A Level Maths: The Chain Rule
Introduction to the Chain Rule
Consider the following expression: y=(2 x+5)^3
We want to differentiate it. What can we do?
One option is that we could expand the brackets which would give us the following:
8 x^3+60 x^2+150 x+125Now we are in a position to differentiate each term and this would give:
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} & =\left(8 \times 3 x^2\right)+(60 \times 2 x)+150 \\ & =24 x^2+120 x+150 \end{aligned}You could go a step further with this answer and perhaps factorise it which would give an answer of:
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x} & =24 x^2+120 x+150 \\ & =6\left(4 x^2+20 x+25\right) \\ & =6(2 x+5)^2 \end{aligned}This is a perfectly valid option but what if the power was higher than 3? What if it was 30 or even 300?!?. What would we do then?
What is needed is a clever way of differentiating such questions which means there needs to be a more effective method. And it follows the use of the standard result.
\begin{aligned} y & =x^n \\ \frac{d y}{d x} & =n x^{n-1} \end{aligned}If we apply the standard result to our expression then we have:
\begin{aligned} & y=(2 x+5)^3 \\ & \frac{d y}{d x}=3(2 x+5)^2 \end{aligned}Which is not quite the answer that we obtained previously. Somehow there needs to be a 6 in front of the bracket. But how to obtain this 6 from what we have?
Well if you look inside the bracket you have 2x + 5 and if you differentiate this you get 2. And 3 x 2 = 6. So this means that in order to obtain a full and final answer, we need to multiply by the derivative of what is inside the bracket.
\begin{aligned} \therefore \frac{d y}{d x} & =3(2 x+5)^2(2) \\ & =6(2 x+5)^2 \end{aligned}This method of differentiation is known as the Chain Rule and it is a technique that you will be using a lot during your 2 years of A Level maths.
Now there is a more formal technique for these types of questions and here we will show the implementation of the chain rule itself.
\begin{aligned} & y=(2 x+5)^3 \\ & u=2 x+5 \quad \quad y=u^3 \\ & \frac{d u}{d x}=2 \quad \frac{d y}{d u}=3 u^2 \\ & \therefore \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x} \\ & =3 u^2 \times 2 \\ & =6 u^2 \\ & \\ & =6(2 x+5)^2 \end{aligned}What we have done here is to take a form of substitution. You can see that u=2 x+5 which has then been differentiated and y=u^3
Once these have both been differentiated a connection can then be made using the Chain Rule. This more formal method is acceptable and there will be many instances where this technique is more than acceptable especially when dealing with more complicated expressions.
However, if the expression is more straightforward, then following the process of using the standard result is much more simplistic and also much faster.
We will take a look at a few examples and see how the chain rule is applied. In each example both techniques will be used.
Chain Rule - Example
Method 1 – Following the standard result
\begin{aligned} y & =\left(3 x^2-4\right)^7 \\ \frac{d y}{d x} & =7\left(3 x^2-4\right)^6(6 x) \\ & =42 x\left(3 x^2-4\right)^6 \end{aligned}
Method 2 – Setting up the chain rule
\begin{aligned} & y=\left(3 x^2-4\right)^7 \\ & u=3 x^2-4 \quad y=u^7 \\ & \begin{aligned} \frac{d u}{d x} & =6 x \quad \frac{d y}{d u}=7 u^6 \\ \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =7 u^6(6 x) \\ & =42 x\left(3 x^2-4\right) \end{aligned} \end{aligned}It is important to put the question back in terms of x rather than to have u within the answer.
Chain Rule - Example
Method 1 – Following the standard result
\begin{aligned} y & =\sqrt{1+x^2} \\ & =\left(1+x^2\right)^{\frac{1}{2}} \\ \frac{d y}{d x} & =\frac{1}{2}\left(1+x^2\right)^{-\frac{1}{2}}(2 x) \\ & =x\left(1+x^2\right)^{-\frac{1}{2}} \\ & =\frac{x}{\sqrt{1+x^2}} \end{aligned}Method 2 – Setting up the chain rule
\begin{aligned} y & =\sqrt{1+x^2} \\ & =\left(1+x^2\right)^{\frac{1}{2}} \\ u & =1+x^2 \quad y=u^{\frac{1}{2}} \\ \frac{d u}{d x} & =2 x \quad \frac{d y}{d u}=\frac{1}{2} u^{-\frac{1}{2}} \\ \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =\frac{1}{2} u^{-\frac{1}{2}} \times 2 x \\ & =x\left(1+x^2\right)^{-\frac{1}{2}} \\ & =\frac{x}{\sqrt{1+x^2}} \end{aligned}Whichever method you choose to use it is important that you are getting the expression to differentiate in the form xn. If this is not done then it will not be possible to differentiate so it is important to have a strong understanding of the rules of indices.
Finally we will consider a past exam question relating to the chain rule.
Chain Rule Exam Question
For part (a) of the question it is important to rewrite the question in a form which can then be differentiated:
y=3(2 x+1)^{-1}To differentiate this expression you need to now use the chain rule for A Level Maths.
\begin{aligned} \frac{d y}{d x} & =-3(2 x+1)^{-2} \cdot 2 \\ & =-6(2 x+1)^{-2} \end{aligned}You are given that x=1 which if you substitute in will give you the value of the gradient of the tangent at this particular point. Since we want the equation of the normal we need to take the negative reciprocal.And because we are finding the equation of a straight line which is what a normal is, you will also need to know the value of y when x=1.
\text { when } x=1 \quad \begin{aligned} \frac{d y}{d x} & =-6(3)^{-2} \\ & =-\frac{2}{3} \\ y & =3(3)^{-1} \\ & =1 \end{aligned}This means that the gradient of the normal is \frac{3}{2}
So at the point (1, 1) the equation of the normal is:
\begin{gathered} y=\frac{3}{2} x+c \\ 1=\frac{3}{2}+c \\ c=-\frac{1}{2} \\ y=\frac{3}{2} x-\frac{1}{2} \end{gathered}For part (b) of the question you need to solve simultaneously the equation of the normal and the equation of the curve. Because they are both equal to y , these equations can be equated.
\begin{array}{rl} \frac{3}{2 x+1}= & \frac{3}{2} x-\frac{1}{2} \\ \frac{6}{2 x+1}= & 3 x-1 \\ 6= & (3 x-1)(2 x+1) \\ 6 & 6 x^2+3 x-2 x-1 \\ 0= & 6 x^2+x-7 \\ 0 & (6 x+7)(x-1) \\ & x=-\frac{7}{6} \quad x=1 \end{array}This means that the other point will have x=-\frac{7}{6}
The question is asking for coordinates so the y value needs to be determined as well:
\begin{aligned} y & =\frac{3}{2}\left(-\frac{7}{6}\right)-\frac{1}{2} \\ & =\frac{-21}{12}-\frac{1}{2} \\ & =-\frac{9}{4} \end{aligned}The other coordinate is \left(-\frac{7}{6}, \frac{-9}{4}\right).
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