AS Maths Differentiation Techniques | 9 Amazing Questions
Differentiation Techniques – Introduction
This is the third article in this particular section where the focus is on differentiation techniques in particular for students who are in year 12 or are working towards their AS Level Maths exam.
It has been mentioned that the whole topic of differentiation at A Level is quite vast and if you are doing differentiation at year 12 level, then you will know that it interlinks into many other topic areas such as sketching curves, polynomials and also straight line geometry where you can be asked to determine the equation of a tangent and normal.
As well as this there is the area of increasing and decreasing functions as well being able to determine the nature of a turning point. All these skills are essential for both AS and A Level.
For this article we will consider a mix of 9 questions in total where the approach will be explained and a detailed solution will be provided. Quite often you are having to use the rules of indices in the question in order to obtain an expression that can be differentiated.
Differentiation Techniques - Questions 1 - 3.
Q1
For this question it is a case of using the rules of indices to rewrite the second term of the expression. Once this has been done it should not cause any problems but do be careful with the arithmetic.
\begin{aligned} y=2 x^3+\frac{3}{x^2} & \Rightarrow y=2 x^3+3 x^{-2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \times 3 x^2+3 \times\left(-2 x^{-3}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^2-6 x^{-3} . \end{aligned}Q2.
This initial part of the question is asking you to write the expression in index form. Simply expand the numerator and divide each term by the denominator as shown here:
\begin{aligned} f(x) & =\frac{(3-4 \sqrt{x})^2}{\sqrt{x}} \\ & =\frac{9-24 x^{\frac{1}{2}}+16 x}{x^{\frac{1}{2}}} \\ & =9 x^{-\frac{1}{2}}+16 x^{\frac{1}{2}}-24 . \end{aligned}b) \text { Find } \mathrm{f}^{\prime}(x) \text {. }
For the second part of the question you are now asked to differentiate. Note the use of the different notation that has been used.
\begin{aligned} f(x)=9 x^{-\frac{1}{2}}+16 x^{\frac{1}{2}}-24 & \Rightarrow \mathrm{f}^{\prime}(x)=9 \times\left(-\frac{1}{2} x^{-\frac{3}{2}}\right)+16 \times \frac{1}{2} x^{-\frac{1}{2}}+0 \\ & \Rightarrow \mathrm{f}^{\prime}(x)=-\frac{9}{2} x^{-\frac{3}{2}}+8 x^{-\frac{1}{2}} . \end{aligned}Q3
This is a standard question requiring standard use of differentiation techniques and the solution is shown here:
\begin{aligned} y=x^4+x^{\frac{1}{3}}+3 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 x^3+\frac{1}{3} x^{-\frac{2}{3}}+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 x^3+\frac{1}{3} x^{-\frac{2}{3}} . \end{aligned}Differentiation Techniques - Questions 4 - 6.
Q4
For this question the expression given is not in a form that can be differentiated. First the numerator needs to be expanded and then it is necessary to divide by the denominator. Then you are in a position to find the derivative.
The solution is shown here:
\begin{aligned} y=\frac{(x+3)(x-8)}{x} & \Rightarrow y=\frac{x^2-5 x-24}{x} \\ & \Rightarrow y=x-5-24 x^{-1} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=1+0-24 \times\left(-x^{-2}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=1+24 x^{-2} . \end{aligned}Q5.
It is important to be comfortable with the use of the rules of indices. Sometimes a question will ask you to do this as we have seen but here you are directly asked to find the first derivative.
Again using the rules of indices and then differentiating gives the following result:
\begin{aligned} & y=8 x^3-4 \sqrt{x}+\frac{3 x^2+2}{x} \\ \Rightarrow & y=8 x^3-4 x^{\frac{1}{2}}+3 x+2 x^{-1} \\ \Rightarrow & \frac{\mathrm{d} y}{\mathrm{~d} x}=8 \times 3 x^2-4 \times \frac{1}{2} x^{-\frac{1}{2}}+3+2 \times\left(-x^{-2}\right) \\ \Rightarrow & \frac{\mathrm{d} y}{\mathrm{~d} x}=24 x^2-2 x^{-\frac{1}{2}}+3-2 x^{-2} . \end{aligned}Q5.
Here there is an example of the use of expanding triple brackets. Take care when performing any expansions especially if there are any negative signs.
The expansion and the process for finding the derivative is shown below:
\begin{aligned} &y=(x+1)(x+3)^2=(x+1)\left(x^2+6 x+9\right)\\ &\begin{array}{c|ccc} & x^2 & +6 x & +9 \\ \hline x & x^3 & +6 x^2 & +9 x \\ +1 & +x^2 & +6 x & +9 \end{array} \end{aligned}\begin{aligned} y=x^3+7 x^2+15 x+9 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+7 \times 2 x+15 \times 1+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+14 x+15 . \end{aligned}
Q6.
Here is an example of finding the value of the gradient at a particular point. It is important to note that when you find the value of a gradient at a particular point that is also the gradient of the tangent at that particular point.
\begin{aligned} & y=x^3+4 x+1 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2+4 \times 1+0=3 x^2+4 \\ & x=3 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \times 3^2+4=3 \times 9+4=27+4=\underline{\underline{31}} . \end{aligned}
Differentiation Techniques – Questions 7 – 9.
Here you simply need to expand taking care with any negative signs.
Once this is done it is then possible to differentiate as shown below:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x}\left[(1-2 x)^2\right] & =\frac{\mathrm{d}}{\mathrm{d} x}\left(1-4 x+4 x^2\right) \\ & =0-4 \times 1+4 \times 2 x \\ & =-4+8 x . \end{aligned}Part b) \text { (b) } \frac{x^5+6 \sqrt{x}}{2 x^2} \text {. }
Here you need to rewrite this expression using the rules of indices as shown here before performing the actual differentiation:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{x^5+6 \sqrt{x}}{2 x^2}\right] & =\frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{x^5+6 x^{\frac{1}{2}}}{2 x^2}\right] \\ & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{2} x^3+3 x^{-\frac{3}{2}}\right) \\ & =\frac{1}{2} \times 3 x^2+3 \times\left(-\frac{3}{2} x^{-\frac{5}{2}}\right) \\ & =\frac{3}{2} x^2-\frac{9}{2} x^{-\frac{5}{2}} . \end{aligned}Q8.
This is a standard question using rules of indices prior to performing any differentiation.
\begin{aligned} y=3 x^2+6 x^{\frac{1}{3}}+\frac{2 x^3-7}{3 \sqrt{x}} & \Rightarrow y=3 x^2+6 x^{\frac{1}{3}}+\frac{2 x^3-7}{3 x^{\frac{1}{2}}} \\ & \Rightarrow y=3 x^2+6 x^{\frac{1}{3}}+\frac{2}{3} x^{\frac{5}{2}}-\frac{7}{3} x^{-\frac{1}{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \times 2 x+6 \times \frac{1}{3} x^{-\frac{2}{3}}+\frac{2}{3} \times \frac{5}{2} x^{\frac{3}{2}}-\frac{7}{3} \times\left(-\frac{1}{2} x^{-\frac{3}{2}}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x+2 x^{-\frac{2}{3}}+\frac{5}{3} x^{\frac{3}{2}}+\frac{7}{6} x^{-\frac{3}{2}} . \end{aligned}Q9.
Here you are asked to find the value of the gradient at a specific point and to leave your answer in a particular form.
The solution is used here using the rules of indices and suitable differentiation techniques:
\begin{aligned} y=\sqrt{x}+\frac{4}{\sqrt{x}}+4 & \Rightarrow y=x^{\frac{1}{2}}+4 x^{-\frac{1}{2}}+4 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2} x^{-\frac{1}{2}}+4 \times\left(-\frac{1}{2} x^{-\frac{3}{2}}\right)+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2} x^{-\frac{1}{2}}-2 x^{-\frac{3}{2}} \end{aligned}\begin{aligned} x=8 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2} \times 8^{-\frac{1}{2}}-2 \times 8^{-\frac{3}{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 \times 8^{\frac{1}{2}}}-\frac{2}{8^{\frac{3}{2}}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{2 \sqrt{8}}-\frac{2}{8 \sqrt{8}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4}{16 \sqrt{2}}-\frac{2}{16 \sqrt{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2}{16 \sqrt{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{8 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1}{16} \sqrt{2} . \end{aligned}You would probably find that you would not be able to obtain the value of “a” if you just used your calculator. It is important therefore you are not over-reliant on the use of your calculator to perform such calculations.
The maths for the last question is not overly difficult but it shows that you need to be confident in a number of areas such as that of surds, indices and also differentiation. Practising questions like these are very important if you are looking to achieve a high grade at AS and A level.