Differentiating simple algebraic expressions - Differentiation - Higher Maths Revision - BBC Bitesize (2024)

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:

The general rule for differentiation is:

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

When calculating the rate of change or the of a 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:

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

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

Example

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

Solution

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 \(x = 3\),

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

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:

Example

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

Solution

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

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

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

\(-2x^{\frac{-3}{2}}\)

\(=\frac{-2}{x^{\frac{3}{2}}}\)

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