© 2025 Perplex Learning Incorporated. All rights reserved.
All content on this website has been developed independently from and is not endorsed by the International Baccalaureate Organization. International Baccalaureate and IB are registered trademarks owned by the International Baccalaureate Organization.
Discussion
Take the derivative of the derivative of f(x)=x2.
To find the derivative of the derivative of f(x)=x2, we first differentiate f(x):
The formula for the derivative of a power function is:
Applying this to x2:
Now, differentiate f′(x)=2x again:
So, the derivative of the derivative of f(x)=x2 is 2.
The derivative of the derivative of a function is called the second derivative, we can write f′′(x)=2. The second derivative gives us the rate of change of the slope of a function, which helps us further interpret graphs.
Second Derivative
The derivative of the derivative of a function is its second derivative:
Discussion
Let's look at some examples with the second derivative to understand its significance graphically.
First, we already saw that if f(x)=x2, then f′′(x)=2. Following a similar calculation, we can show that if
g(x)=−x2, then g′′(x)=−2.
Powered by Desmos
Use these familiar functions to distinguish between a positive and negative second derivative. Focus on how the slope changes from left to right.
For f(x)=x2 or a positive second derivative:
Its graph is an upward-facing parabola with vertex at (0,0).
– To the left of the vertex the curve slopes downwards, so f′(x)<0.
– At the vertex the slope is zero.
– To the right of the vertex the curve slopes upwards, so f′(x)>0.
As we move from left to right, f′(x) goes from negative through zero to positive. In other words, the slope is increasing as x increases.
For g(x)=−x2 or a negative second derivative:
Its graph is a downward-facing parabola with vertex at (0,0).
– To the left of the vertex it slopes upwards, so g′(x)>0.
– At the vertex the slope is zero.
– To the right of the vertex it slopes downwards, so g′(x)<0.
Here g′(x) goes from positive through zero to negative as x increases, so the slope is decreasing.
Finally, recall from stationary-point analysis that if the slope changes from decreasing to increasing we get a minimum, and if it changes from increasing to decreasing we get a maximum. Since f′′>0 means slope goes from negative to positive (decreasing→increasing), f has a minimum at its stationary point; since g′′<0 means slope goes from positive to negative (increasing→decreasing), g has a maximum at its stationary point.
We call graphs (or intervals of graphs) concave up when the second derivative is positive. The upward-facing
U-shape of f(x)=x2 is typical of concave up curves which bend upward to the left and right of minimum values. Similarly, we call graphs and intervals concave down when the second derivative is negative. The downward-facing U-shape of g(x)=−x2 is typical of concave down curves which bend downward to the left and right of maximum values.
Concavity
We determine concavity by the sign of f′′:
Powered by Desmos
Checkpoint
State when f(x)=4x3−5x2+6x−1
is concave up.
is concave down.
Select the correct option
Discussion
Suppose f′(a)=0. Classify the stationary point if
f′′(a)>0.
f′′(a)<0.
Part (a) (i)
Since f′′(a)>0, f′ is increasing at x=a. With f′(a)=0, this means for x just less than a,
so f is decreasing on the left of a, and for x just greater than a,
so f is increasing on the right of a. Hence the curve goes from decreasing to increasing at x=a, and the stationary point at a is a local minimum.
Part (a) (ii)
Since f′′(a)<0, f′ is decreasing at x=a. With f′(a)=0, for x<a close to a we have
so f is increasing just before a, and for x>a close to a
so f is decreasing just after a. Hence the stationary point at x=a is a local maximum.
Classifying stationary points using the second derivative
At a stationary point (f′(a)=0),
If f′′(a)>0, then f has a local minimum at x=a.
If f′′(a)<0, then f has a local maximum at x=a.
Using the second derivative to classify a stationary point is often called the second derivative test.
Powered by Desmos
Checkpoint
It is given that f′(x)=x2−6x.
Determine the value of x at which f has a local maximum.
Select the correct option
Discussion
A collection of functions are shown below. Each function has a single point highlighted.
Powered by Desmos
What do the graphs have in common at the highlighted points?
At the highlighted point on each graph, the concavity of the function changes from concave down to concave up. This means the second derivative changes sign from negative to positive at that point. Points where the concavity changes from down to up (or vice versa) called a point of inflexion.
For f(x), the graph transitions from bending downward (concave down) to bending upward (concave up) at the highlighted point. This is the classic inflection point at the origin.
For g(x), the curve bends downward (concave down) for 0<x<π/2, and bends upward (concave up) for π/2<x<π. The highlighted point at x=π/2 is where the concavity changes.
For h(x), although the curve does not have a clear U-shape, you can observe that to the left of the highlighted point, the slope is positive but decreasing, and to the right, the slope is increasing again. This change in the behavior of the slope also indicates a change in concavity from down to up.
In summary, at the highlighted point on each curve, the second derivative changes from negative to positive.
Inflexion Points
Inflexion points occur when f′′(x)=0 and f′′(x) changes sign. 🚫
Powered by Desmos
Now that we have fully explored the graphical interpretation of both the first and second derivative, we can determine a lot about the graph of f from the graphs of f′ and f′′:
Graphs of f, f' and f''
Powered by Desmos
When f′ crosses the x-axis f has a maximum (f′′<0) or minimum (f′′>0)
When f′′ crosses the x-axis, f has an inflexion point.
Chat