## 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).