# 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}