x\to0}\frac{\Delta v}{\Delta x}\right)+\left(\lim_{\Delta x\to0}v\right)\left(\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}\right)+\left(\lim_{\Delta x\to0}\Delta u\right)\left(\lim_{\Delta x\to0}\frac{\Delta v}{\Delta x}\right).\end{align}\]
We note that as $\Delta x\to0$, $\Delta u\to0$ and $\Delta v\to0$. Also \[\lim_{\Delta x\to0}u=\lim_{\Delta x\to0}u(x+\Delta x)=u(x),\] which is simply denoted by $u$. Similarly ${\displaystyle \lim_{\Delta x\to0}v=v(x)}$. Thus
\[\begin{align} \frac{dy}{dx} & =u\frac{dv}{dx}+v\frac{du}{dx}+0\frac{dv}{dx}\\ & =u\frac{dv}{dx}+v\frac{du}{dx}\\ \phi'(x) & =f(x)g'(x)+g(x)f'(x).\end{align}\]

(d) Again $y=\phi(x),u=f(x)$, and $v=g(x)$. Thus \[y=\frac{u}{v}\qquad(v\neq0).\] Step 1: Assume $y,u$, and $v$ take on $\Delta y$, $\Delta u$, and $\Delta v$, respectively, when $x$ changes to $x+\Delta x$. Thus \[y+\Delta y=\frac{u+\Delta u}{v+\Delta v}.\] Step 2: Computing
$\Delta y$
\[\begin{align} \Delta y & =\frac{u+\Delta u}{v+\Delta v}-\frac{u}{v}\\ & =\frac{(u+\Delta u)v-u(v+\Delta v)}{v(v+\Delta v)}\\ & =\frac{v\Delta u-v\Delta v}{v(v+\Delta v)}\end{align}\]
Step 3: Dividing $\Delta y$ by $\Delta x$
\[\frac{\Delta y}{\Delta x}=\frac{v\dfrac{\Delta u}{\Delta x}-v\dfrac{\Delta v}{\Delta x}}{v(v+\Delta v)}\]
Step 4:
\[\begin{align} \frac{dy}{dx} & =\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}\\ & =_{}\frac{v{\displaystyle \lim_{\Delta x\to0}}\dfrac{\Delta u}{\Delta x}-v{\displaystyle \lim_{\Delta x\to0}}\dfrac{\Delta v}{\Delta x}}{{\displaystyle \lim_{\Delta x\to0}}v\;{\displaystyle \lim_{\Delta x\to0}}(v+\Delta v)}\\ & =\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v\times v}\\ \phi'(x) & =\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}.\end{align}\]

(e) An increment $\Delta x$ determines an increment $\Delta u$, and this in turn determines an increment $\Delta y$; that is,
\[\begin{align} \Delta u & =g(x+\Delta x)-g(x),\\ \Delta y & =f(u+\Delta u)-f(u).\end{align}\]
Then evidently
\[\frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x},\]
provided $\Delta u\neq 0$. Because $u=g(x)$ is a continuous function $\Delta u\to0$ as $\Delta x\to0$. Thus
\[\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim_{\Delta u\to0}\frac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x};\]
that is, \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx},\] or \[\phi'(x)=f'(g(x))g'(x).\]

What if Δu =0?

In fact, we may have $\Delta u=0$ for infinitely many values of $\Delta x$. If $\Delta u=0$, we cannot multiply and divide $\Delta y/\Delta x$ by $\Delta u$. To include this case as well, let’s define a new function $\alpha$:
$$\alpha(\Delta u)=\begin{cases}
\dfrac{f(u+\Delta u)-f(u)}{\Delta u}-f'(u) & \text{if }\Delta u\neq0\\[9pt] 0 & \text{if }\Delta u=0
\end{cases}.\tag{i}$$ Notice that in this definition $\alpha$ is a function of $\Delta u$ and $u$ is considered as a constant, so $f'(u)$ is just a number. Obviously $\alpha$ is a continuous function when $\Delta u\ne0$. Also $\alpha$ is continuous at $\Delta u=0$ because
$$
\begin{align*}
\lim_{\Delta u\to0}\alpha(\Delta u) & =\lim_{\Delta u\to0}\left(\frac{f(u+\Delta u)-f(u)}{\Delta u}-f'(u)\right)\\
& =\lim_{\Delta u\to0}\frac{f(u+\Delta u)-f(u)}{\Delta u}-\lim_{\Delta u\to0}f'(u)&&{\small \text{(Limit laws)}}\\
& =f'(u)-f'(u)=0=\alpha(0).&&{\small \text{(Definition of derivative)}}
\end{align*}
$$
It follows from (i) that
$$
f(u+\Delta u)-f(u)=\Delta u\left[\alpha(\Delta u)+f'(u)\right].\tag{ii}
$$
This equation is valid whether or not $\Delta u=0.$ Now we are ready to investigate the derivative of $\phi(x)$:
$$
\begin{align*}
\lim_{\Delta x\to 0}\frac{\phi(x+\Delta x)-\phi(x)}{\Delta x}=& =\lim_{\Delta x\to0}\frac{f\big(\overbrace{g(x+\Delta x)}^{u+\Delta u}\big)-f\big(\overbrace{g(x)}^{u}\big)}{\Delta x} &&{\small \text{($\phi=f\circ g$)}}\\[9pt] & =\lim_{\Delta x\to0}\frac{f(u+\Delta u)-f(u)}{\Delta x}\\[9pt] & =\lim_{\Delta x\to0}\frac{\Delta u\left[\alpha(\Delta u)+f'(u)\right]}{\Delta x}&&{\small \text{(Use (ii))}}\\[9pt] & =\left(\lim_{\Delta x\to0}\frac{\Delta u}{\Delta x}\right)\left(\lim_{\Delta x\to0}\alpha(\Delta u)+\lim_{\Delta x\to0}f'(u)\right)&&{\small \text{(Limit laws)}}
\end{align*}
$$
Since as $\Delta x\to0$, $\Delta u\to0$, and as $\Delta u\to 0$, $\alpha(\Delta u)\to0$, we get
$$
\phi'(x)=\underbrace{g'(x)}_{du/dx}f'(u)=g'(x)f’\big(\underbrace{g(x)}_{u}\big).
$$
This completes the proof of the Chain Rule.

Expressing the chain rule as $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ when $y=f(u)$ and $u=g(x)$ is sloppy. To be more precise, we have to write the chain rule as \[ \bbox[#F2F2F2,5px,border:2px solid black]{\left(\frac{dy}{dx}\right)_{x_0}=\left(\frac{dy}{du}\right)_{g(x_0)}\left(\frac{du}{dx}\right)_{x_0}.}\] This notation tells us that the derivative of $y$ with respect to $x$ at $x_0$ is equal to the derivative of $y$ with respect to $u$ at $g(x_0)$ multiplied by the derivative of $u$ with respect to $x$ at $x_0$.

Read more about Theorem 2

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 \[y’=f_{1}'(x)+f_{2}'(x)+\cdots+f_{n}'(x).\]

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 \[\begin{align}y’=&\left[f_{2}(x)f_{3}(x)\cdots f_{n}(x)\right]\frac{df_{1}}{dx}+[f_{1}(x)f_{3}(x)\cdots f_{n}(x)]\frac{df_{2}}{dx}\\&+\cdots+[f_{1}(x)f_{2}(x)\cdots f_{n-1}(x)]\frac{df_{n}}{dx}.\tag{a}\end{align}\]

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 \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}=f'(g(h(x)))g'(h(x))h'(x).\]

Example 1

Given $y=x^{2}$, differentiate $y$ with respect to $x$.

Solution

We write $y=f(x)g(x)$ where $f(x)=x$ and $g(x)=x$. Thus
\[\begin{align} y’ & =f'(x)g(x)+f(x)g'(x)\\ & =1\cdot x+x\cdot1\\ & =2x.\end{align}\]

Example 2

Differentiate $y=x^{n}$ with respect to $x$ where $n>0$ is an integer.

Solution

Using the Product Rule, we can write
\[\begin{align} y’ & =(\underbrace{xx\cdots x}_{n-1\text{ times}})\frac{dx}{dx}+\cdots+(\underbrace{xx\cdots x}_{n-1\text{ times}})\frac{dx}{dx}\\ & =\underbrace{x^{n-1}\cdot1+\cdots+x^{n-1}\cdot1}_{n\text{ times}}\\ & =nx^{n-1}.\end{align}\]

Example 3

Find the derivative of $y$ with respect to $x$ if \[y=4x^{3}-2x^{2}+6x+5.\]

Solution

\[\begin{align} y’ & =4\times3x^{2}-2\times2x+6+0.\\ & =12x^{2}-4x+6.\end{align}\]

Example 4

Prove ${\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}}$, where $n$ is a negative integer.

Solution

Let $n=-m$, where $m$ is a positive integer. Thus \[x^{n}=x^{-m}=\frac{1}{x^{m}}.\] Using the Quotient Rule, we can write
\[\begin{align} \frac{d}{dx}\frac{1}{x^{m}} & =\frac{0\times x^{m}-mx^{m-1}\times1}{\left(x^{m}\right)^{2}}\\ & =\underbrace{-m}_{=n}\frac{x^{m-1}}{x^{2m}}\\ & =nx^{m-1-2m}\\ & =nx^{-m-1}\\ & =nx^{n-1}.\tag{$-m=n$}\end{align}\]
That is, \[\frac{d}{dx}x^{n}=nx^{n-1}\]

Example 5

Given ${\displaystyle y=\frac{1}{x^{3}}}$, find $\dfrac{dy}{dx}$.

Solution

Because $y=x^{-3}$,
\[\begin{align} \frac{d}{dx}y & =-3x^{-3-1}\\ & =-3x^{-4}\\ & =-\frac{3}{x^{4}}.\end{align}\]

Example 6

Given $y=\sqrt{x}$ ($x>0$), find $\dfrac{dy}{dx}$.

Solution

Let $f(x)=\sqrt{x}$ and $g(x)=f(x)f(x)=x$.
\[\begin{align} g'(x) & =f'(x)f(x)+f(x)f'(x)\\ & =2f(x)f'(x)\end{align}\]
We know $g'(x)=1$ (Theorem 1). Thus
\[\begin{align} 1 & =2f(x)f'(x)\\ & =2\sqrt{x}f'(x)\end{align}\]
Finally \[f'(x)=\frac{1}{2\sqrt{x}}.\]

Example 7

Evaluate ${\displaystyle \frac{d}{dx}\sqrt{1-3x^{2}}}$.

Solution

Let $u=1-3x^{2}$ and $y=\sqrt{u}$. Then
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =\left(\frac{1}{2\sqrt{u}}\right)(-6x)\\ & =\frac{-6x}{2\sqrt{1-3x^{2}}}\\ & =\frac{-3x}{\sqrt{1-3x^{2}}}.\end{align}\]

Example 8

Find $\dfrac{dy}{dx}$ if $y=x^{1/n}$ where $n$ is an integer.

Solution

Let $f(x)=x^{1/n}$ and $g(x)=[f(x)]^{n}=\underbrace{f(x)\cdots f(x)}_{n\text{ times}}=x^{n/n}=x$. Using the Product Rule, we have
\[\begin{align} g'(x) & =\underbrace{[f(x)]^{n-1}f'(x)+\cdots+[f(x)]^{n-1}f'(x)}_{n\text{ times}}\\ & =n[f(x)]^{n-1}f'(x)\\ 1 & =n\left[x^{1/n}\right]^{n-1}f'(x).\end{align}\]
Thus
\[\begin{align} f'(x) & =\frac{1}{nx^{(n-1)/n}}\\ & =\frac{1}{nx^{1-1/n}}\\ & =\frac{1}{n}x^{\frac{1}{n}-1}.\end{align}\]
That is \[\frac{d}{dx}x^{\frac{1}{n}}=\frac{1}{n}x^{\frac{1}{n}-1}.\]

Example 9

Evaluate \[\frac{d}{dx}\sqrt[5]{4x^{2}-\frac{3}{x}}.\]

Solution

Let $y=u^{1/5}$ and $u=4x^{2}-\frac{3}{x}$. Then
\[\begin{align} \frac{dy}{du} & =\frac{1}{5}u^{\frac{1}{5}-1}\\ & =\frac{1}{5}u^{-\frac{4}{5}}\\ & =\frac{1}{5}\left(4x^{2}-\frac{3}{x}\right)^{-\frac{4}{5}}\end{align}\]
\[\begin{align} \frac{du}{dx} & =8x-3\frac{d}{dx}\left(x^{-1}\right)\\ & =8x-3\left(-1\cdot x^{-2}\right)\\ & =8x+\frac{3}{x^{2}}.\end{align}\]
Thus
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =\frac{1}{5}\left(4x^{2}-\frac{3}{x}\right)^{-\frac{4}{5}}\left(8x+\frac{3}{x^{2}}\right)\\ & =\left(\frac{8}{5}x+\frac{3}{5x^{2}}\right)\frac{1}{\sqrt[5]{\left(4x^{2}-\frac{3}{x}\right)^{4}}}.\end{align}\]

Example 10

Differentiate $f(x)=\dfrac{x^{2}-1}{x^{3}+1}$.

Solution

\[\begin{align} \dfrac{df}{dx} & =\frac{2x(x^{3}+1)-3x^{2}(x^{2}-1)}{(x^{3}+1)^{2}}\\ & =\frac{-x^{4}+3x^{2}+2x}{(x^{3}+1)^{2}}.\end{align}\]

Example 11

Prove $\dfrac{d}{dx}u^{n}=nu^{n-1}\dfrac{du}{dx}$, where $n$ is a positive integer.

Solution

Let $y=u^{n}$ and $u=u(x)$. Then
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =nu^{n-1}\frac{du}{dx}.\end{align}\]

Example 12

Prove $\dfrac{d}{dx}x^{m/n}=\frac{m}{n}x^{\frac{m}{n}-1}$, where $m$ and $n$ are two integers.

Solution

Let $y=u^{m}$ and $u=x^{1/n}$. Then
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =\left(mu^{m-1}\right)\left(\frac{1}{n}x^{\frac{1}{n}-1}\right)\\ & =\frac{m}{n}\left(x^{1/n}\right)^{m-1}x^{\frac{1-n}{n}}\\ & =\frac{m}{n}x^{\frac{m-1}{n}}x^{\frac{1-n}{n}}\\ & =\frac{m}{n}x^{\frac{m-n}{n}}\\ & =\frac{m}{n}x^{\frac{m}{n}-1}.\end{align}\]

The Power Rule

In general,

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}x^r=r x^{r-1}\qquad r\in\mathbb{R}.}\]

So far we have proved the above formula only when $r$ is a rational number (= a number that can be written as the ratio of two integers). However, it will be shown that this formula holds true for any value of $r$. This is called the Power Rule. We shall make use of this general result from now on.

The function $f(x)=x^r$ is differentiable at $x=0$ if $r$ is a number such that $x^{r-1}$ is defined on an interval containing 0.

If $u$ is a differentiable function of $x$, then it follows from the Chain Rule and the Power Rule that

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}u^{r}=r u^{r-1}\frac{du}{dx}.}\]

The derivative of a function with a constant exponent is equal to the product of the exponent, the function with the exponent diminished by unity, and the derivative of the function.

Example 13

Differentiate $y$ with respect to $x$ given
\[y=\left(\frac{1}{x^{2}}\right)^{2}+3\left(\frac{1}{x^{2}}\right)+1.\]

Solution

Let $u=1/x^{2}$, then $y=u^{2}+3u+1.$
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =(2u+3)\frac{d}{dx}x^{-2}\\ & =(2u+3)\left(-2x^{-3}\right)\\ & =\left(\frac{2}{x^{2}}+3\right)\left(\frac{-2}{x^{3}}\right)\\ & =\frac{-4}{x^{5}}-\frac{6}{x^{3}}.\end{align}\]

Example 14

If $y=(x+1)\sqrt{x^{2}+1}$, find $dy/dx.$

Solution

\[\begin{align} \frac{dy}{dx} & =\frac{d(x+1)}{dx}\sqrt{x^{2}+1}+(x+1)\frac{d\sqrt{x^{2}+1}}{dx}\\ & =1\times\sqrt{x^{2}+1}+(x+1)\frac{d}{dx}(x^{2}+1)^{1/2}\\ & =\sqrt{x^{2}+1}+(x+1)\left[\frac{1}{2}(x^{2}+1)^{-1/2}\frac{d}{dx}(x^{2}+1)\right]\\ & =\sqrt{x^{2}+1}+(x+1)\left[\frac{1}{2}(x^{2}+1)^{-1/2}\times2x\right]\\ & =\sqrt{x^{2}+1}+\frac{2x(x+1)}{2\sqrt{x^{2}+1}}\\ & =\frac{x^{2}+1}{\sqrt{x^{2}+1}}+\frac{x(x+1)}{\sqrt{x^{2}+1}}\\ & =\frac{2x^{2}+x+1}{\sqrt{x^{2}+1}}.\end{align}\]

Example 15

If ${\displaystyle y=\sqrt[5]{\frac{x}{x^{5}+1}}}$, find $\dfrac{dy}{dx}$.

Solution

We write \[y=\left(\frac{x}{x^{5}+1}\right)^{1/5}.\] Let $u=x/(x^{2}+1)$. Then $y=u^{1/5}$ and
\[\begin{align} \frac{dy}{dx} & =\frac{1}{5}u^{1-\frac{1}{5}}\ \frac{du}{dx}\\ & =\frac{1}{5}\left(\frac{x}{x^{5}+1}\right)^{-4/5}\frac{d}{dx}\left(\frac{x}{x^{5}+1}\right)\\ & =\frac{1}{5}\left(\frac{x}{x^{5}+1}\right)^{-4/5}\times\frac{1\times(x^{5}+1)-5x^{4}\times x}{(x^{5}+1)^{2}}\\ & =\frac{1}{5x^{4/5}(x^{5}+1)^{-4/5}}\frac{1-4x^{5}}{(x^{5}+1)^{2}}\\ & =\frac{1}{5x^{4/5}}\frac{1-4x^{5}}{(x^{5}+1)^{2-4/5}}\\ & =\frac{1}{5x^{4/5}}\frac{1-4x^{5}}{(x^{5}+1)^{6/5}}.\\\end{align}\]

Example 16

If $f(x)=\sqrt[3]{x^{2}-1}$ and $g(x)=x^{4}-1$, find the derivative of $g\circ f(x)$ and the value of its derivative at $x=3$.

Solution

In the first method, we find $g\circ f(x)$ and the differentiate it. In the second method, we find $f’$ and $g’$, and evaluate them at $x=3$ and $f(3)$ respectively; finally we multiply the results. As we can see the second method is much easier.

Method a: Let $u=f(x)=\sqrt[3]{x^{2}-1}$ and $y=g(u)=u^{4}-1$. Then

\[\frac{d}{dx}g\circ f(x)=g'(u)f'(x)\] or
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =4u^{3}\frac{du}{dx}.\tag{i}\end{align}\]
To find $du/dx$, let $u=v^{1/3}$ where $v=x^{2}-1$. Thus
\[\begin{align} \frac{du}{dx} & =\frac{du}{dv}\frac{dv}{dx}\\ & =\frac{1}{3}v^{\frac{1}{3}-1}(2x)\\ & =\frac{1}{3}(x^{2}-1)^{-2/3}2x\\ & =\frac{2x}{3\sqrt[3]{(x^{2}-1)^{2}}}\tag{ii }\end{align}\]
Placing (ii) in (i), we get
\[\begin{align} \frac{dy}{dx} & =4u^{3}\frac{2x}{3\sqrt[3]{(x^{2}-1)^{2}}}\\ & =4\left(\sqrt[3]{x^{2}-1}\right)^{3}\frac{2x}{3\sqrt[3]{(x^{2}-1)^{2}}}\\ & =\frac{8x(x^{2}-1)}{3\sqrt[3]{(x^{2}-1)^{2}}}\\ & =\frac{8}{3}x\sqrt[3]{(x^{2}-1)^{2}}\end{align}\]
And
\[\begin{align} \left.\frac{dy}{dx}\right|_{x=3} & =\frac{8}{3}\times3\times\sqrt[3]{(9-1)^{2}}\\ & =8\times8^{2/3}\\ & =8\times(2^{3})^{2/3}=16.\end{align}\]

Method b:

\[f(x)=(\underbrace{x^2-1}_u)^{1/3}\Rightarrow f^\prime(x)=\frac{1}{3}(x^2-1)^{-2/3}\underbrace{2x}_{u^\prime}\]

\[g(x)=x^4-1\Rightarrow g'(x)=4x^3\]

Since $f(3)=(3^2-1)^{1/3}=8^{1/3}=2$, we have
\begin{align*}
(g\circ f)'(3)=&g'(\underbrace{2}_{f(3)})f'(3)\\
=& \left(4x^3\big|_2 \right)\left(\frac{2x}{3}(x^2-1)^{-2/3}\big|_3\right)\\
=&4\times 8\times \frac{2\times 3}{3}\frac{1}{(3^2-1)^{2/3}}\\
=&16.
\end{align*}

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The Power Rule