Differentiation For AS Maths
Differentiation For AS Maths – Introduction
In this article we are going to consider 7 questions and look at their solutions where the focus topic is differentiation for AS Maths. Differentiation as a topic is introduced initially in GCSE Further Maths, but not all students cover this particular topic and there is no requirement for this in order to pursue A Level Maths.
Therefore in most cases, the majority of students will meet differentiation for the very first time when they are in year 12 in their chosen school or college.
The topic of differentiation is a vast topic and it is extended in year 13. There are many different techniques that you will learn and there is quite often interlinking with other topics such as polynomials and geometry. So it is very important that you are comfortable and confident in dealing with the topic of differentiation.
Differentiation For AS Maths – Vocabulary
There are a number of terms that you need to recognise that mean differentiation. These include:
- The first derivative
- The derivative
- The second derivative
- Gradient function
- Turning points and/or stationary points
- Rate of change of y with respect to x
Differentiation For AS Maths – How to to start
One thing that is important when you are faced with a question that involves differentiation is that you have it in the correct format. Which is:
y=x^nOnce your expression is in this term you are then able to use the standard result which is:
\frac{d y}{d x}=n x^{n-1}As well as this notation you may also see f(x) and the derivative of this is represented by f'(x).
The notation for the second derivative is as follows:
\frac{d^2 y}{d x^2}Differentiation For AS Maths - Questions 1 - 2
Q1.
This is a standard question and it is simply asking you to determine the first and second derivative. This should not cause any issues.
Solution
\begin{aligned} y=5 x^3+7 x+3 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=5 \times 3 x^2+7 \times 1+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=15 x^2+7 \end{aligned}
Solution
\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=15 \times 2 x=30 xQ2.
Before you can perform the actual differentiation you need to rewrite the expression in a form that can first be differentiated. It is important that you are fully aware of the rules of indices as these will be needed to help you simplify any expressions.
\begin{aligned} & y=4 x^2+\frac{5-x}{x} \\ & y=4 x^2+5 x^{-1}-1 \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=4 \times 2 x+5 \times\left(-x^{-2}\right)+0 \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=8 x-5 x^{-2} . \end{aligned}x=1
\frac{\mathrm{d} y}{\mathrm{~d} x}=8 \times 1-5 \times 1^{-2}=8-5=3
Differentiation For AS Maths - Questions 3 - 5
Q3.
Again this is a standard question but you first need to use the rules of indices to rewrite the expression in a form that can be differentiated.
Q4.
As per the question above, this is quite a standard question and one that you may see at the test of a test or exam. It is a case of using indices to obtain the expression in a form which can be differentiated.
\begin{aligned} y=2 x^2-\frac{6}{x^3} & \Rightarrow y=2 x^2-6 x^{-3} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \times 2 x-6 \times\left(-3 x^{-4}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 x+18 x^{-4} . \end{aligned}Q5.
To answer this particular question you first need to expand the brackets before you can differentiate.
\begin{array}{c|cc} & x^2 & -4 \\ \hline x & x^3 & -4 x \\ -1 & -x^2 & +4 \end{array}\begin{aligned} y=x^3-x^2-4 x+4 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2-2 x-4 \times 1+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 x^2-2 x-4 . \end{aligned}
Differentiation For AS Maths – Questions 6 – 7
Solution
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{(x+4)^2}{x}\right) & =\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^2+8 x+16}{x}\right) \\ & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x+8+16 x^{-1}\right) \\ & =1+0+16 \times\left(-x^{-2}\right) \\ & =1-16 x^{-2} . \end{aligned}Both parts of this question require the use of indices like the previous questions before it. Take care if you need to add or subtract any powers as it is easy to make errors with any arithmetic.
Q7.
Solution
\begin{aligned} y=4 x^3-1+2 x^{\frac{1}{2}} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 \times 3 x^2+0+2 \times \frac{1}{2} x^{-\frac{1}{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=12 x^2+x^{-\frac{1}{2}} \end{aligned}Here the expression has a mixture of whole number powers and fractional powers. With fractional be careful when subtracting one from any fractional power and it is generally best and neater to leave any fractional powers as an improper fraction.
The overall process of differentiation is not difficult. As mentioned, the topic is introduced formally in year 12 and it is expanded in year 13. You are still expected to know all the techniques and methods introduced in year 12 differentiation for year 13.