 ## 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.