Second Derivatives

For certain functions, it is possible to differentiate twice and find the derivative of the derivative. It is often denoted as f or \frac{d^2y}{dx^2}. For example, given that f(x)=x^7-x^5 then the derivative is f and the second derivative is given by f.

How to classify stationary points:

The second derivative can tell us something about the nature of a stationary point. Suppose that we have found the x-coordinates of all of the stationary points by solving f(x)=0. For a minimum, the gradient changes from negative to 0 to positive, i.e. the gradient is increasing. If the gradient is increasing then the gradient of the gradient is positive, i.e. f. For a maximum, the gradient changes from positive to 0 to negative, i.e. the gradient is decreasing. If the gradient is decreasing then the gradient of the gradient is negative, i.e. f. By putting the x-coordinates of the stationary points into f, we can classify whether they are minima or maxima by determining whether the second derivative is positive or negative at those x-coordinates.

ExampleFind and classify the stationary points of f(x)=x^3-2x^2+x-5.

We locate the stationary points by solving f. f'(x) is given by

f

We can solve f'(x)=0 by factorising:

(3x-1)(x-1)=0
which gives x=1/3 or x=1. The corresponding y coordinates are \left(\frac{1}{3}\right)^3-2\left(\frac{1}{3}\right)^2+\frac{1}{3}-5=-\frac{131}{27} (don’t be afraid of strange fractions) and (1)^3-2(1)^2+1-5=-5. Hence, the stationary points are at (1/3,-131/27) and (1,-5). We can classify the stationary points by substituting the x coordinate of the stationary point into the second derivative and seeing if it is positive or negative. Differentiating a second time gives
f. It follows that f which is less than 0, and hence (1/3,-131/27) is a MAXIMUM. f and (1,-5) is a MINIMUM.

Stationary Points

Stationary points are the points on a curve where the gradient is 0. This means that at these points the curve is flat. Usually, the gradient of a curve is always changing and so the gradient is only 0 instantaneously (unless the curve is a flat line, in which case, the gradient is always 0).

Recall the graph of y=x^2. The vertex at the bottom of the curve is a STATIONARY POINT. In this case, there is a stationary point at (0,0).

Now consider the general positive quadratic in the form y=ax^2+bx+c where a\textgreater 0. There is a stationary point at the bottom of the curve; this is called a MINIMUM. Now consider a negative quadratic of the form y=ax^2+bx+c where a\textless 0. There is a stationary point at the top of the curve; this is called a MAXIMUM.

A stationary point can be found by solving \frac{dy}{dx}=0, i.e. finding the x coordinate where the gradient is 0. dy/dx is found by differentiating.

Example – Find the stationary points on the curve y=\frac{2}{3}x^3-5x^2+8x-4.

Start by solving \frac{dy}{dx}=0. \frac{dy}{dx}=2x^2-10x+8=0 i.e. x^2-5x+4=0. Factorising gives (x-4)(x-1)=0 and so the x coordinates of the stationary points are x=4 and x=1. Substituting these into the y equation gives the coordinates of the stationary points as (4,-28/3) and (1,-1/3).