Differentiation For AS Maths
Differentiation AS Maths – Introduction
This article is going to continue looking at the topic of Differentiation AS Maths by examining a number of questions and discussing the methods that are needed. From the previous article the importance of being able to correctly use and apply the rules of indices has been stressed and with the questions that will be looked at in this article, this still holds true.
In fact, you never know what skills you are going to need when it comes to doing maths. This is why it is important to be very competent in all areas of mathematics and knowing how to apply the knowledge that you have acquired.
Differentiation AS Maths - Things To Remember
The main thing to remember is the standard result which is as follows:
\begin{aligned} & y=x^n \\ & \frac{d y}{d x}=n x^{n-1} \end{aligned}Also be aware of the second derivative, \frac{d^2 y}{d x^2}, which means that you are just differentiating a term twice.
We will consider another 6 questions and discuss how to answer the question and to provide a detailed clear solution.
Differentiation AS Maths - Questions 1 - 2
Q1.
A standard question, just be careful when dealing with fractional powers.
\begin{aligned} y=4 x+3 x^{\frac{3}{2}}-2 x^2 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 \times 1+3 \times \frac{3}{2} x^{\frac{1}{2}}-2 \times 2 x \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4+\frac{9}{2} x^{\frac{1}{2}}-4 x \end{aligned}Q2.
This should not cause any real concerns but you need to rewrite x in suitable index form before you can perform the actual differentiation.
Solution
\begin{aligned} y=3 x^2+4 \sqrt{x} & \Rightarrow y=3 x^2+4 x^{\frac{1}{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \times 2 x+4 \times \frac{1}{2} x^{-\frac{1}{2}} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x+2 x^{-\frac{1}{2}} . \end{aligned}
Here we simply need to find the second derivative which means that the answer to part a) needs to differentiated once again to give the following result:
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x+2 x^{-\frac{1}{2}} & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=6 \times 1+2 \times\left(-\frac{1}{2} x^{-\frac{3}{2}}\right) \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=6-x^{-\frac{3}{2}} \end{aligned}
Differentiation AS Maths – Questions 3 – 5
Here you need to write this single algebraic fraction in terms of two terms so the rules of indices need to be used. This is shown as follows:
\frac{2 \sqrt{x}+3}{x}=\frac{2 x^{\frac{1}{2}}+3}{x}=2 x^{-\frac{1}{2}}+3 x^{-1}
Once again with this type of question you are having to use the rules of indices in order to obtain an expression that can first be differentiated. This is shown below as is the process of differentiation:
\begin{aligned} y=5 x-7+\frac{2 \sqrt{x}+3}{x} & \Rightarrow y=5 x-7+2 x^{-\frac{1}{2}}+3 x^{-1} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=5 \times 1+0+2 \times\left(-\frac{1}{2} x^{-\frac{3}{2}}\right)+3 \times\left(-x^{-2}\right) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=5-x^{-\frac{3}{2}}-3 x^{-2} \end{aligned}Q4.
The notation f(x) is sometimes used to represent “y” and the notation f'(x) is used to represent \frac{d y}{d x}.
It is just another way to write the same thing.
The process for differentiation here is straightforward and there is no need to rewrite the expression as shown in previous examples:
\mathrm{f}(x)=3 x+x^3 \Rightarrow \mathrm{f}^{\prime}(x)=3 \times 1+3 x^2 \Rightarrow \mathrm{f}^{\prime}(x)=3+3 x^2.
Q5.
This constant is simply asking you to differentiate where the value of a constant is unknown. In such cases, differentiate as you would normally as shown below:
\begin{aligned} y=k x^3-x^2+x-5 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=k \times 3 x^2-2 x+1+0 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=3 k x^2-2 x+1 . \end{aligned}Differentiation AS Maths – Final Question
This question is asking for the use of indices as seen in a previous example. It is important that you are fully aware of the rules of indices as they appear quite a lot in all areas of differentiation and integration.
\frac{2 x^2-x^{\frac{3}{2}}}{\sqrt{x}}=\frac{2 x^2-x^{\frac{3}{2}}}{x^{\frac{1}{2}}}=2 x^{\frac{3}{2}}-x\begin{array}{r} p=\frac{3}{2} \\ q=1 \end{array}
You simply need to replace the algebraic fraction with the answer that you obtained in part a) and the differentiation is then very easy indeed as shown here:
\begin{aligned} y=5 x^4-3+\frac{2 x^2-x^{\frac{3}{2}}}{\sqrt{x}} & \Rightarrow y=5 x^4-3+2 x^{\frac{3}{2}}-x \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=5 \times 4 x^3+0+2 \times \frac{3}{2} x^{\frac{1}{2}}-1 \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=20 x^3+3 x^{\frac{1}{2}}-1 . \end{aligned}
We will continue to create blog posts that cover the topic of differentiation. Look at the examples above and see if you are able to do them on your own. There are many techniques for differentiation and further techniques are developed in year 13 of the A Level course.