Second Order Derivatives – A Level Maths Revision

If you’re looking for revision info on second order derivatives, then you’ve come to the right place. The derivative of a function (if it can be differentiated) is a function that describes the rate of change at any point on the original function. As the derivative is a function, it can also be differentiated, giving the second derivative.

If any of these terms aren’t familiar to you, check out our blogs on Differentiation from First Principles and Differentiation of Functions of the Form xn.

If you’d like to download this blog in PDF or PowerPoint formats, click here. To practice some of the prior knowledge required for this topic try these multiple-choice questions.


The second derivative is found by differentiating the derivative. It can be written in two ways – in function notation, the second derivative of f(π‘₯) is written as f’’(π‘₯). With Leibniz’s notation, the second derivative of 𝑦 with respect to π‘₯ is given by \frac{d^2y}{dx^2} (said β€œd-two-y by d-x-squared”).

Since the second derivative is the derivative of the gradient function, it gives the rate of change of the gradient.


Example Question 1

Given that 𝑦 = 3π‘₯5 – 4π‘₯2 + 5π‘₯ – 3, find \frac{d^2y}{dx^2} .

First, differentiate to get \frac{dy}{dx} . Remember, multiply by the power then reduce the power by 1.

\frac{dy}{dx} = 15x^4-8x+5

Now, just differentiate again:

\frac{d^2y}{dx^2} = 60x^3-8


Example Question 2

Given that \text{f}(x) = \sqrt{x}+\frac{1}{x} , find f”(π‘₯).

This time, you need to write the function using index notation before differentiating:

\text{f}(x)=x^{\frac{1}{2}}+x^{-1}

This example is almost identical to example 1 but with different notation. Use the same process – we differentiate twice.

\begin{aligned} &\text{f}'(x)=\frac{1}{2}x^{-\frac{1}{2}}-x^{-2} \\ &\text{f}''(x)=-\frac{1}{4}x^{-\frac{3}{2}}+2x^{-3}\end{aligned}

You can leave the result in this format or simplify it:

\text{f}''(x) = -\frac{1}{4\sqrt{x^3}}+\frac{2}{x^3}


Practice Questions

1. Find the second derivative of the following functions:

a. f(π‘₯) = 8π‘₯2 – 3π‘₯ + 4
b. f(π‘₯) = 2π‘₯3 – 15π‘₯
c. 𝑦 = π‘₯4 + 4π‘₯3 – π‘₯2

2. Find the second derivative for each of the functions below.

a. \text{f}(x) = \frac{3x^2+x^5+3}{x}

b. 𝑦 = (2π‘₯ + 1)(π‘₯ – 5)(π‘₯ + 1)

3. Given that f(π‘₯) = 4π‘₯3 – 2π‘₯ + 5, find each of the values below.

a. f(3)

b. f'(3)

c. f”(3)

4. Find the π‘₯-coordinates at which the second derivatives of the following function equals 16.

\text{f}(x) = (\frac{1}{x} + x)(\frac{1}{x}-x)

5. Given that 𝑦 = pπ‘₯4 – 2pπ‘₯3 + 3pπ‘₯2 – π‘₯ + 2, and when π‘₯ = -2, \frac{d^2y}{dx^2} = 39, find the value of 𝑝.


Answers

1. Find the second derivative of the following functions:

a. f(π‘₯) = 8π‘₯2 – 3π‘₯ + 4

\boldsymbol{\begin{aligned}&\textbf{f}\,'(x) = 16x-3\\&\textbf{f}\,''(x)=16\end{aligned}}

b. f(π‘₯) = 2π‘₯3 – 15π‘₯

\boldsymbol{\begin{aligned}&\textbf{f}\,'(x) = 6x^2-15\\&\textbf{f}\,''(x)=12x\end{aligned}}

c. 𝑦 = π‘₯4 + 4π‘₯3 – π‘₯2

\boldsymbol{\begin{aligned}&{\frac{dy}{dx}} = 4x^3+12x^2-12x\\&\frac{d^2y}{dx^2}=12x^2+24x-2\end{aligned}}

2. Find the second derivative for each of the functions below.

a. \text{f}(x) = \frac{3x^2+x^5+3}{x}

\boldsymbol{\begin{aligned}\textbf{f}(x) &= 3x +x^4+\frac{3}{x}\\&=3x+x^4+3x^{-1}\end{aligned}}
\boldsymbol{\begin{aligned}&\textbf{f}\,'(x)&=3+4x^3-x^{-2}\end{aligned}}
\boldsymbol{\begin{aligned}\textbf{f}\,''(x)&=12x^2+6x^{-3}\\&=12x^2+\frac{6}{x^3}\end{aligned}}

b. 𝑦 = (2π‘₯ + 1)(π‘₯ – 5)(π‘₯ + 1)

\boldsymbol{\begin{aligned}&y=2x^3-7x^2-14x-5\\&{\frac{dy}{dx}} = 6x^2-14x-14\\&\frac{d^2y}{dx^2}=12x-14\end{aligned}}

3. Given that f(π‘₯) = 4π‘₯3 – 2π‘₯ + 5, find each of the values below.

a. f(3)

\boldsymbol{\begin{aligned}\textbf{f}(3) &= 4(3)^3-2(3)+5\\&=107\end{aligned}}

b. f'(3)

\boldsymbol{\begin{aligned}\textbf{f}\,'(x) &= 12x^2-2\\\textbf{f}\,'(3)&=12(3)^2-2\\&=106\end{aligned}}

c. f”(3)

\boldsymbol{\begin{aligned}\textbf{f}\,''(x) &= 24x\\\textbf{f}\,''(3)&=24(3)\\&=72\end{aligned}}

4. Find the π‘₯-coordinates at which the second derivatives of the following function equals 16.

\text{f}(x) = (\frac{1}{x} + x)(\frac{1}{x}-x)

\boldsymbol{\begin{aligned}\textbf{f}(x) &= \frac{1}{x^2}-x^2\\&=x^{-2}-x{2}\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{f}\,'(x) = -2x^{-3}-2x\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{f}\,''(x) = 6x^{-4}-2\end{aligned}}

\boldsymbol{\begin{aligned} &6x^{-4}-2=16\\&6x^{-4}=18\\&x^{-4}=3\\&x^4=\frac{1}{3}\\&x=\sqrt[4]{\frac{1}{3}}=0.760 \textbf{ (3s.f.)}\end{aligned}}

5. Given that 𝑦 = pπ‘₯4 – 2pπ‘₯3 + 3pπ‘₯2 – π‘₯ + 2, and when π‘₯ = -2, \frac{d^2y}{dx^2} = 39, find the value of 𝑝.

\boldsymbol{\begin{aligned}&{\frac{dy}{dx}}=4px^3-6px^3+6px-1\\&\frac{d^2y}{dx^2}=12px^2-12px+6p\end{aligned}}

\boldsymbol{\begin{aligned}&\textbf{When }x=-2:\\&12(-2)^2-12p(-2)+6p=39\\&78p=39\\&p=\frac{1}{2}\end{aligned}}


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