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.

(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:
\[\begin{align} \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & =\lim_{\Delta x\to0}\frac{c-c}{\Delta x}\\ & =0.\end{align}\]
(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. If \(\phi(x)=f(x)+g(x)\), then \(\phi(x)\) has a derivative \[\phi'(x)=f'(x)+g'(x).\]
  2. If \(\phi(x)=af(x)\), where \(a\) is a constant, then \(\phi(x)\) has a derivative \[\phi'(x)=af'(x).\]
  3. If \(\phi(x)=f(x)g(x),\) then \(\phi(x)\) has a derivative \[\phi'(x)=f'(x)g(x)+f(x)g'(x).\]
  4. 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
    \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=f'(u)g'(x)=f'(g(x))g'(x).\]
    Similarly if \(y=f(u)\) where \(u=g(v)\) and \(v=h(x)\), then \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}=f'(g(h(x)))g'(h(x))h'(x).\]

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
\[\frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x},\]
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 \[\phi'(x)=f'(g(x))g'(x).\] The proof of the case where \(y=\phi(u),u=f(v)\), and \(v=g(x)\) is similar and is left for practice.

We can generalize this theorem 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.
  • Note 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).
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 (a) 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 (d) in Theorem 2, 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 \(y=\sqrt{x}\qquad\) (\(x>0\)), 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 (a), 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
\[\begin{align} \frac{dy}{dx} & =\frac{dy}{du}\frac{du}{dx}\\ & =\left(mu^{m-1}\right)\left(\frac{1}{n}x^{\frac{1}{n}-1}\right)\\ & =\frac{m}{n}\left(x^{1/n}\right)^{m-1}x^{\frac{1-n}{n}}\\ & =\frac{m}{n}x^{\frac{m-1}{n}}x^{\frac{1-n}{n}}\\ & =\frac{m}{n}x^{\frac{m-n}{n}}\\ & =\frac{m}{n}x^{\frac{m}{n}-1}.\end{align}\]

In general

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\frac{d}{dx}u^{\alpha}=\alpha u^{\alpha-1}\frac{du}{dx}.}\]
So far we have proved the above formula only when \(\alpha\) 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 \(\alpha\). We shall make use of this general result from now on.

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

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
\[\begin{align} \left.\frac{dy}{dx}\right|_{x=3} & =\frac{8}{3}\times3\times\sqrt[3]{(9-1)^{2}}\\ & =8\times8^{2/3}\\ & =8\times(2^{3})^{2/3}=16.\end{align}\]

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