Differentiating simple algebraic expressions
Watch this video to learn about simple differentiation.
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:
\(f(x) = a{x^n} \rightarrow f\textquotesingle(x)= na{x^{n - 1}}\)
Question
Differentiate \(y = {x^5}\)
\(\frac{{dy}}{{dx}} = 5{x^4}\)
Question
Find the derivative of \(f(x) = 4{x^3}\)
\(f\textquotesingle(x)= 12{x^2}\)
When calculating the rate of change or the gradientIn a graph, the gradient is the steepness of the line. The greater the gradient, the greater the rate of change. of a tangentA straight line that just touches a point on a curve. A tangent to a circle is perpendicular to the radius which meets the 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:
\({a^{ - n}} = \frac{1}{{{a^n}}}\)
\({a^{\frac{m}{n}}} = \sqrt[n]{{{a^m}}}\)
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 \(x = 3\),
\(f\textquotesingle(3) = \frac{{ - 8}}{{{3^3}}} = \frac{{ - 8}}{{27}}\)
Question
Find the gradient of the tangent to the curve with equation \(y = 3{x^{\frac{2}{3}}}\) at the point when \(x = 8\).
\(\frac{{dy}}{{dx}}=\frac{{2}}{{3}}\times3{x^{\frac{-1}{3}}} \)
\(= 2{x^{\frac{-1}{3}}}\)
\(= \frac{2}{{{x^{\frac{1}{3}}}}}\)
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
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.
\(\frac{{dy}}{{dx}} = - \frac{1}{2} \times 4{x^{ - \frac{3}{2}}}\)
\(-2x^{\frac{-3}{2}}\)
\(=\frac{-2}{x^{\frac{3}{2}}}\)
\(= \frac{{ - 2}}{{\sqrt[2]{{{x^3}}}}}\)
Question
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.
\(f(x) = \frac{{{x^2}}}{x} + \frac{5}{x}\)
\(f(x) = x + 5{x^{ - 1}}\)
\(f\textquotesingle(x)= {x^{0}} - 5{x^{ - 2}}\)
\(f\textquotesingle(x)= 1 - \frac{5}{{{x^2}}}\)
Question
Find the derivative of \(y = (x + 1)(x - 3)\)
Remove the brackets first, then differentiate.
\(y = {x^2} - 2x - 3\)
\(\frac{{dy}}{{dx}} = 2{x^{1}} - 2{x^{0}}\)
\(\frac{{dy}}{{dx}} = 2x - 2\)
