Differentiation in Mathematics:
Differentiation – Introduction
Differentiation is a fundamental concept in mathematics that involves finding the gradient of a curve at a specific point. It is taught at both AS Level and A Level Mathematics to help students understand how to analyse and describe the behaviour of functions.
Understanding Differentiation:
A Level Maths Differentiation allows us to calculate the rate at which a function is changing at a particular point. It involves finding the derivative of a function, which represents the gradient of the curve at that point. By determining the rate of change, we can gain insights into the behaviour of the function, such as whether it is increasing, decreasing, or reaching a maximum or minimum point.
History of Differentiation:
The concept of differentiation can be traced back to the ancient Greeks, particularly to the work of mathematicians such as Archimedes and Euclid. However, it was Sir Isaac Newton and Gottfried Wilhelm Leibniz who independently developed the modern foundations of differentiation in the late 17th century. Their work laid the groundwork for calculus, with differentiation being one of its key components.
Applications of Differentiation:
Differentiation has numerous practical applications in various fields, including physics, economics, engineering, and biology. Here are a few examples:
Physics: Differentiation helps us analyse the motion of objects by calculating velocities and accelerations. It plays a crucial role in understanding concepts like velocity-time graphs and determining the instantaneous velocity of an object at a specific point.
Economics: Differentiation is used to analyse and optimise economic models. It helps economists calculate marginal cost, marginal revenue, and elasticity of demand, which are essential in making informed business decisions.
Engineering: Differentiation is vital in engineering disciplines such as civil, mechanical, and electrical engineering. It helps engineers analyse the behaviour of systems, optimise designs, and calculate rates of change in various physical quantities.
Biology: Differentiation is used in biology to model and understand population dynamics, growth rates, and the behaviour of biological systems. It helps biologists analyse how populations change over time and predict future trends.
Conclusion:
In summary, differentiation is a fundamental concept in mathematics that allows us to determine the gradient of a curve at a specific point. It has a rich history dating back to ancient times and was further developed by Newton and Leibniz. Differentiation finds wide applications in fields such as physics, economics, engineering, and biology, helping us analyse and understand various real-world phenomena.
Differentiation - Basic Examples
Here we will discuss some examples of questions and demonstrate the solution in detail. Some of the questions will involve the use of index rules as well as finding the second derivative.
Example 1
For this question you are using the standard result for differentiation.
\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}
Here you are being asked to find the second derivative and all that is required to differentiate once again.
\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=15 \times 2 x=\underline{\underline{30 x}} .Example 2
Before you can differentiate you need to write the equation in a form that can be differentiated. Once this has been done you can then perform the differentiation.
\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}Once the differentiation has been done it is simply a case of substituting in x = 1 into the gradient function as shown below.
When x = 1:
\frac{\mathrm{d} y}{\mathrm{~d} x}=8 \times 1-5 \times 1^{-2}=8-5=\underline{\underline{3}}Example 3
First the surd needs to be written in terms of a fractional power and then the differentiation can be done successfully as shown below:
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} x}\left(x^4+6 \sqrt{x}\right) & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^4+6 x^{\frac{1}{2}}\right) \\ & =4 x^3+6 \times \frac{1}{2} x^{-\frac{1}{2}} \\ & =4 x^3+3 x^{-\frac{1}{2}} . \end{aligned}
For the above first expand the numerator and then divide by the denominator to obtain an expression that then can be differentiated as shown below:
\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}Example 4
As you can see there are a lot of questions that rely on the use of indices. Remember the rules of indices that you were taught at GCSE. The rules of indices are required in both parts of this particular question.
\frac{2 \sqrt{x}+3}{x}=\frac{2 x^{\frac{1}{2}}+3}{x}=\underline{\underline{2 x^{-\frac{1}{2}}+3 x^{-1}}} .
Differentiation - Harder Examples
Example 5
\mathrm{f}(x)=\frac{(3-4 \sqrt{x})^2}{\sqrt{x}}, x>0
Here you are asked to find the value of two constants. First you need to expand the numerator and then divide by the denominator changing it to a fractional power first. Again you need to make sure that you are using the rules of indices correctly and to perform any arithmetic correctly. Note that for the first part of the question you are just asked to obtain the function in a form that can be differentiated.
\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) find \mathbf{f}^{\prime}(x)
It is this part of the question where you are being asked to differentiate the expression obtained in part a)
\begin{aligned} \mathrm{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}Example 6
Again expand the numerator and divide by the numerator before differentiating.
\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}Example 7
Here there are two terms that need to be rewritten before the differentiation can take place and it is important that you do this correctly.
\begin{aligned} & y=8 x^3-4 \sqrt{x}+\frac{3 x^2+2}{x} \\ & y=8 x^3-4 x^{\frac{1}{2}}+3 x+2 x^{-1} \\ & \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) \\ & \frac{\mathrm{d} y}{\mathrm{~d} x}=24 x^2-2 x^{-\frac{1}{2}}+3-2 x^{-2} . \end{aligned}
Differentiation – Exam Style Questions
Here are some actual past exam questions. Using the techniques from above see if you can find the solutions before looking at the answers. If you understand the techniques from above then you ought to find these questions straightforward.
Part a)
\begin{aligned} & y=2 x-5+3 x^{-1} \\ & \frac{d y}{d x}=2-3 x^{-2} \end{aligned}Part b)
When x = 3:
\begin{aligned} \frac{d y}{d x} & =2-3(3) \\ & =\frac{5}{3} \end{aligned}
When x = 1 we have:
\begin{aligned} \frac{d y}{d x} & =6(1)^2+10(1)-7 \\ & =9 \end{aligned}When x = 1:
\begin{aligned} &y=2(1)^3+5(1)^2-7(1)+10\\ &y=10 \end{aligned}\begin{aligned} & y=9 x+c \\ & 10=9(1)+c \\ & c=1 \\ & y=9 x+1 \end{aligned}You can see from these examples alone that there is a lot of work that can cover the topic of differentiation and we have only just scratched the surface of this particular topic. We will be posting more content in the very near future that explores all the necessary topics and areas that you need to be aware of to be successful at this course at both AS and A Level.
It is important that you practise these questions especially being comfortable with the indices. If you need to recap these skills then we do have a blog topic on working in index form which you should take a look at.
Stay posted for further posts on differentiation.