A Level Maths: Quotient Rule
Introduction
So far in other articles we have discussed other techniques for differentiation which were the Chain Rule and the Product Rule.
In this article we are going to focus on the Quotient Rule.
During your A Level Maths Course you will discover a range of differentiation techniques and you will see how these are applied to the areas of exponentials as well as to the area of trigonometry.
For the quotient rule you are essentially have a function divided by another function:
y=\frac{f(x)}{g(x)}And just as with the Product Rule for A Level Maths, there is also a formula for the Quotient Rule which is as follows:
\begin{aligned} y & =\frac{u}{v} \\ \frac{d y}{d x} & =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2} \end{aligned}You can also write this as:
\begin{aligned} & y=\frac{u}{v} \\ & y=u v^{-1} \end{aligned}Example:
Differentiate y=\frac{x^3}{1+2 x^3}, simplifying your answer.
Solution
\begin{aligned} u & =x^3 \quad v=1+2 x^3 \\ \frac{d u}{d x} & =3 x^2 \quad \frac{d v}{d x}=6 x^2 \\ \frac{d y}{d x} & =\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2} \\ & =\frac{\left(1+2 x^3\right)\left(3 x^2\right)-x^3\left(6 x^2\right)}{\left(1+2 x^3\right)^2} \\ & =\frac{3 x^2+6 x^5-6 x^5}{\left(1+2 x^3\right)^2} \end{aligned}\therefore \frac{d y}{d x}=\frac{3 x^2}{\left(1+2 x^3\right)^2}Exam Question 1
Solution
\begin{aligned} & \frac{d y}{d x}=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2} \\ & u=e^x \quad v=2 x+1 \\ & \frac{d u}{d x}=e^x \quad \frac{d v}{d x}=2 \\ & \frac{d y}{d x}=\frac{e^x(2 x+1)-2 e^x}{(2 x+1)^2} \end{aligned}The question has not asked you to simplify the answer and so therefore there is no need to go any further than what has been shown. Make sure that you are comfortable with the differentiation of exponentials.
b)
\begin{aligned} u & =\ln \left(x^2+1\right) \quad v=3 x+2 \\ \frac{d u}{d x} & =\frac{2 x}{x^2+1} \quad \frac{d v}{d x}=3 \\ & \frac{\frac{2 x(3 x+2)}{x^2+1}-3 \ln \left(x^2+1\right)}{(3 x+2)^2} \\ & \frac{2 x(3 x+2)-3\left(x^2+1\right) \ln \left(x^2+1\right)}{(3 x+2)^2\left(x^2+1\right)} \end{aligned}Here you are using the quotient rule as well as the chain rule. If you are not sure about the use of any standard results then you can always set up and use the chain rule.
Exam Question 2
Solution
\begin{aligned} & y=\frac{3 x}{2 x-1} \\ & u=3 x \quad v=2 x-1 \\ & \frac{d u}{d x}=3 \quad \frac{d v}{d x}=2 \\ & \frac{d y}{d x}=\frac{3(2 x-1)-2(3 x)}{(2 x-1)^2} \\ & =\frac{6 x-3-6 x}{(2 x-1)^2} \\ & =\frac{-3}{(2 x-1)^2} \\ & \end{aligned}The first step is to find the gradient function and the method has been shown above, again you are having to use the quotient rule.
When x = 1:
\begin{aligned} \frac{d y}{d x} & =-\frac{3}{1} \\ & =-3 \\ y & =\frac{3}{1} \\ & =3 \\ y & =-3 x+c \\ 3 & =-3(1)+c \\ 6 & =c \\ \therefore y & =-3 x+6 \end{aligned}Above we have found the gradient value at the specified point and then the equation of the tangent can then be found.
Exam Question 3
Solution
The first part that needs to be done is to find the gradient function. Here, you can check with what has been done below:
\begin{aligned} y & =\frac{e^x+1}{e^x+3} \\ u & =e^x+1 \quad \quad v=e^x+3 \\ \frac{d u}{d x} & =e^x \quad \frac{d v}{d x}=e^x \\ \frac{d y}{d x} & =\frac{e^x\left(e^x+3\right)-e^x\left(e^x+1\right)}{\left(e^x+3\right)} \\ & =\frac{e^{2 x}+3 e^x-e^{2 x}-e^x}{\left(e^x+3\right)^2} \\ & =\frac{2 e^x}{\left(e^x+3\right)^2} \end{aligned}We then need to find the value of the gradient function at the specified point as well as the value of y. Remember when you substitute in a value you obtain the gradient of the tangent. You then need to use the value of the negative reciprocal to obtain the gradient of the normal. This is shown below.
\begin{aligned} & \frac{d y}{d x}=\frac{2}{16}=\frac{1}{8} \\ & y=\frac{1+1}{1+3}=\frac{2}{4}=\frac{1}{2} \end{aligned}The gradient of the normal is therefore -8
\begin{aligned} y & =-8 x+c \\ \frac{1}{2} & =c \\ \therefore y & =-8 x+\frac{1}{2} \end{aligned}Take care when having to work out negative reciprocal gradients as sometimes the signs can be missed or it can just be forgotten to do.