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Power Rule for Differentiation

In this section we learn how to differentiate, find the derivative of, any power of \(x\).
That's any function that can be written: \[f(x)=ax^n\] We'll see that any function that can be written as a power of \(x\) can be differentiated using the power rule for differentiation.
In particular we learn how to differentiate when:

  • the power is a positive integer like \(f(x) = 3x^5\).
  • the power is a negative number, this means that the function will have a "simple" power of \(x\) on the denominator like \(f(x) = \frac{2}{x^7}\).
  • the power is a fraction, this means that the function will have an \(x\) under a root like \(f(x) = 5\sqrt{x}\).
We start by learning the formula for the power rule.

Power Rule

Given a function which is a power of \(x\), \(f(x)=ax^n\), its derivative can be calculated with the power rule: \[\text{if} \quad f(x)=ax^n \quad \text{then} \quad f'(x)=n\times ax^{n-1}\] We can also write this: \[\text{if} \quad y=ax^n \quad \text{then} \quad \frac{dy}{dx}=n\times ax^{n-1}\]

Tutorial 1: Power Rule for Differentiation

In the following tutorial we illustrate how the power rule can be used to find the derivative function (gradient function) of a function that can be written \(f(x)=ax^n\), when \(n\) is a positive integer.

Example 1

Find the derivative of the function defined by: \[f(x) = 2x^4\]

Detailed Solution

Comparing the function \(f(x) = 2x^4\) to the generic "power function" \(f(x) = ax^n\), we can see that: \[a = 2 \quad \text{and} \quad n = 4\] The power rule for differentiation: \[f'(x) = n\times ax^{n-1}\] therfore leads to: \[\begin{aligned} f'(x) & =4\times 2x^{3-1} f'(x) & = 8x^2 \end{aligned}\] The derivative is therefore: \[f'(x) = 8x^2\]

Exercise 1

Use the power rule for differentiation to find the derivative function of each of the following:

  1. \(f(x) = 6x^3\)
  2. \(y = x^4\)
  3. \(f(x) = -2x^6\)
  4. \(y = \frac{x^2}{2}\)
  5. \(f(x) = 3x\)
  6. \(y = \frac{2}{5}x^{10}\)
  7. \(f(x) = -6x^3\)
  8. \(y = 4x^4\)

Answers Without Working

  1. For \(f(x) = 6x^3\) we find: \[f'(x) = 18x^2\]
  2. For \(y = x^4\) we find: \[\frac{dy}{dx} = 4x^3\]
  3. For \(f(x) = -2x^6\) we find: \[f'(x) = -12x^5\]
  4. For \(y = \frac{x^2}{2}\) we find: \[\frac{dy}{dx} = x\]
  5. For \(f(x) = 3x\) we find: \[f'(x) = 3\]
  6. For \(y = \frac{2}{5}x^{10}\) we find: \[\frac{dy}{dx} = 4x^9\]
  7. For \(f(x) = -6x^3\) we find: \[f'(x) = -18x^2\]
  8. For \(y = 4x^4\) we find: \[\frac{dy}{dx}= 16x^3 \]

Negative Exponents

The power rule also works for negative exponents.
Remember that: \[\frac{a}{x^m} = ax^{-m}\] this allows us to use the power rule to differentiate any function that can be written: \[f(x)=\frac{a}{x^m}\]

Tutorial 2: Negative Exponents

In the following tutorial we illustrate how the power rule can be used to find the derivative function (gradient function) of a function that can be written \(f(x)=\frac{a}{x^m}\), when \(m\) is a positive integer.

We use the fact that \(\frac{a}{x^m} = a.x^{-m}\) to then use the power rule.

Exercise 2

Differentiate each of the following:

  1. \(f(x) = \frac{3}{x^2}\)
  2. \(f(x) = \frac{1}{x}\)
  3. \(y = \frac{5}{x^3}\)
  4. \(f(x) = -\frac{5}{x^3}\)
  5. \(y = \frac{6}{x^4}\)
  6. \(f(x) = - \frac{2}{x}\)
  7. \(y = \frac{3}{4x^2}\)
  8. \(f(x) = -\frac{2}{3x^3}\)

Solution

  1. For \(f(x) = \frac{3}{x^2}\), we find: \[f'(x) = -6.x^{-3}\] Which we can also write: \[f'(x) = -\frac{6}{x^3}\]
  2. For \(f(x) = \frac{1}{x}\) we find: \[f'(x) = -1.x^{-2}\] Which can/should be written: \[f'(x) = -\frac{1}{x^2}\]
  3. For \(y = \frac{5}{x^3}\) we find: \[\frac{dy}{dx} = -15.x^{-4}\] which can/should be written: \[\frac{dy}{dx} = - \frac{15}{x^4}\]
  4. For \(f(x) = -\frac{5}{x^3}\) we find: \[f'(x) = 15.x^{-4}\] which can/should be written: \[f'(x) = \frac{15}{x^4}\]
  5. For \(y = \frac{6}{x^4}\) we find: \[\frac{dy}{dx} = -24.x^{-4}\] which can/should be written: \[\frac{dy}{dx} = -\frac{24}{x^5}\]
  6. For \(f(x) = - \frac{2}{x}\) we find: \[f'(x) = 2.x^{-2}\] which can/should be written: \[f'(x) = \frac{2}{x^2}\]
  7. For \(y = \frac{3}{4x^2}\) we find: \[\frac{dy}{dx} = -\frac{3}{2}.x^{-3}\] which can/should be written: \[\frac{dy}{dx} = -\frac{3}{2x^3}\]
  8. For \(f(x) = -\frac{2}{3x^3}\) we find: \[f'(x) = 2.x^{-4}\] which can/should be written: \[f'(x) = \frac{2}{x^4}\]

Fractional Exponents

The power rule for differentiation also works for any fraction.

Remembering that: \[\sqrt[n]{x^m}=x^{\frac{m}{n}} \quad \text{and} \quad \frac{1}{\sqrt[n]{x^m}} = x^{-\frac{m}{n}}\] we can differentiate any function that can be written: \[f(x)=a.\sqrt[n]{x^m} \quad \text{and} \quad f(x)=\frac{a}{\sqrt[n]{x^m}}\]

Exercise 3

Differentiate each of the following:

  1. \(f(x)=3.\sqrt{x}\)
  2. \(y=5.\sqrt[3]{x^4}\)
  3. \(y = 12.\sqrt[6]{x}\)
  4. \(f(x) = 10.\sqrt[5]{x^3}\)
  5. \(f(x) = -4.\sqrt{x^5}\)
  6. \(y = 9.\sqrt[3]{x^2}\)
  7. \(f(x)=\frac{4}{\sqrt{x}}\)
  8. \(y = - \frac{8}{\sqrt{x^5}}\)

Solution

Each of these can be differentiated using the power rule.

  1. For \(f(x)=3\sqrt{x}\) we find: \[f'(x) = \frac{3}{2}.x^{-\frac{1}{2}}\] which we can write: \[f'(x) = \frac{3}{2\sqrt{x}}\]
  2. For \(y=5\sqrt[3]{x^4}\) we find: \[\frac{dy}{dx} = \frac{20}{3}x^{\frac{1}{3}}\] which we can write: \[\frac{dy}{dx} = \frac{20}{3}\sqrt[3]{x}\]
  3. For \(f(x)=\frac{4}{\sqrt{x}}\) we find: \[f'(x) = -2.x^{-\frac{3}{2}}\] which we can write: \[f'(x) = -\frac{2}{\sqrt{x^3} }\]
  4. For \(f(x) = 10.\sqrt[5]{x^3}\) we find: \[f'(x) = 6.x^{-\frac{2}{5}}\] which we can write: \[f'(x) = \frac{6}{\sqrt[5]{x^2}}\]
  5. For \(f(x) = -4.\sqrt{x^5}\) we find: \[f'(x) = -10.x^{\frac{3}{2}}\] which we can also write: \[f'(x) = - 10 \sqrt{x^3}\]
  6. For \(y = 9.\sqrt[3]{x^2}\) we find: \[\frac{dy}{dx} = 6.x^{-\frac{1}{3}}\] which we can also write: \[\frac{dy}{dx} = \frac{6}{\sqrt[3]{x}}\]
  7. For \(f(x)=\frac{4}{\sqrt{x}}\) we find: \[f'(x) = -2.x^{-\frac{3}{2}}\] which we can also write: \[f'(x) = - \frac{2}{\sqrt{x^3}}\]
  8. For \(y = - \frac{8}{\sqrt{x^5}}\) we find: \[\frac{dy}{dx} = 20.x^{-\frac{7}{2}}\] which we can also write: \[\frac{dy}{dx} = \frac{20}{\sqrt{x^7}}\]