A Level Maths: Which Rule Of Differentiation
Introduction
When it comes to doing differentiation at A Level maths, you will have met a number of different rules: Chain Rule, Product Rule and Quotient Rule. But the biggest problem that many students face is knowing which rule of differentiation to use and when to use it.
In this article we are going to discuss just that. When do you know which rule of differentiation to use when doing A Level maths questions?
A Level Maths: The Chain Rule
You use the chain rule when you have a function of a function.
An example of this would be:
\begin{aligned} & y=(x+7)^{21} \\ & y=\cos 2 x \end{aligned}In both of these cases you have a function of a function.
For instance x + 7 is a function. The raising to the power 21 is another function. So this is indeed a function of a function.
2x is also a function as is cos2x, so again we have a function of a function and so the chain rule is to be used.
With the trig term you cannot separate the terms. What this means is that you cannot have “cos” and “2x” and then try to differentiate using the product rule – it just does not work.
A Level Maths: The Product And Quotient Rule
You use the product rule when you have one function that is being multiplied by something else.
In other words: f(x) \times y(x)
An example of this would be y=(x+3)^{10} \times 5 x^3
Quite often in A Level maths differentiation you need to use more than one technique to differentiate a term. Here, even though the main rule is the product rule, you also need to use the chain rule to differentiate (x+3)^{10}.
The formula for the product rule is \frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}
Quotient Rule
To use the quotient rule you are essentially looking for a fraction. So an expression of the form:
\frac{f(x)}{g(x)}You can rewrite these expressions so you have something of the form f(x)[g(x)]^{-1}
And you can then use the product rule to differentiate. However you need to be extra careful with this technique as it involves negative powers or even fractional powers which are negative, and this just increases the chance of making mistakes.
The formula for the quotient rule is:
\frac{d y}{d x}=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2}
Examples – which rule of differentiation?
a) \text { Differentiate } y=\left(3 x^2-4\right)^7 \text {. }
We want to differentiate the above and the first task is to determine which rule of differentiation is needed.
You can see that what is written above is definitely a function of a function and so the chain rule must be used.
There are two methods to using the chain rule.
Long method – here you are making a substitution and then differentiating both in order to set up a chain rule. It is important to put your answer back in terms of x.
\begin{aligned} u & =3 x^2-4 \quad y=u^7 \\ \frac{d u}{d x} & =6 x \quad \frac{d y}{d u}=7 u^6 \\ \therefore & \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}=7 u^6 \times 6 x=42 x\left(3 x^2-4\right)^6 \end{aligned}When you are setting up the chain rule it is important to treat what you have on the right side as a set of fractions, where the diagonals will cancel. In our case, we have du appearing in the diagonal, and according to the rule of fractions, these would indeed cancel.
Whilst this is a longer technique it is a method that you need to know especially if you are differentiating a more complicated expression.
Using the same example we will now differentiate this using a technique based on the standard result for differentiation.
\begin{aligned} & \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}With this technique you are multiplying by the derivative of the bracket and the same answer has been obtained for both techniques.
Obviously the second method is much faster, but this may not always be possible to use if differentiating something more complicated.
b) y=x^2 \sqrt{1+x} \text { find } \frac{d y}{d x}
With this question you can see that you have two terms that are being multiplied together and so the product rule must be used.
\begin{aligned} & y=x^2 \times \sqrt{1+x} \\ & u=x^2 \quad v=(1+x)^{\frac{1}{2}} \\ & \frac{d u}{d x}=2 x \quad \frac{d v}{d x}=\frac{1}{2}(1+x)^{-\frac{1}{2}} \\ & \therefore \frac{d y}{d x}=\frac{1}{2} x^2(1+x)^{-\frac{1}{2}}+2 x(1+x)^{\frac{1}{2}} \end{aligned}c) Differentiate y=\frac{x^2}{1+2 x^3}
Here you can clearly see that two functions have been divided. Or you can see that you have an algebraic fraction. In this case, the quotient rule can be used. Note: you can use the product rule by taking the denominator to the top, but this will involve a negative power which increases the risk of making an error.
\begin{array}{ll} u=x^2 & v=1+2 x^3 \\ \frac{d u}{d x}=2 x & \frac{d v}{d x}=6 x^2 \\ \Rightarrow & \frac{d y}{d x}=\frac{\left(1+2 x^3\right)(2 x)-\left(x^2\right)\left(6 x^2\right)}{\left(1+2 x^3\right)^2} \end{array}Factorising the gradient function.
You will have come across turning points or stationary points and depending on the question, these need to be found. If your gradient function is complicated it can be helpful to factorise the gradient function which makes finding any turning points more easier.
Here is an example which highlights this:
First, which rule of differentiation needs to be obtained. You have a function multiplied by another function, so the product rule is needed, and it is more than likely that the chain rule is also needed.
Part a)
\begin{aligned} & u=x^3 \quad v=(1+3 x)^{\frac{1}{3}} \\ & \frac{d u}{d x}=3 x^2 \quad \frac{d v}{d x}=\frac{1}{3}(1+3 x)^{-\frac{2}{3}}(3) \\ & =(1+3 x)^{-\frac{2}{3}} \\ & \therefore \frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x} \\ & =x^3(1+3 x)^{-\frac{2}{3}}+3 x^2(1+3 x)^{\frac{1}{3}} \\ & \end{aligned}Here we do have an answer for the gradient function but it is not what the answer wants. We need to obtain an answer for the gradient function as shown. In order to do this, you need to factorise. Clearly x^2 is common and so are the terms with the brackets. For the terms with the brackets always take the term with the negative bracket out as a factor as follows:
\begin{aligned} & \frac{d y}{d x}=x^2(1+3 x)^{-\frac{2}{3}}[x+3(1+3 x)] \\ & \frac{d y}{d x}=x^2(1+3 x)^{-\frac{2}{3}}(10 x+3) \end{aligned}Which is the required answer.
Part b) Now that you have the gradient function in factorised form you can now found the x-coordinates of the turning points:
\begin{aligned} & \frac{d y}{d x}=0 \\ & x^2(1+3 x)^{-\frac{2}{3}}(10 x+3)=0 \\ & x^2=0 \rightarrow x=0 \quad 10 x+3=0 \rightarrow x=-\frac{3}{10} \\ & 1+3 x=0 \rightarrow x=-\frac{1}{3} \end{aligned}As a final example of A Level Maths differentiation we want to find the derivative of x^2 \sqrt{2 x+1}
And we want to simplify the gradient function.
Again, first you need to determine the rule that is needed. You can see that there are two functions being multiplied together so the product rule needs to be used in this instance as well as using the chain rule:
\begin{array}{rl} u=x^2 & v=(2 x+1)^{1 / 2} \\ \frac{d u}{d x} & =2 x \quad \frac{d v}{d x}=\frac{1}{x}(2 x+1)^{-\frac{1}{2}}(x)=(2 x+1)^{-\frac{1}{2}} \\ \therefore \frac{d y}{d x} & =x^2(2 x+1)^{-\frac{1}{2}}+(2 x+1)^{\frac{1}{2}}(2 x) \\ & =x(2 x+1)^{-\frac{1}{2}}[x+2(2 x+1)] \\ \frac{d y}{d x} & =x(2 x+1)^{-\frac{1}{2}}(5 x+2) \end{array}In terms of determining what is the common term, it is clear that x is common and you will see two terms contained within a bracket. One has a negative power and the other a positive power. Take the term with the negative power out as a common factor.
Deciding which rule of differentiation for A Level Maths to use can often be a stumbling block but if you follow the guidance in this article then knowing how to actually start a question can now be done with much confidence. What you do need to be careful about is obtaining the gradient function in a particular form and so you need to be careful with the factorising that is involved.