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.

1. The Sum Rule: If $$\phi(x)=f(x)+g(x)$$, then $$\phi(x)$$ has a derivative $\phi'(x)=f'(x)+g'(x).$
2. The Constant Multiple Rule: If $$\phi(x)=af(x)$$, where $$a$$ is a constant, then $$\phi(x)$$ has a derivative $\phi'(x)=af'(x).$
3. The Product Rule: If $$\phi(x)=f(x)g(x),$$ then $$\phi(x)$$ has a derivative $\phi'(x)=f'(x)g(x)+f(x)g'(x).$
4. The Quotient Rule: If $$\phi(x)=\frac{f(x)}{g(x)}$$ and $$f'(x)\neq0$$, then $$\phi(x)$$ has a derivative $\phi'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}.$
5. 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 $x_0$.

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*}