# Second Derivative

Sub Topics
Generally a derivative means the first derivative of a continuous function. It gives the instantaneous rate of change of a function at any point.

The second derivative of a function is the derivative of the first derivative of the same function.

Hence, it depicts the instantaneous rate of change of the derivative at the same point.

The concept of the second derivative can be better understood with a real life example.

When you are driving a car, the distance from the staring point changes with time. At any point, the instantaneous rate of change of distance with respect to time is called the instantaneous velocity at that point.

In other words, the instantaneous velocity is the derivative of the distance. The velocity can not be constant normally. Because you tend to accelerate or decelerate at times and hence the velocity changes accordingly. It means, the acceleration is a derivative of the velocity and in turn the second derivative of the distance with respect to time.

## Definition of Second Derivative

$(\frac{d^{2}y}{dx^{2}})$, known as Leibnitz’s notation and f ''(x) is known as Lagrange’s notation.

## Second Derivative Notation

The most common notations of a second derivative are

($\frac{d^{2}y}{dx^{2}}$), known as Leibnitz’s notation and f “(x) known as Lagrange’s notation.

Some times a function may vary with respect to more than one variable. In such a case, the derivative of the function with respect to any one of the variables is called a partial derivative. The partial derivative to the first partial derivative either with respect to the same variable or even with respect to the remaining variable is called the second partial derivative.

The most common notations of a second partial derivative are:

($\frac{\partial^{2} z}{\partial x^{2}}$), known as Leibnitz’s notation and fx “(x) known as Lagrange’s notation.

## Finding the Second Derivative

The second derivative of a function is the derivative of the first derivative of the same function. Hence the only method to find the second derivative of a function is to first find the first derivative and once again differentiate that.

For example, let us consider the function f(x) = $x^{3}$

The derivative of $x^{3}$  is $3x^{2}$, which is the first derivative of the function.

Now, the derivative of $3x^{2}$ is 6x, which is same as the second derivative of $x^{3}$.

## Second Derivative of ex

We have a unique function in calculus. That is the function f(x) = $e^{x}$. The peculiarity and the uniqueness of this function is it is the only function for which the derivative is the function itself. That is f'(x) = $e^{x}$. Consequently therefore, the second derivative of f(x) is also ex.

Somewhat close to this, are the second derivatives of sin (x) and cos (x). In these cases, the second derivative is the negative of the function itself.

For example, the derivative of sin (x) is cos (x) and the derivative of cos (x), which is the second derivative of sin (x), is –sin(x).

## Second Partial Derivative

As mentioned earlier, a function might vary with respect to more than one variable. In such a case, the derivative of the function with respect to any one of the variables is called a partial derivative. The partial derivative to the first partial derivative either with respect to the same variable or even with respect to the remaining variable is called the second partial derivative.

That is, if z = f(x, y) the first partial derivatives are, ($\frac{\partial z}{\partial x}$) = fx , and also, ($\frac{\partial z}{\partial y}$) = fy

The derivatives, ($\frac{\partial^{2} z}{\partial x^{2}}$), ($\frac{\partial^{2} z}{\partial y^{2}}$), [$\frac{\partial^{2} z}{(\partial x \partial y)}$] and [$\frac{\partial^{2} z}{(\partial y \partial x)}$] are all called second partial derivatives. For a continuous function, [$\frac{\partial^{2} z}{(\partial x \partial y)}$] = [$\frac{\partial^{2} z}{(\partial y \partial x)}$]
For example,  if z = x3 + y3 + x2y + xy2,

($\frac{\partial z}{\partial x}$) = 3x2 +2xy + y2 and ($\frac{\partial z}{\partial y}$) = 3y2 + x2 +2xy

($\frac{\partial^{2} z}{\partial x^{2}}$) = 6x + 2y and ($\frac{\partial^{2} z}{\partial y^{2}}$) = 6y + 2x

[$\frac{\partial^{2} z}{(\partial x \partial y)}$] = 2x + 2y and [$\frac{\partial^{2} z}{(\partial y \partial x)}$] = 2x + 2y

It may be seen [$\frac{\partial^{2} z}{(\partial x \partial y)}$] = [$\frac{\partial^{2}z}{(\partial y \partial x)}$] in this case.

## Second Derivative Test

As we mentioned earlier, when a function reaches a local maximum or minimum, the first derivative at that point is 0. On the other hand, we can say that when the first derivative of a function is 0 at a point, then that point must be a local maximum or minimum. But how will you make out which one is correct? The second derivative helps us here. Let us study in detail how it does this.

We have explained that when a function attains a local maximum, the function changes its state from increasing to decreasing.

Increasing means a positive slope and thereby a positive derivative at all points of increasing of function. Similarly decreasing means a negative slope and thereby negative derivative at all points of increasing of function.

Therefore we can say that at the point of the local maximum the first derivative changes its sign from positive to negative. It means the slope of the derivative becomes negative at that point. In other words, the second derivative is negative at the point of the local maximum.

By the same argument we can conclude that the second derivative is positive at the point of local minimum.

Let us illustrate the concept with an example.

Let, f(x) = x3 – 3x2 – 45x + 5

f’(x) = 3x2 – 6x – 45 = 3(x2 – 2x – 15) = 3(x – 5)(x + 3)

Now, f’(x) is 0 when x = 5 and when x = -3. At this stage we are at an uncertainty because,

Both the points may be local maximums, both the points may be local minimums or one point may be local maximum and the other may be local minimum. To resolve this, let us take the second derivative.

These points of uncertainty are called as critical points.

f’’(x) = derivative of 3x2 – 6x – 45 = 6x – 6 = 6(x – 1)

At x = 5, the second derivative f’’(5) = 6(5 – 1) = 24, which is positive. Therefore, the point x = 5 is a local minimum.

At x = -3, the second derivative f’’(5) = 6(- 3 – 1) = -24, which is negative. Therefore, the point x = -3 is a local maximum.
This is the concept of second derivative test and it is used to identify the local maxima or local minima when the critical points are known.

## Second Derivative Graph

Look at the above diagram. The figure (i) shows the graph of a function f(x) in an interval [-6, 6]. The graph indicates that the function is increasing in the interval [-6,-3] and again in the interval [5, 6]. The function is decreasing in the interval (-3, 5). It means the local maximum is at x = -3 and the local minimum is at x = 5.

The figure (ii) is the graph of the first derivative of the same function which is correctly shown to be positive in the intervals [-6, -3] and [5, 6]. It also correctly shows the first derivative to be negative in the interval (-3, 5).

The figure (iii) is the graph of the second derivative of the same function. The second derivative is negative at x = -3 confirming that x = -3 is a point of local maximum. Similarly, at x = 5, which is a local minimum, the second derivative is positive, confirming our conclusion.

Now we notice that the second derivative is 0 at x = 1. But at that point the function is in not changing its state and also the derivative is not 0 and shows negative in this case.

## Second Derivative Concavity

In the last section we had seen that at a particular point the second derivative was 0 but the graphs of the function and its first derivative did not indicate any criticality there. But if you keenly observe, the concavity of the graph changes at that point from concave down to concave up. Such a point is called the point of inflection.

In some cases, it may so happen that the first derivative of a function and the second derivative of the function may both be 0. An example is illustrated below.

The first graph represents a particular function. The second and third graphs represent the first derivative and the second derivative respectively, of the same function.

The graph of the function suggests that the function is increasing all through. But at point P the instantaneous rate of increase is 0 and this is truly reflected in the second graph. Had we not seen the graph how do we conclude about point P? Also the second derivative graph shows that the second derivative is also 0 at the point P. Hence the point P is neither a maximum nor a minimum but the concavity of the function changes there. Hence it is the point of inflection.

Now comparing the first and the third graphs, one can conclude that when the function is concave down, the second derivative is negative, and when the function is concave up; the second derivative is positive, and the second derivative is 0 at the point of inflection.

## Second Derivative of Parametric Equations

A function may be represented in the parametric form. That is, in terms of a set of parametric equations.

Let y = f(x) be a function and be represented by the set of parametric equations as,

x = x(t) and y = y(t)

The first derivative ($\frac{dy}{dx}$) = ($\frac{dy}{dt}$)*($\frac{dt}{dx}$), as per chain rule.

The right side can be rewritten in the quotient form. That is,

($\frac{dy}{dx}$) = $\frac{[(\frac{dy}{dt})]}{[(\frac{dx}{dt}]}$ = $\frac{[y'(t)]}{[x'(t)]}$

Now,     ($\frac{d^{2}y}{dx^{2}}$) = [$\frac{d{(\frac{dy}{dx})}}{dx}$]

= [$\frac{d{(\frac{dy}{dx})}}{dt}$]*[$\frac{dt}{dx}$]

= $\frac{[\frac{d{(\frac{dy}{dx})}}{dt}]}{[\frac{dx}{dt}]}$

[$\frac{d{(\frac{dy}{dx})}}{dt}$] = [$\frac{d{(\frac{y'(t)}{x'(t)})}}{dt}$] = $\frac{[x'(t)*y"(t)-x"(t)*y'(t)]}{[x'(t)]^{2}}$, by quotient rule and $\frac{dx}{dt}$ = x’(t)

Therefore,  ($\frac{d^{2}y}{dx^{2}}$) = $\frac{[\frac{d{(\frac{dy}{dx})}}{dt}]}{[\frac{dx}{dt}]}$

= ${\frac{[x'(t)*y''(t) – x''(t)*y'(t)]}{[x'(t)]^{2}}}$*${\frac{1}{x’(t)}}$

= $\frac{[x'(t)*y''(t) – x''(t)*y'(t)]}{[x'(t)]^{3}}$.