Differentiating simple algebraic expressions

Differentiation is used in maths for calculating rates of change.

For example in mechanics, the rate of change of displacement (with respect to time) is the velocity. The rate of change of velocity (with respect to time) is the acceleration.

The rate of change of a function f(x) with respect to x can be found by finding the derived function f\textquotesingle(x).

For an equation beginning y =, the rate of change can be found by differentiating y with respect to x. In its notation form this is written as \frac{{dy}}{{dx}}. This is also known as 'Leibniz Notation'.

There are many ways a question can ask you to differentiate:

  • Differentiate the function...
  • Find f\textquotesingle(x)
  • Find \frac{{dy}}{{dx}}
  • Calculate the rate of change of...
  • Find the derivative of…
  • Calculate the gradient of the tangent to the curve...

The general rule for differentiation is:

f(x) = a{x^n} \rightarrow f\textquotesingle(x)=  na{x^{n - 1}}

  • In other words, you bring the power down to the front to multiply and subtract 1 from the power.


Differentiate y = {x^5}

\frac{{dy}}{{dx}} = 5{x^4}


Find the derivative of f(x) = 4{x^3}

f\textquotesingle(x)= 12{x^2}

When calculating the rate of change or the gradient of a tangent to a curve, we are required to write the final answer to the differentiated expression without negative or fractional powers. Doing so makes it much easier to evaluate for specific values without a calculator.

To remove negative and fractional powers, we need to recall the laws of indices. The two that will be useful here are:

{a^{ - n}} = \frac{1}{{{a^n}}}

{a^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}


Find the rate of change of f(x) = 4{x^{ - 2}} at x = 3.


Using f(x) = a{x^n} \rightarrow f\textquotesingle(x)=  na{x^{n - 1}}, we find that:

f\textquotesingle(x)= - 8{x^{ - 3}}

This is very difficult to evaluate when x = 3 without a calculator, so we need to use our laws of indices to change this into a positive power.

f\textquotesingle(x)= \frac{{ - 8}}{{{x^3}}}

Now when x = 3,

f\textquotesingle(3) = \frac{{ - 8}}{{{3^3}}} = \frac{{ - 8}}{{27}}


Find the gradient of the tangent to the curve with equation y = 3{x^{\frac{2}{3}}} at the point when x = 8.


= 2{x^{\frac{-1}{3}}}

= \frac{2}{{{x^{\frac{1}{3}}}}}

Now when x = 8,

\frac{{dy}}{{dx}} = \frac{2}{{\sqrt[3]{8}}} = \frac{2}{2} = 1

Therefore the gradient of the tangent to the curve is 1.

The previous examples have very simple expressions. Sometimes we aren't able to differentiate all expressions in their current form as we require the expression to be sums and/or differences of terms of the form a{x^n}.

Before differentiating:

  • Remove brackets
  • Separate 'top heavy' fractions
  • Change terms involving roots into fractional powers
  • Change terms with x on the denominator to negative powers


Differentiate y = \frac{4}{{\sqrt x }}


y = \frac{4}{{{x^{\frac{1}{2}}}}} = 4{x^{ - \frac{1}{2}}}

Now we have it in the correct form we can differentiate.

\frac{{dy}}{{dx}} =  - \frac{1}{2} \times 4{x^{ - \frac{3}{2}}}



= \frac{{ - 2}}{{\sqrt[2]{{{x^3}}}}}


Find the derivative of f(x) = \frac{{{x^2} + 5}}{x}

We need to separate the top heavy fraction first then use our laws of indices to get into the correct form to be able to differentiate.

f(x) = \frac{{{x^2}}}{x} + \frac{5}{x}

f(x) = x + 5{x^{ - 1}}

f\textquotesingle(x)= {x^{0}} - 5{x^{ - 2}}

f\textquotesingle(x)= 1 - \frac{5}{{{x^2}}}


Find the derivative of y = (x + 1)(x - 3)

Remove the brackets first, then differentiate.

y = {x^2} - 2x - 3

\frac{{dy}}{{dx}} = 2{x^{1}} - 2{x^{0}}

\frac{{dy}}{{dx}} = 2x - 2