Differentiating Functions of the Form x^n – A Level Maths Revision

Differentiating Functions of the Form x^n

Differentiation is a major new topic at A Level. While you need to know how to differentiate a function from first principles, that is a slow process so it is useful to know some shortcuts. The following guide explains how to quickly differentiate functions of the form y = xn, for example:

y = 2x^5 \\ y = 3x^2 + 4x^5 - \frac{x^2}{2} \\ y = 3 \sqrt{x} \\ y = \frac{5}{x^2}

This guide is also available in PDF or presentable formats, here, along with multiple-choice prior knowledge questions to get you ready for the topic. If you’re comfortable with the topic, try this differentiation puzzle or these progression questions, which get harder as you go on.

The derivative of a function describes the rate of change of a graph. In other words, it is a function that describes how the gradient changes with x – you might also see it referred to as the gradient function of the graph.

The process of finding the derivative of a function is called differentiation. Differentiation and
its inverse process, integration, are collectively known as calculus.

There are two common versions of notation for differentiation. The first is written as:


This is said d-y-by-d-x and means ‘differentiate y with respect to x’, or differentiate the function y = … by taking into account the variable x. This is the notation developed by Gottfried Leibniz, who (along with Isaac Newton) is credited with the discovery of calculus. Consequently, this is known as Leibniz’s notation.

The second was made popular by Joseph Legrange and is known as Legrange’s notation. The derivative is written as:


This is said f-prime-x or f-dash-x.

You should be able to differentiate from first principles but, even for simple functions, this can take a long time. There are a number of rules that can speed up this process.

The following table shows various functions and their derivatives:


In each of these examples, the power of x in the derivative is 1 less than the power of x in the original function (remember: x = x1 and 1 = x0) and the coefficient of the derivative is the power of x in the original function. In general:

\boldsymbol{f(x) = y =  x^a}

\boldsymbol{f'(x) = \frac{dy}{dx} = ax^{a\,-\,1}}

For example, consider x5; to differentiate this, we multiply by the power then subtract 1 from the power:


This version of Leibniz’s notation can be read as “differentiate x5 with respect to x”.

The same process applies if the term has a coefficient, such as 2x3:


If an expression or equation is the sum (or difference) of 2 or more terms, simply differentiate each term separately.

Example Question 1

Find the derivative of y = x5 – 2x3 + 1

You have already seen how to differentiate the first two terms:



For the final term, remember x0 = 1. When you multiply by the power, you are multiplying by 0 so the result is 0.


So, for y = x5 – 2x3 + 1:


While it is possible to differentiate expressions where expressions are multiplied or divided, it is more complicated. Where possible, simplify to make them easier to differentiate.

Example Question 2

Find the derivative of the function f(x) = (3x + 2)(x + 5)

First, expand the brackets:

f(x) = 3x^2 + 17x + 10

Next, multiply by the power then subtract one from the power:

f'(x) = 3(2x^{2\,-\,1}) + 17(x^{1\,-\,1})+10(0)

f'(x) = 6x + 17

Example Question 3

Find the derivative of the function:
f(x) = \frac{2x^2\,+\,5x\,+\,3}{2x\,+\,3}
(where x ≠1.5)

This time, start by factorising the numerator:

f(x) = \frac{(x\,+\,1)(2x\,+\,3)}{2x\,+\,3}

Now, simplify the expression:

f(x) = x + 1

Now, differentiate each term as normal – multiply by the power then subtract one from the power:

f'(x) = x^{1\,-\,1} + 0

f(x) = 1

In some situations, fractional and negative indices can be used to make a function easier to simplify:

Example Question 4

Find the derivative of the function:
y = \frac{10}{\sqrt{x}}
(where x ≠ 0)

Start by writing this in index form. Remember:

\frac{1}{x} = x^{-1}
\sqrt{x} = x^{\frac{1}{2}}

This gives:

y = \frac{10}{\sqrt{x}} = 10x^{-\frac{1}{2}}

Now, differentiate as before. Be particularly careful with negative signs:

\frac{dy}{dx} = 10(-\frac{1}{2}x^{-\frac{1}{2}\,-\,1})

\frac{dy}{dx} = -5x^{-\frac{3}{2}}

\frac{dy}{dx} = -\frac{5}{\sqrt{x^3}}

Differentiation is used frequently in a number of fields, including science, engineering and economics. It will also come up in the context of mechanics as part of the A level course. As such, it is important to remember that the derivative and second derivative are not abstract concepts, but refer to the rate of change of one variable with respect to another. Here are some real-life examples:

  • The derivative of displacement with respect to time is velocity.
  • The derivative of velocity with respect to time is acceleration.
  • The derivative of electrical charge with respect to time is current.
  • The derivative of population size with respect to time is population growth.
  • The derivative of cost with respect to items produced is marginal cost.

Differentiating Functions of the Form xn
Practice Questions

Test out what you’ve learnt about differentiating functions of the form xn with these exam-style questions.

Find the derivative of the following functions, with respect to x:

1) y = 4x^3

2) y = 6x^5

3) y = \frac{3x^8}{4}

4) y = \frac{2}{x^3}

5) y = 5\sqrt{x}

6) y = \sqrt[3]{8x}

7) y = 6x^5 - 4x^{-3} + 2x^{-2}

8) y = 3(2x - 3)^2

9) y = \frac{6x^2\,-\,x\,-\,2}{2x\,+\,1}

10) y = \frac{(3x\,+\,1)(x\,+\,2)^2}{(3x^2\,+\,7x\,+\,2)}

\text{1) } \frac{dy}{dx} = 4(3x^{3-1}) = 12x^2

\text{2) } \frac{dy}{dx} = 6(5x^{5-1}) = 30x^4

\text{3) } \frac{dy}{dx} = \frac{3}{4}(8x^{8-1}) = 6x^7

\text{4) } y = 2x^{-3}; \frac{dy}{dx} = 2(-3x^{-3-1}) = -6x^{-4}

\text{5) } y = 5x^{\frac{1}{2}}; \frac{dy}{dx} = 5(\frac{1}{2}x^{\frac{1}{2}\,-\,1}) = \frac{5}{2}x^{-\frac{1}{2}} = \frac{5}{2\sqrt{x}}

\text{6) } y = 2x^{\frac{1}{3}}; \frac{dy}{dx} = 2(\frac{1}{3}x^{\frac{1}{3}\,-\,1}) = \frac{2}{3}x^{-\frac{2}{3}} = \frac{2}{3\sqrt[3]{x^2}}

\text{7) } \begin{aligned} \frac{dy}{dx} &= 6(5x^{5-1}) - 4(-3x^{-3-1}) + 2(-2x^{-2-1}) \\ &= 30x^4 +12x^{-4} - 4x^{-3} \end{aligned}

\text{8) } \begin{aligned} y &= 3(4x^2 - 12x + 9) = 12x^2 - 36x + 27 \\ \frac{dy}{dx} &= 12(2x^{2-1} - 36x^{1-1} + 27(0) = 24x - 36 \end{aligned}

\text{9) } \begin{aligned} y &= \frac{(3x\,-\,2)(2x\,+\,1)}{2x\,+\,1} = 3x - 2\\ \frac{dy}{dx} &= 3(x^{1-1}) -2(0) = 3 \end{aligned}

\text{10) } \begin{aligned} y &= \frac{(3x\,+\,1)(x\,+\,2)^2}{(3x\,+\,1)(x\,+\,2)} = x + 2\\ \frac{dy}{dx} &= x^{1-1} + 2(0) = 1 \end{aligned}

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