Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Rules of Differentiation

We have applied Definition 4.5 to find the derivative function. However, the similarities among certain types of functions allow for generalizations and for the development of rules or formulas for differentiation, instead of the direct application of Definition 4.3 or Definition 4.5.

Since the derivative is the slope of the tangent line, we can see that the graph of the constant function $g\left(x\right)=c$ is a horizontal line which coincides with its tangent line. Therefore, the derivative is zero; that is,

${\displaystyle {\displaystyle \frac{d}{dx}}c=0\text{.}}$

Also, for a linear function $g\left(x\right)=mx$, the tangent line coincides with the function, and the slope of this line is $m$. Hence,

${\displaystyle {\displaystyle \frac{d}{dx}}mx=m\text{.}}$

From Examples 4.14 and 4.16, we have

${\displaystyle {\displaystyle \frac{d}{dx}}\left(x^{1∕2}\right)={\displaystyle \frac{d}{dx}}\sqrt{x}={\displaystyle \frac{1}{2\sqrt{x}}}={\displaystyle \frac{1}{2}}\left(x^{-1∕2}\right)\text{ and }{\displaystyle \frac{d}{dx}}{x}^3=3x^2\text{.}}$

These two examples suggest a generalization of the form

${\displaystyle {\displaystyle \frac{d}{dx}}{x}^r=r{x}^{r-1}\text{.}}$

If this generalization is indeed workable, then the derivative of $f\left(x\right)=x^{-1}$ should be $f^{\prime }\left(x\right)=-\left(x^{-2}\right)\text{.}$ [Why?]

Let us see if this surmise is true by applying Definition 4.3:

${\displaystyle \underset{x\to a}{\lim }{\displaystyle \frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{a}}}{x-a}}=\underset{x\to a}{lim}{\displaystyle \frac{a-x}{\left(xa\right)\left(x-a\right)}}=\underset{x\to a}{lim}-{\displaystyle \frac{1}{xa}}=-{\displaystyle \frac{1}{a^2}}=-a^{-2}\text{.}}$

Since this statement is true for any number $a$, we conclude that $f^{\prime }\left(x\right)=-\left(x^{-2}\right)$, as we suspected.

Pages 132 and 133 of the textbook give a proof (explanation) of why, for any positive integer $n$,

${\displaystyle {\displaystyle \frac{d}{dx}}{x}^n=n{x}^{n-1}\text{.}}$

The proof for any real number $r$ is more complex, and is not covered in this course. Instead, we simply present the power rule, shown below.

The Power Rule:   For any real number $r$,   ${\displaystyle {\displaystyle \frac{d}{dx}}{x}^r=r{x}^{r-1}\text{.}}$

To apply the power rule, we must identify the power $r$.

Example 4.22. In the function

${\displaystyle f\left(x\right)={\displaystyle \frac{1}{x^{3∕2}}}\text{,}}$

the power is $r=-3∕2$, since $f\left(x\right)=x^{-3∕2}$.

So, applying the power rule, we find that

${\displaystyle f^{\prime }\left(x\right)=\left(-{\displaystyle \frac{3}{2}}\right)x^{\left(-3∕2\right)-1}=\left(-{\displaystyle \frac{3}{2}}\right)x^{-5∕2}=-{\displaystyle \frac{3}{2x^{5∕2}}}\text{.}}$

Example 4.23. In the function

${\displaystyle f\left(x\right)={\displaystyle \frac{1}{\sqrt{x}}}\text{,}}$

the power is $r=-1∕2$. [Why?]

Then,

${\displaystyle f^{\prime }\left(x\right)=\left(-{\displaystyle \frac{1}{2}}\right)x^{\left(-1∕2\right)-1}=\left(-{\displaystyle \frac{1}{2}}\right)x^{-3∕2}=-{\displaystyle \frac{1}{2x^{3∕2}}}\text{.}}$

Example 4.24. ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{x^{\sqrt{3}}}}={\displaystyle \frac{d}{dx}}{x}^{-\sqrt{3}}=-\sqrt{3}{x}^{-\sqrt{3}-1}=-{\displaystyle \frac{\sqrt{3}}{x^{\sqrt{3}+1}}}\text{.}}$

By the laws of limits (Theorem 3.23), we derive the sum and constant multiple rules shown below.

The Sum Rule:   ${\displaystyle {\displaystyle \frac{d}{dx}}\left[f\left(x\right)\pm g\left(x\right)\right]={\displaystyle \frac{d}{dx}}f\left(x\right)\pm {\displaystyle \frac{d}{dx}}g\left(x\right)}$

The Constant Multiple Rule:   ${\displaystyle {\displaystyle \frac{d}{dx}}cf\left(x\right)=c{\displaystyle \frac{d}{dx}}f\left(x\right)}$

Observe how these rules are applied in the differentiation of the functions below.

Example 4.25. We indicated by a geometric argument that the derivative of $f\left(x\right)=mx$ is $m\text{.}$ Using the rules of differentiation we will confirm this conclusion, using $m=5$.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}5x& =5{\displaystyle \frac{d}{dx}}x\hfill & \text{constant multiple rule}\hfill \\ \hfill & =\mathrm{5 }\left(1\right)x^{1-1}\hfill & \text{power rule}\hfill \\ \hfill & =5x^0=5\hfill & \text{as we indicated above}\hfill \end{array}}$

Example 4.26.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{3}{4x}}+\sqrt[3]{x}& ={\displaystyle \frac{d}{dx}}{\displaystyle \frac{3}{4}}{x}^{-1}+{\displaystyle \frac{d}{dx}}{x}^{1/3}\hfill & \text{sum rule}\hfill \\ \hfill & ={\displaystyle \frac{3}{4}}{\displaystyle \frac{d}{dx}}{x}^{-1}+{\displaystyle \frac{d}{dx}}{x}^{1/3}\hfill & \text{constant multiple rule}\hfill \\ \hfill & ={\displaystyle \frac{3}{4}}\left(-1\right)x^{-2}+\left({\displaystyle \frac{1}{3}}\right)x^{\left(1/3\right)-1}\hfill & \text{power rule}\hfill \\ \hfill & =-{\displaystyle \frac{3}{4x^2}}+{\displaystyle \frac{1}{3x^{2/3}}}\text{.}\hfill & \hfill \end{array}}$

Example 4.27.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}6x^{12}& =6{\displaystyle \frac{d}{dx}}{x}^{12}\hfill & \text{constant multiple rule}\hfill \\ \hfill & =6\left(12\right)x^{11}\hfill & \text{power rule}\hfill \\ \hfill & =72x^{11}\hfill & \hfill \end{array}}$

In practical terms, we see from Examples 4.26 and 4.27 that

${\displaystyle {\displaystyle \frac{d}{dx}}c{x}^r=\left(cr\right)x^{r-1}\text{.}}$

Example 4.28.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{2x^4}{3}}& +6x^2-\sqrt{5}\hfill & \hfill \\ \hfill & ={\displaystyle \frac{2}{3}}{\displaystyle \frac{d}{dx}}{x}^4+6{\displaystyle \frac{d}{dx}}{x}^2-{\displaystyle \frac{d}{dx}}\sqrt{5}\hfill & \text{constant multiple and sum rules}\hfill \\ \hfill & ={\displaystyle \frac{8}{3}}{x}^3+12x\hfill & \text{power and constant rules}\hfill \end{array}}$

Let us use Definition 4.2 to find the derivative of the reciprocal function ${\displaystyle f\left(x\right)={\displaystyle \frac{1}{g\left(x\right)}}\text{.}}$

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{g\left(x\right)}}& =\underset{x\to a}{\lim }{\displaystyle \frac{{\displaystyle \frac{1}{g\left(x\right)}}-{\displaystyle \frac{1}{g\left(a\right)}}}{x-a}}\hfill & \text{Definition 4}\text{.2}\hfill \\ \hfill & =\underset{x\to a}{\lim }{\displaystyle \frac{g\left(a\right)-g\left(x\right)}{g\left(x\right)g\left(a\right)\left(x-a\right)}}\hfill & \text{doing the operations}\hfill \\ \hfill & =\underset{x\to a}{\lim }{\displaystyle \frac{-\left(g\left(x\right)-g\left(a\right)\right)}{x-a}}\underset{x\to a}{\lim }{\displaystyle \frac{1}{g\left(x\right)g\left(a\right)}}\hfill & \text{law of limits}\hfill \\ \hfill & =-{\displaystyle \frac{g^{\prime }\left(a\right)}{g{\left(a\right)}^2}}\hfill & \hfill \end{array}}$

From this operation, we derive the reciprocal rule.

The Reciprocal Rule:   ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{g\left(x\right)}}=-{\displaystyle \frac{g^{\prime }\left(x\right)}{g{\left(x\right)}^2}}}$

Example 4.29. To apply the reciprocal rule in the function

${\displaystyle f\left(x\right)={\displaystyle \frac{1}{4x^3-\sqrt{x}}}\text{,}}$

we identify $g\left(x\right)$ as $g\left(x\right)=4x^3-\sqrt{x}$.

Since we need $g^{\prime }\left(x\right)$, we differentiate it, using the sum and multiple constant rules:

${\displaystyle g^{\prime }\left(x\right)=12x^2-{\displaystyle \frac{1}{2\sqrt{x}}}\text{.}}$

Therefore,

${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{4x^3-\sqrt{x}}}=-{\displaystyle \frac{12x^2-{\displaystyle \frac{1}{2\sqrt{x}}}}{{\left(4x^3-\sqrt{x}\right)}^2}}\text{.}}$

The slope of the tangent line at the point ${\displaystyle \left(1,f\left(1\right)\right)=\left(1,{\displaystyle \frac{1}{3}}\right)}$ is

${\displaystyle f^{\prime }\left(1\right)=-{\displaystyle \frac{12-{\displaystyle \frac{1}{2}}}{{\left(3\right)}^2}}=-{\displaystyle \frac{23}{18}}\text{,}}$

and the equation of the tangent line is

${\displaystyle y-{\displaystyle \frac{1}{3}}=-{\displaystyle \frac{23}{18}}\left(x-1\right)\text{ or }y=-{\displaystyle \frac{23}{18}}x+{\displaystyle \frac{29}{18}}\text{,}}$

Example 4.30.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{30}{4x^4-2x+6}}& ={\displaystyle \frac{30}{2}}{\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{2x^4-x+3}}\hfill & \text{multiple constant rule}\hfill \\ \hfill & =\left(-15\right){\displaystyle \frac{{\displaystyle \frac{d}{dx}}\left(2x^4-x+3\right)}{{\left(2x^4-x+3\right)}^2}}\hfill & \text{reciprocal rule}\hfill \\ \hfill & =\left(-15\right){\displaystyle \frac{8x^3-1}{{\left(2x^4-x+3\right)}^2}}\hfill & \text{power and constant rules}\hfill \\ \hfill & =-{\displaystyle \frac{-120x^3+15}{{\left(2x^4-x+3\right)}^2}}\text{.}\hfill & \hfill \end{array}}$

The slope of the tangent line at the point $\left(0,f\left(0\right)\right)=\left(0,5\right)$ is

${\displaystyle f^{\prime }\left(0\right)={\displaystyle \frac{15}{9}}={\displaystyle \frac{5}{3}}\text{,}}$

and the equation of the tangent line at this point is

${\displaystyle y-5={\displaystyle \frac{5x}{3}}\text{ or }y={\displaystyle \frac{5x}{3}}+5\text{.}}$

Unfortunately, the derivative of the product of two functions is not the product of the derivatives. You can see why in the proof of the product rule on page 136 of the textbook. The product rule is given below.

The Product Rule:   ${\displaystyle {\displaystyle \frac{d}{dx}}f\left(x\right)g\left(x\right)=f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)}$

Example 4.31. For the derivative of the function $g\left(x\right)={\left(f\left(x\right)\right)}^2=f\left(x\right)f\left(x\right)$, we can use the product rule to determine that

$g^{\prime }\left(x\right)=f\left(x\right)f^{\prime }\left(x\right)+f^{\prime }\left(x\right)f\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)\text{.}$

Moreover, we can use this result to find

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\left(f\left(x\right)\right)}^3& ={\displaystyle \frac{d}{dx}}{\left(f\left(x\right)\right)}^{2}f\left(x\right)\hfill & \hfill \\ \hfill & =2f\left(x\right)f^{\prime }\left(x\right)f\left(x\right)+{\left(f\left(x\right)\right)}^2{f}^{\prime }\left(x\right)\hfill & \text{product rule}\hfill \\ \hfill & =3{\left(f\left(x\right)\right)}^2{f}^{\prime }\left(x\right)\hfill & \hfill \end{array}}$

Example 4.32. Since the derivative of the function $f\left(x\right)=3x^2+5x$ is $f^{\prime }\left(x\right)=6x+5$, we can use Example 4.31 to differentiate the function

$g\left(x\right)={\left(3x^2+5x\right)}^2=f\left(x\right)f\left(x\right)=\left(3x^2+5x\right)\left(3x^2+5x\right)\text{.}$

Hence,

$g^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)=2\left(3x^2+5x\right)\left(6x+5\right)\text{.}$

We can also go further and find the derivative of

$h\left(x\right)={\left(3x^2+5x\right)}^3=g\left(x\right)f\left(x\right)={\left(3x^2+5x\right)}^2\left(3x^2+5x\right)\text{.}$

By Example 4.31,

$h^{\prime }\left(x\right)=3{\left(f\left(x\right)\right)}^2{f}^{\prime }\left(x\right)=3{\left(3x^2+5x\right)}^2\left(6x+5\right)\text{.}$

What do you observe in this example? What do you think the derivative of the function $u\left(x\right)={\left(3x^2+5x\right)}^4$ is?

Observe that the product rule involves the derivatives of both functions. Using the reciprocal and product rules, we can get the rule for the quotient of two functions, as is shown below.

${\displaystyle \begin{array}{lll}\hfill {\displaystyle \frac{d}{dx}}{\displaystyle \frac{f\left(x\right)}{g(x)}}& ={\displaystyle \frac{d}{dx}}f\left(x\right)\left({\displaystyle \frac{1}{g\left(x\right)}}\right)\hfill & \hfill \\ \hfill & =\left({\displaystyle \frac{d}{dx}}f\left(x\right)\right){\displaystyle \frac{1}{g\left(x\right)}}+f\left(x\right){\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{g\left(x\right)}}\hfill & \text{product rule}\hfill \\ \hfill & =f^{\prime }\left(x\right){\displaystyle \frac{1}{g\left(x\right)}}+f\left(x\right)\left[-{\displaystyle \frac{g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}\right]\hfill & \text{reciprocal rule}\hfill \\ \hfill & ={\displaystyle \frac{f^{\prime }\left(x\right)}{g\left(x\right)}}-{\displaystyle \frac{f\left(x\right)g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}\hfill & \hfill \\ \hfill & ={\displaystyle \frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}\hfill & \hfill \end{array}}$

And so we have found the quotient rule.

The Quotient Rule:   ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{f\left(x\right)}{g\left(x\right)}}={\displaystyle \frac{f^{\prime }\left(x\right)g\left(x\right)-f\left(x\right)g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}\text{.}}$

Warning: Observe that in the quotient rule, the numerator must be $f'(x)\,g(x) - f(x)\,g'(x)$. It is a common mistake to change the order to $f(x)\,g'(x) - f'(x)\,g(x)$.

You must now practise so that you can apply these rules without making mistakes. When you start to use these rules, you should do as many steps as necessary, so that you eventually memorize them. See how we differentiate the following functions. We will not simplify the resulting function at this point.

Example 4.33. For the function

${\displaystyle f\left(x\right)={\displaystyle \frac{4\sqrt{x}+6x^4-4x}{5x^3-x}}\text{,}}$

we must apply the quotient rule. Since we need the derivative of the numerator and denominator functions, we obtain these first. By the power rule,

${\displaystyle {\displaystyle \frac{d}{dx}}\left(4\sqrt{x}+6x^4-4x\right)={\displaystyle \frac{2}{\sqrt{x}}}+24x^3-4}$

and

${\displaystyle {\displaystyle \frac{d}{dx}}5x^3-x=15x^2-1\text{.}}$

Now we can apply the quotient rule

${\displaystyle \begin{array}{ll}\hfill {\displaystyle \frac{d}{dx}}& {\displaystyle \frac{4\sqrt{x}+6x^4-4x}{5x^3-x}}\hfill \\ \hfill & ={\displaystyle \frac{\left({\displaystyle \frac{2}{\sqrt{x}}}+24x^3-4\right)\left(5x^3-x\right)-\left(4\sqrt{x}+6x^4-4x\right)\left(15x^2-1\right)}{{\left(5x^3-x\right)}^2}}\text{.}\hfill \end{array}}$

Example 4.34. We could use the quotient rule to differentiate the function

${\displaystyle f\left(x\right)={\displaystyle \frac{3x^4-\sqrt{3x}-5}{x^2}}\text{,}}$

but it is easier if we simplify it first.

Hence,

$f\left(x\right)=3x^2-\sqrt{3}{x}^{-3∕2}-5x^{-2}\text{,}$

and by the power rule,

${\displaystyle f^{\prime }\left(x\right)=6x+{\displaystyle \frac{3\sqrt{3}}{2x^{5∕2}}}+{\displaystyle \frac{10}{x^3}}\text{.}}$