A Level Maths: The Second Derivative
Introduction
In this article we are going to consider the second derivative which is just the first derivative differentiated again. Using the second derivative helps us to determine the nature of turning points much quicker compared to testing the sign of the gradient either.
You will have met \frac{d y}{d x}
and this is the “rate of change of y with respect to x”. This is also referred to as the first derivative.
The second derivative simply means that we differentiate the first derivative once again. And this is shown as follows:
\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^2 y}{d x^2}The second derivative is essentially the rate of change of gradient.
It should also be noted that \frac{d^2 y}{d x^2} DOES NOT EQUAL \left(\frac{d y}{d x}\right)^2
As well as the second derivative we will also look at increasing and decreasing functions which also forms a part of A Level maths.
Example
Find the second derivative of y=x^4+x^3-2 x^2-3 x+1
Solution
\begin{aligned} & \frac{d y}{d x}=4 x^3+3 x^2-4 x-3 \\ & \frac{d^2 y}{d x^2}=12 x^2+6 x-4 \end{aligned}Maximum & Minimum Stationary Points
You will know that stationary points occur when \frac{d y}{d x}=0. What is also important is that we are able to determine the nature of that turning point.
Here we have a maximum turning point where the gradient initially increases and then decreases. Because it is decreasing this means that the second derivative is less than zero.
Here we have a minimum turning point and you will see that the curve initially decreases and then increases. This means that the second derivative is greater than zero.
We can use these results to help us to determine the nature of the turning point of a curve.
One thing that you need to be careful about when doing A Level maths revision questions is when the second derivative is equal to zero. In this instance you need to test the gradient on either side of the turning point.
Example
Given y=x^4-8 x^2
determine the nature of any turning points of the curve.
Solution
\frac{d y}{d x}=4 x^3-16 xWe have turning points when \frac{d y}{d x}=0
\begin{aligned} & 4 x^3-16 x=0 \\ & 4 x\left(x^2-4\right)=0 \\ & x=0, x=2, x=-2 \\ & \frac{d^2 y}{d x^2}=12 x^2-16 \end{aligned}When x=0; \frac{d^2 y}{d x^2}=-16<0 which is a max turning point.
\text { When } x=2 ; \quad \frac{d^2 y}{d x^2}=32>0 which is a min turning point.
\text { When } x=-2 ; \frac{d^2 y}{d x^2}=32>0 which is a min turning point.
Having determined the x coordinates and the nature of the turning points we also need to determine the y coordinates because the question is asking us to find the coordinates of any turning points.
When x = 0, y = 0 🡪 (0, 0) Maximum turning point
When x = 2, y = -16 🡪 (-2, -16) Minimum turning point
When x = 2, y = – 16 🡪 (2, -16) Minimum turning point
Increasing and decreasing functions
A function is increasing when \frac{d y}{d x}>0
A function is decreasing when \frac{d y}{d x}<0
Example
Is y=x^3+x an increasing function?
Solution
\frac{d y}{d x}=3 x^2+1Is this > 0 for it to be an increasing function?
Yes it is because anything square is always positive and by adding one it is still positive. So the expression is an increasing function.
Example
What are the values of x for which y=x^2-6 x is it a decreasing function?
Solution
\frac{d y}{d x}=2 x-6For this to be a decreasing function 2 x-6<0 \text { i.e. } 2 x<6 \therefore x<3
Challenge Question
Here is an example of you having to find the second derivative of an expression which involves an exponential. There are two stages to this question. The first requires finding the first derivative and the second requires finding the second derivative.
In the first stage you need to use the chain rule and in the second stage you need to use the product rule as well and the chain rule. This is a good A Level Maths question as it shows you how different skills, knowledge and techniques are needed to be able to answer these questions.
Given: y=e^{3 x^2} \text { find } \frac{d^2 y}{d x^2}
First stage solution is as follows:
\begin{aligned} u & =3 x^2 \quad y=e^u \\ \frac{d u}{d x} & =6 x \quad \frac{d y}{d u}=e^u \\ \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =e^u \times 6 x \\ \therefore \frac{d y}{d x} & =6 x e^{3 x^2} \end{aligned}Here is the second stage to find the second derivative:
\begin{aligned} u & =6 x \quad \quad \quad \quad v=e^{3 x^2} \\ \frac{d u}{d x} & =6 \quad \frac{d v}{d x}=6 x e^{3 x^2} \\ \frac{d^2 y}{d x^2} & =u \frac{d v}{d x}+v \frac{d u}{d x} \\ & =6 x\left(6 x e^{3 x^2}\right)+e^{3 x^2}(6) \\ & =36 x^2 e^{3 x^2}+6 e^{3 x^2} \\ \therefore \frac{d^2 y}{d x^2} & =6 e^{3 x^2}\left(6 x^2+1\right) \end{aligned}There are no shortcuts to doing these types of questions and you need to be showing all your work in order to maximise marks.