A Level Maths: Understanding Differentiation
Introduction
When you embark upon your A Level Maths journey you are going to meet a variety of topics. Some you will have met and GCSE and there will be some new topics as well. One of these topic areas is that of Calculus which has two subtopics. The first is differentiation and the other is integration. It is essential that you understand differentiation.
In this article we are going to look at differentiation in terms of what it is and why it is used.
Understanding differentiation for A Level Maths
When we are differentiating we are actually determining the gradient function of a curve. We do this because the gradient of a curve varies unlike the gradient of a straight line which is consistent.
The rule for differentiation is as follows:
\text { If } y=x^n \text { then } \frac{d y}{d x}=n x^{n-1}\frac{d y}{d x} is pronounced “dee y by dee x” and it is “the rate of change of y with respect to x”.
When doing A Level Maths you will meet a number of standard results and this is the standard result for differentiation. It is not provided in a formula booklet so it is important that this is a result that you remember.
Example
Find \frac{d y}{d x} when:
a) y=10
b) y=x^5
c) y=12 x^4
Solution
a) Here you will see that there is no “x” term. So what we can do for this example is to write y=10 \text { as } y=10 x^0 ..
If we follow the rule for differentiation i.e. bring the power to the front and to reduce the power by 1, then we have the sum “10 x 0” which is clearly 0. And this is our answer. In fact any constant that is differentiated will always give an answer of 0.
b) Applying the rule for differentiation here we have \frac{d y}{d x}=5 x^5
c) Applying the rule for differentiation here we have \frac{d y}{d x}=48 x^3. In this example, the power has been brought forward and then multiplied by 12 which is the current coefficient.
Differentiating with fractional powers and negative powers
Whether we have positive powers, fractional powers or even negative powers the rules for differentiation are always the same i.e. bring the power to the front and reduce the power by 1. Many of the examples below require you to use the rule of indices and this is quite a common calculation that is required in many A Level Maths questions.
Let us look at the following example.
Suppose we want to differentiate y=\frac{1}{x}. In order to differentiate this we need to have something of the form x^n. In order for \frac{1}{x} the x term must be brought to the top and this can be done using one of the rules of indices.
So y=\frac{1}{x}=x^{-1} which can now be differentiated:
\frac{d y}{d x}=-1 x^{-2}=-\frac{1}{x^2}Example
Find \frac{d y}{d x} \text { when } y=\sqrt[5]{x}
Solution
First we need to rewrite \sqrt[5]{x} in a form that can be differentiated and this will be x^{\frac{1}{5}}:
So, y=\sqrt[5]{x}=x^{\frac{1}{5}} \rightarrow \frac{d y}{d x}=\frac{1}{5} x^{-\frac{4}{5}}
Example
Find the value of \frac{d y}{d x} when y=4 x^{-2} \text { at } x=4
Solution
Here we can differentiate as needed and there is no need to rearrange because what we have is in the required form.
\frac{d y}{d x}=-8 x^{-3}. This is the gradient function.
In order to determine the gradient value we simply need to know substitute x=4 \text { into } \frac{d y}{d x}.
Rather than just enter the values into a calculator it can be better to rearrange the gradient function so there are no negative powers. i.e. \frac{d y}{d x}=-\frac{8}{x^3} simply because we want to ensure that we get an exact answer rather than to have an answer that has been rounded.
So when we substitute x=4 \text { into } \frac{d y}{d x} \text { we have } \frac{d y}{d x}=-\frac{8}{4^3}=-\frac{1}{8}
Differentiating more than one term
Example
Find \frac{d y}{d x} \text { if } y=\sqrt{x}\left(\frac{1}{4} x^3-\frac{1}{x^2}+3 x^4+2\right)
Solution
The first thing that needs to be done here is to expand the brackets and to ensure that are all terms are of the form x^n.
\begin{aligned} & y=\sqrt{x}\left(\frac{1}{4} x^3-\frac{1}{x^2}+3 x^4+2\right) \\ & y=\sqrt{x}\left(\frac{1}{4} x^3-x^{-2}+3 x^4+2\right) \end{aligned}We can now expand the brackets as follows:
y=x^{\frac{7}{2}}-x^{-\frac{3}{2}}+3 x^{\frac{9}{2}}+2 x^{\frac{1}{2}}The above is now an expression which can be differentiated:
\frac{d y}{d x}=\frac{7}{2} x^{\frac{5}{2}}+\frac{3}{2} x^{-\frac{5}{2}}+\frac{27}{2} x^{\frac{7}{2}}+x^{-\frac{1}{2}}It is essential that when doing A level maths differentiation you are effectively able to use the rules of indices. You were introduced to indices whilst doing GCSE so the topic itself is not new.
Example
Find \frac{d y}{d x} \text { if } y=\frac{x^3-2 x^2+16 x-1}{x^6}
Solution
Again, before we can differentiate we need to make sure that each term is of the form x^n
In this example we have a single algebraic fraction where the common denominator is x^6. The single algebraic fraction can be re-written as follows:
y=\frac{x^3}{x^6}-\frac{2 x^2}{x^6}+\frac{16 x}{x^6}-\frac{1}{x^6}And using rules of indices and ensuring that all x terms are at the top we have:
y=x^{-3}-2 x^{-4}+16 x^{-5}-x^{-6}And we are now in a position where we are able to differentiate which gives us:
\frac{d y}{d x}=-3 x^{-4}+8 x^{-5}-80 x^{-6}+6 x^{-7}Finding gradient values at specific points
It is important to remember that when we have differentiated we have determined the gradient function. The gradient value at a specific point is the gradient value at that point only.
Example
Determine where the curve y=x^2(x-6) meets the x axis and determine the gradient value at each point.
Solution
The curve will meet the x axis when y = 0.
\begin{aligned} & 0=x^2(x-6) \\ & x-6=0 \therefore x=6 \text { and } x^2=0 \therefore x=0 \end{aligned}Next we need to actually find the gradient function and the brackets needed to be expanded first.
\begin{aligned} & y=x^3-6 x^2 \\ & \frac{d y}{d x}=3 x^2-12 x \\ & x=0: \frac{d y}{d x}=3(0)^2-12(0)=0 \\ & x=6: \frac{d y}{d x}=3(6)^2-12(6)=36 \end{aligned}Example
For the curve y=x^3+6 x^2+5 determine the coordinates where \frac{d y}{d x}=-9
Solution
Here we are given a numerical value of the gradient function. So we should also know what \frac{d y}{d x} actually is.
Differentiating we obtain:
\frac{d y}{d x}=3 x^2+12 xAnd we are told that \frac{d y}{d x}=-9 which means:
3 x^2+12 x=-9 and the highest power here is telling you that you now have a quadratic equation.
Solving the quadratic equation:
3 x^2+12 x+9=0 [we can make the equation more straightforward by dividing throughout by 3]
\begin{aligned} & x^2+4 x+3=0 \\ & (x+1)(x+3)=0 \\ & \therefore x=-1 \text { and } x=-3 \end{aligned}Having determined the x co-ordinates we also need to determine the y co-ordinates and this is done by using the original expression for y NOT \frac{d y}{d x}.
x=-1 ; y=(-1)^3+6(-1)^2+5=10 \text { i.e. }(-1,10)x=-3 ; y=(-3)^3+6(-3)^2+5=10 \text { i.e. }(-3,32)Example
Determine the points where the curve y=x^2+3 x-1 is parallel to the straight line y=5 x-1.
Solution
If we take the straight line into account for a moment, we know that the gradient is 5. So we need to find where on our curve \frac{d y}{d x}=5. This is very similar to the previous example but it has introduced a straight line.
Differentiating the curve we have: \frac{d y}{d x}=2 x+3
We know that:
\begin{aligned} & \frac{d y}{d x}=5 \therefore 5=2 x+3 \\ & 2=2 x \text { and so } x=1 \end{aligned}It is important to also find the y coordinate: x=1 ; y=(1)^2+3(1)-1=3.
The point is (1, 3)
As you can see there is a lot of content within calculus and it is essential understanding differentiation and there is still more to know! You will meet some practical problems involving differentiation in year 12 and in year 13 your knowledge of differentiation is extended further.