# Differentiation of Standard Functions – A Level Maths

Recognise a positive rate of change in your learning capacity by tapping into the expert-mined nuggets of knowledge found in this post on differentiation of standard functions.

Here, we’ll take you through each element of the topic and put the theory into practice with worked examples that highlight the methodology behind the maths (madness!).

Are you ready? Of course you are…

### Differentation of Standard Functions – What is Differentiation?

Differentiation is a form of calculus, and is one of the bigger new mathematical topics to appear at A Level. In this text, we’ll help to explain how to differentiate functions that include an $x^n$ term, from simple ones, like $3x^5$, to the more complex, like $\frac{2x^3\,-\,3}{10x^6\,-\,15x^2}$.

If you want to practise some of the earlier skills needed for this topic, have a look at this prior knowledge test. To download the explanation below (as a PDF or PowerPoint) along with questions and answers, look here, or for a progression pack filled with questions with increasing difficulty, look here.

The derivative of a function describes the rate of change of the graph. In other words, it is a function that describes the gradient for different values of 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 $\frac{dy}{dx}$ (said d-y-by d-x). This is the notation use 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 $f'(x)$ (said f-prime-x or f-dash-x).

We can differentiate from first principles but, even for simple functions, that can take a long time. There are a number of rules that we can use to 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 original power of x. In general, we can say that:

If:

$f(x) = y = x^a$ (where $a$ is a real number)

Then:

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

Consider $x^5$: to differentiate this, we multiply by the power then subtract 1 from the power:

$\frac{d}{dx}(x^5) = 5x^{5\,-\,1}=5x^4$

This version ($\frac{d}{dx}(x^5)$) of Leibniz’s notation can be read as “differentiate $x^5$ with respect to $x$“.

“With respect to $x$” simply means that $x$ is the variable we are considering, in this case.

The same process applies if the term has a coefficient (other than 1), such as $2x^3$:

$\frac{d}{dx}(2x^3) = 2(3x^{3\,-\,1})=6x^2$

When we have an expression or equation that is the sum (or difference) of two or more terms, we simply differentiate each term separately:

Example Question 1
Find the derivative of $\boldsymbol{y = x^5 - 2x^3 + 1}$

We have already looked at how to differentiate the first two terms:

$\frac{d}{dx}(x^5) = 5x^4$

$\frac{d}{dx}(2x^3) = 6x^2$

For the final term, remember $x^0 = 1$ for any value of $x$. When we multiply the constant term by the power, we are multiplying by 0, so the result is 0.

$\frac{d}{dx}(1) = 0$

So, if $y = x^5 - 2x^3 + 1$ then $\boldsymbol{\frac{dy}{dx} = 5x^4 - 6x^2}$

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

Example Question 2
Find the derivative of the function $\boldsymbol{f(x)=(3x+2)(x+5)}$

First, expand the brackets:

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

Now, multiply each term by its power then subtract one from that power:

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

$\boldsymbol{f'(x) = 6x + 17}$

Example Question 3
Find the derivative of the function $\boldsymbol{f(x)=\frac{2x^2\,+\,5x\,+\,3}{2x\,+\,3}}$ (where x ≠ -1.5)

This time, we will start by factorising the numerator:

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

Now, simplify the expression:

$f(x) = x + 1$

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

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

$\boldsymbol{f'(x) = 1}$

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

Example Question 4
Find the derivative of the function $\boldsymbol{y=\frac{10}{\sqrt{x}}}$ (where x ≠ -0)

We need to write this function in index form. Remember, $\frac{1}{x} = x^{-1}$ and $\sqrt{x} = x^{\frac{1}{2}}$

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

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

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

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

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

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.

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