2.1 Power Rule

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We learn how to find the derivative of a power function.

1 The Power Rule

In this section we present the derivatives power functions. A power function is a function of the form $f(x) = x^\text {n}$ .

Power Rule For any number n, the derivative of the power function $f(x) = x^\text {n}$ is given by $f'(x) = \text {n}x^{\text {n} -1}$ , i.e., $\frac {d}{dx} \left (x^\text {n}\right ) = \text {n}x^{\text {n}-1}.$

We will investigate this rule in the case where n is a whole number. The key is to use the Binomial Theorem, which gives a formula for expanding powers of $x+h$ . Here are some examples of the theorem:

\begin{eqnarray*} (x+h)^{\bf 2} &=&x^2 +{\bf 2}xh + h^2\\ (x+h)^{\bf 3} &=& x^3 +{\bf 3}x^2h + 3xh^2 + h^3\\ (x+h)^{\bf 4} &=& x^4 +{\bf 4}x^3h+ 6x^2h^2 + 4xh^3 + h^4\\ (x+h)^{\bf 5} &=& x^5 +{\bf 5}x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5 \end{eqnarray*}

The coefficients can be determined using a device called Pascal’s Triangle. Notice that the coefficient of the second term on the right hand side always matches the power on the left hand side. In general,

$(x+h)^{\bf \text {n}} = x^\text {n} + {\bf \text {n}}x^{\text {n}-1}h + \dots + \text {n}xh^{\text {n}-1} + h^\text {n}.$

Also observe that as the powers of $x$ decrease, the powers of $h$ increase.

We will now use the definition of the derivative to discover the power rule. Let $f(x) = x^\text {n}$ , where n is a whole number. Then

\begin{eqnarray*} f'(x) &=& \lim _{h\to 0} \frac{f(x+h)-f(x)}{h}\\[10 pt] &=& \lim _{h\to 0} \frac{(x+h)^\text{n}-x^\text{n}}{h}\\[10 pt] &=& \lim _{h\to 0} \frac{\left (x^\text{n} + \text{n}x^{\text{n}-1}h + \dots + \text{n}xh^{\text{n}-1} + h^\text{n}\right )-x^\text{n}}{h}\\[10 pt] &=& \lim _{h\to 0} \frac{\text{n}x^{\text{n}-1}h + \dots + \text{n}xh^{\text{n}-1} + h^\text{n}}{h} \\[10 pt] &=& \lim _{h\to 0} \left (\text{n}x^{\text{n}-1} + \dots + \text{n}xh^{\text{n}-2} + h^{\text{n}-1}\right ) \\[10 pt] &=& \text{n}x^{\text{n}-1}. \end{eqnarray*}

example 1 Find the derivative of $f(x) = x^4$ .
We can use the power rule with $n = 4$ to obtain $f'(x) = 4x^3.$

(problem 1a) Compute $\frac {d}{dx} \left (x^3\right ).$

Use the power rule with $n = 3$

The derivative of $f(x) = x^3$ is $f'(x) = \answer {3x^2}.$

(problem 1b) Compute $\dd {x} \left (x^5\right ).$

Use the power rule with $n = 5$

The derivative of $f(x) = x^5$ is $f'(x) = \answer {5x^4}.$

(problem 1c) Find the slope of the tangent line to the graph of $f(x) = x^6 \;\; \text {at} \;\; x = -2.$

The derivative gives the slope of the tangent line

Use the power rule with $n = 6$

The slope is $\answer {-192}.$

example 2 Find the derivative of $f(x) = x^{100}$ .
We can use the power rule with $n = 100$ to obtain $f'(x) = 100x^{99}.$

(problem 2) Compute $\frac {d}{dx} \left (x^{250}\right ).$

Use the power rule with $n = 250$

The derivative of $x^{250}$ is $\answer {250x^{249}}.$

example 3 Find the derivative of $f(x) = 1$ .
We can rewrite the function as a power function, since $x^0 = 1$ and then we can use the power rule with $n=0$ . We get $f'(x) = 0x^{-1} = 0$ .

We can also arrive at this answer using a geometric understanding of the derivative. The graph of the constant function $f(x) = 1$ is a horizontal line, which has slope 0. Since the derivative is the slope of the tangent line, the derivative is 0.

In general, the graph of a constant function, $f(x) = c$ is a horizontal line, which has slope 0. Hence, the derivative of any constant is 0, as stated in the following proposition.

Constant Functions If $c$ is any constant, then $\frac {d}{dx} \left (c\right ) = 0.$

The graphs of $y = -2$ , $y = 1/2$ and $y = 3$ are shown below. (To see $y=3$ , zoom out by clicking on the minus in the graphing window or by placing the cursor in the graphing window and using the scroll wheel on your mouse). $\graph {y = 3, y = 1/2, y = -2}$

example 4 If $f(x) = 5, -7, \frac 23$ or $4.15$ , then $f'(x) = 0$ because these are all constants. Notice that none of the expressions contain the variable $x$ .

example 5 If $f(x) = -\pi , e^3, \sqrt 2$ or $\ln (8)$ , then $f'(x) = 0$ because these are all constants. Notice that none of the expressions contain the variable $x$ .

(problem 5) Compute $\frac {d}{dx} \left (94.1\right ) = \answer {0}$

$\frac {d}{dx} \left (\pi ^2\right ) = \answer {0}$

$\frac {d}{dt} \left ( \sqrt 5 \right ) = \answer {0}$

example 6 Find the derivative of $f(x) = x$ .
We can use the power rule with $n=1$ to obtain $f'(x) = 1x^0 = 1$ .

This answer can also be arrived at by interpreting the derivative as slope. The graph of the function $f(x) = x$ is a straight line with slope $1$ , so the derivative is $1$ .

(problem 6) $\frac {d}{dt}\left (t\right ) = \answer {1}$

The graph of a function $f(x) = mx$ is a line with slope $m$ , so the derivative of $mx$ is $m$ . This is stated in the proposition below.

Linear Functions

For any number $m$ , $\frac {d}{dx} (mx) = m.$

The graphs of $y = 2x$ , $y = x/2$ and $y = -x$ are shown below. Which one is which? Click on the double arrows in the upper left hand corner of the graphing window to check your answer. $\graph {2x, x/2, -x}$

example 7 If $f(x) = -3x$ then $f'(x) = -3$ .

(problem 7a) If $f(x) = 4x$ then $f'(x) = \answer {4}$

(problem 7b) If $g(t) = -2t$ then $g'(t) = \answer {-2}$

example 8 If $f(x) = \pi x$ then $f'(x) = \pi$ .

(problem 8a) $\frac {d}{dx} \left (ex\right ) = \answer {e}$

(problem 8b) $\frac {d}{dx} \left (x\ln (5)\right ) = \answer {\ln (5)}$

example 9 If $f(x) = \frac {x}{2}$ then $f'(x) = \frac 12$ .

(problem 9a) $\frac {d}{dx} \left (\frac {x}{3}\right ) = \answer {1/3}$

(problem 9b) $\frac {d}{dx} \left (-\frac {2x}{5}\right ) = \answer {-2/5}$

We now consider examples where the exponent $n$ is a fraction. Recall the definition of rational exponents: $x^{\frac {m}{n}} = \sqrt [n]{x^m}.$

example 10 Find the derivative of $f(x) = \sqrt x$ .
We can rewrite $\sqrt x$ as $x^{1/2}$ and use the power rule with $n = 1/2$ to obtain $f'(x) = (1/2)x^{-1/2}$ . This can then be rewritten in radical form : $(1/2)x^{-1/2} = \frac {1}{2}\cdot \frac {1}{x^{1/2}} = \frac {1}{2\sqrt x}.$ This result is used frequently, so it is best to remember it as $\frac {d}{dx}\left (\sqrt x\right ) = \frac {1}{2\sqrt x}.$

example 11 Find the derivative of $f(x) = \sqrt [3] x$ .
We can rewrite $\sqrt [3] x$ as $x^{1/3}$ and use the power rule with $n = 1/3$ to obtain $f'(x) = (1/3)x^{-2/3}$ . This can be rewritten in radical form as $(1/3)x^{-2/3} = \frac {1}{3}\cdot \frac {1}{x^{2/3}} = \frac {1}{3\sqrt [3] {x^2}}.$

(problem 11a) Compute $\frac {d}{dx} \left (\sqrt [4] x\right ).$

Rewrite $\sqrt [4] x$ as $x^{1/4}$

Use the power rule with $n = 1/4$

The derivative of $\sqrt [4] x$ is $\answer {1/4 x^{-3/4}}.$

(problem 11b) Find the slope of the tangent line to the graph of $f(x) = \sqrt [5] x \;\; \text {at} \;\; x = -32.$

The derivative gives the slope of the tangent line

Rewrite $\sqrt [5] x$ as $x^{1/5}$

Use the power rule with $n = 1/5$

The slope is $\answer {1/80}.$

example 12 Find the derivative of $f(x) = x^2\sqrt [4] {x^3}$ .
We first rewrite $f(x)$ as $x^2 \cdot x^{3/4} = x^{11/4}$ (adding exponents) and then we use the power rule with $n = 11/4$ to obtain $f'(x) = (11/4)x^{7/4}$ . This can be rewritten in radical form as $(11/4)x^{7/4} = \frac {11}{4} \sqrt [4] {x^7} = \frac {11}{4} x\sqrt [4] {x^3} .$

(problem 12) Compute $\frac {d}{dx} \left (x\sqrt [3] x\right ).$

Rewrite $x\sqrt [3] x$ as $x^{4/3}$

Use the power rule with $n = 4/3$

The derivative of $x\sqrt [3] x$ is $\answer {(4/3) x^{1/3}}.$

Next, we look at some examples involving negative exponents. Recall the definition of negative exponents: $x^{-n} = \frac {1}{x^n}.$

example 13 Find the derivative of $f(x) = \dfrac {1}{x}$ .
We can rewrite $1/x$ as $x^{-1}$ and use the power rule with $n = -1$ to obtain We get $f'(x) = -1x^{-2}$ . This can be rewritten without the negative exponent as $-1x^{-2} = -\frac {1}{x^2}.$

example 14 Find the derivative of $f(x) = \dfrac {1}{x^3}$ .
We can rewrite $1/x^3$ as $x^{-3}$ and use the power rule with $n = -3$ to obtain $f'(x) = -3x^{-4}$ . This can be rewritten without the negative exponent as $-3x^{-4} = -\frac {3}{x^4}.$

(problem 14a) Compute $\frac {d}{dx} \left (\frac {1}{x^2}\right ).$

Rewrite $\dfrac {1}{x^2}$ as $x^{-2}$

Use the power rule with $n = -2$

The derivative of $f(x) = \dfrac {1}{x^2}$ is $f'(x) = \answer {-2x^{-3}}.$

(problem 14b) Find the slope of the tangent line to the graph of $f(x) = \frac {1}{x^3} \;\; \text {at} \;\; x = 3.$

The derivative gives the slope of the tangent line

Rewrite $\dfrac {1}{x^3}$ as $x^{-3}$

Use the power rule with $n = -3$

The slope is $\answer {-1/27}.$

example 15 Find the derivative of $f(x) = \dfrac {1}{\sqrt x}$ .
We can rewrite $\frac {1}{\sqrt x}$ as $x^{-1/2}$ and use the power rule with $n = -1/2$ to obtain $f'(x) = -\frac 12 x^{-3/2}.$ This can be rewritten without the negative exponent as $(-1/2)x^{-3/2}= -\frac {1}{2} \cdot \frac {1}{x^{3/2}} = -\frac {1}{2\sqrt {x^3}} = -\frac {1}{2x\sqrt x}.$

(problem 15) Compute $\frac {d}{dx} \left (\frac {1}{\sqrt [5] x}\right ).$

Rewrite $1/\sqrt [5] x$ as $x^{-1/5}$

Use the power rule with $n = -1/5$

The derivative of $f(x) = \frac {1}{\sqrt [5] x}$ is $f'(x) = \answer {(-1/5) x^{-6/5}}.$

example 16 Find the derivative of $f(x) = \dfrac {\sqrt [3] {x^2}}{\sqrt [4] x}.$
We can rewrite $f(x)$ as $\frac {x^{2/3}}{x^{1/4}} = x^{5/12}$ (subtracting exponents). Now, we use the power rule with $n = \frac {5}{12}$ to obtain $f'(x) = \tfrac {5}{12}x^{-7/12}.$ This can be rewritten without the negative exponent as $\tfrac {5}{12}x^{-7/12} = \tfrac {5}{12} \cdot \frac {1}{x^{7/12}} = \frac {5}{12\sqrt [12]{x^7}}.$

(problem 16) Compute $\frac {d}{dx} \left (\frac {\sqrt [4] {x^3}}{\sqrt [3] x}\right ).$

Rewrite $\frac {\sqrt [4] {x^3}}{\sqrt [3] x}$ as $\frac {x^{3/4}}{x^{1/3}} = x^{5/12}$

Use the power rule with $n = 5/12$

The derivative of $\frac {\sqrt [4] {x^3}}{\sqrt [3] x}$ is $\answer {(5/12) x^{-7/12}}.$

Below is a graph of $f(x) = x^2$ (in blue) and its derivative, $f'(x) = 2x$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ .

Below is a graph of $f(x) = x^3$ (in blue) and its derivative, $f'(x) = 3x^2$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ .

Below is a graph of $f(x) = \sqrt [3] x$ (in blue) and its derivative, $f'(x) = \frac {1}{3\sqrt [3]x^2}$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ .

2024-09-27 13:53:49