The second derivative

Given a function \(f(x)\), we can differentiate it to obtain \(f'(x)\). It can be useful for many purposes to differentiate again and consider the second derivative of a function.

In functional notation, the second derivative is denoted by \(f''(x)\). In Leibniz notation, letting \(y=f(x)\), the second derivative is denoted by \(\dfrac{d^2 y}{dx^2}\).

The placement of the \(2\)'s in the notation \(\dfrac{d^2 y}{dx^2}\) may appear unusual. We consider that we have applied the differentiation operator \(\dfrac{d}{dx}\) twice to \(y\):

\[ \Bigl(\dfrac{d}{dx}\Bigr)^2 y = \dfrac{d^2}{dx^2}\,y = \dfrac{d^2 y}{dx^2}; \]

or that we have applied the differentiation operator \(\dfrac{d}{dx}\) to \(\dfrac{dy}{dx}\):

\[ \dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\Bigl(\dfrac{dy}{dx}\Bigr). \]

As we will see in the module Applications of differentiation , the second derivative can be very useful in curve-sketching. The second derivative determines the convexity of the graph \(y=f(x)\) and can, for example, be used to distinguish maxima from minima.

The second derivative can also have a physical meaning. For example, if \(x(t)\) gives position at time \(t\), then \(x'(t)\) is the velocity and the second derivative \(x''(t)\) is the acceleration at time \(t\). This is discussed in the module Motion in a straight line .


Find the second derivative of \(f(x) = x^2\).


We have \(f'(x) = 2x\), and so \(f''(x) = 2\).

This example implies that, if you ride your bike and your position is \(x^2\) metres after \(x\) seconds, then your acceleration is a constant 2 m/s\(^2\).


Let \(y = x^7 + 3x^5 + x^{\dfrac{3}{2}}\). Find \(\dfrac{d^2 y}{dx^2}\).


The first derivative is

\[ \dfrac{dy}{dx} = 7x^6 + 15 x^4 + \dfrac{3}{2} x^{\frac{1}{2}}, \]

so the second derivative is

\[ \dfrac{d^2 y}{dx^2} = 42 x^5 + 60 x^3 + \dfrac{3}{4} x^{-\frac{1}{2}}. \]
Exercise 18

Let \(f(x) = (x^2 + 7)^{100}\), as in exercise 11. What is \(f''(x)\)?

Exercise 19

Suppose the position of an object at time \(t\) is given by

\[ x(t) = 1 - 7t + (t-5)^4. \]

Show that \(x''(t) \geq 0\) for all \(t\), so that acceleration is always non-negative.

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