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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:

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: \[\text{if} \quad y=ax^n \quad \text{then} \quad \frac{dy}{dx}=n\times ax^{n-1}\]

Example

Differentiate each of the following:

  1. \(f(x)=x^2\)

  2. \(y=3x^4\)

  3. \(f(x)=-3x^5\)

Solution +

Solution

Using the power rule for differentiation:

  1. The derivative of \(f(x)=x^2\): \[\begin{aligned}f'(x) &=2\times x^{2-1} \\ &= 2x^1 \\ f'(x) &=2x \end{aligned}\]

  2. The derivative of \(y=3x^4\): \[\frac{dy}{dx}= 4\times 3x^{4-1} = 12x^3\]

  3. The derivative of \(f(x)=-3x^5\): \[f'(x)=5\times (-3)x^{5-1} = -15x^4\]

Example

Consider the curve defined by: \[y=2x^2\]

  1. Write down the coordinates of the point along this curve's length with \(x\)-coordinate \(x=2\).

  2. Find the gradient of the curve at this point.

Solution +

Solution

  1. We have the \(x\)-coordinate of the point, \(x=2\). To find the \(y\)-coordinate we replace \(x\) by \(2\) in the curve's equation. That's: \[\begin{aligned} y & =2\times 2^2\\ &= 2 \times 4 \\ y &= 8 \end{aligned}\] So the point along the curve we are interested in is the point \(\begin{pmatrix} 2, 8\end{pmatrix}\).

  2. To find the gradient of the curve at this point we need the drivative function.
    Using the power rule to differentiate, we find the derivative: \[\begin{aligned} \frac{dy}{dx} &= 2\times 2x^{2-1} \\ &= 4x^2 \\ \frac{dy}{dx} &= 4x \end{aligned}\] So at the point \(\begin{pmatrix} 2, 8\end{pmatrix}\), since \(x=2\), the derivative is: \[\frac{dy}{dx} = 4\times 2 = 8\] The gradient of the curve at the point \(\begin{pmatrix} 2, 8\end{pmatrix}\) is therefore \(8\).

Negative Exponents

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

Example

Differentiate each of the following:

  1. \(f(x) = \frac{3}{x^2}\)

  2. \(f(x) = \frac{1}{x}\)

  3. \(y = \frac{5}{x^3}\)

Solution +

Solution

  1. The derivative of \(f(x) = \frac{3}{x^2}\).
    We can start by writing \[f(x) = 3x^{-2}\] Now we use the power rule: \[\begin{aligned} f'(x) &= -2\times 3x^{-2-1} \\ & = -6 x^{-3} \\ f'(x) & = -\frac{6}{x^3} \end{aligned}\]

  2. To differentiate \(f(x) = \frac{1}{x}\) we can start by writing it as a "simple" exponential: \[f(x)=\frac{1}{x} = x^{-1}\] We now use the power rule to differentiate: \[\begin{aligned} f'(x) &= -1\times x^{-1-1} \\ &= -x^{-2} \\ f'(x) &= -\frac{1}{x^2} \end{aligned}\]

  3. Again, start by \(y = \frac{5}{x^3}\) as a simple power of \(x\): \[y = \frac{5}{x^3} = 5x^{-3}\] We can now use the power rule for differentiation: \[\begin{aligned} \frac{dy}{dx} &= -3\times 5x^{-3-1} \\ &= -15 x^{-4} \\ \frac{dy}{dx} &= - \frac{15}{x^4} \end{aligned}\]

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: \[f(x)=a.\sqrt[n]{x^m} \quad \text{and} \quad f(x)=\frac{a}{\sqrt[n]{x^m}}\]

Example

Differentiate each of the following:

  1. \(f(x)=3\sqrt{x}\)

  2. \(y=5\sqrt[3]{x^4}\)

  3. \(f(x)=\frac{4}{\sqrt{x}}\)

Solution +

Solution

Each of these can be differentiated using the power rule.

  1. We start by writing this function in exponential form: \[f(x)=3\sqrt{x} = 3x^{\frac{1}{2}}\] We now use the power rule: \[\begin{aligned} f'(x) &= \frac{1}{2} \times 3x^{\frac{1}{2}-1} \\ & = \frac{3}{2}x^{-\frac{1}{2}} \\ & = \frac{3}{2x^{\frac{1}{2}}} \\ f'(x) &= \frac{3}{2\sqrt{x}} \end{aligned} \]

  2. Again, we start by writing this in exponential form \[y=5\sqrt[3]{x^4} = 5x^{\frac{4}{3}}\] We now use the power rule: \[\begin{aligned} \frac{dy}{dx} &= \frac{4}{3} \times 5x^{\frac{4}{3}-1} \\ &= \frac{20}{3}x^{\frac{1}{3}} \\ \frac{dy}{dx} &= \frac{20}{3}\sqrt[3]x \end{aligned}\]

  3. Once more, we start by writing this function in exponential form: \[f(x)=\frac{4}{\sqrt{x}} = 4x^{-\frac{1}{2}}\] We now use the power rule: \[\begin{aligned} f'(x) &= -\frac{1}{2}\times 4x^{-\frac{1}{2} - 1} \\ &= -\frac{4}{2}x^{-\frac{3}{2}} \\ &= -2x^{-\frac{3}{2}} \\ &= -\frac{2}{x^{\frac{3}{2}}} \\ f'(x) &= -\frac{2}{\sqrt{x^3} } \end{aligned}\]