The derivative of a function is also a function. If f(x) is differentiable then f'(x) is also a function of x. If f'(x) is also differentiable, then we can find its derivative function as well which is denoted by f"(x). Thus the process of differentiation can be continued successively as long as each derivative in the sequence is differentiable. The higher order derivatives so found by repeated differentiation play an important role in the study and application of Calculus

The second derivative is the derivative function of the first derivative and it is denoted by f"(x). Second derivative is used determining the Extreme values of the functions and its concavity.
In general, the second derivative can be interpreted as the rate of change of rate of change. If s (t) the position of a moving particle a function of t, then the velocity v(t) is given by the first derivative s'(t) and the acceleration a(t) by the second derivative s"(t).

The second derivative of an at least twice differentiable function is the function got by differentiating the given function twice in succession.

If f is a differentiable function and if its derivative f' is also differentiable, then the second derivative f" = f=(f')'.

The prime notation used to indicate the second derivative of y = f(x) is either y" or f "(x).

Leibniz notation for the same is $\frac{d^{2}y}{dx^{2}}$

Other alternate notations are D2 f(x) or D2 y.
Using Leibniz notation the definition of second derivative can be shown as
$\frac{d^{2}y}{dx^{2}}$ = $\frac{d}{dx}(\frac{dy}{dx})$

The process of finding the second derivative is rather simple and consists of only two steps.
  1. Find the derivative function f'(x)
  2. Differentiate f'(x) to get the second derivative f"(x).

Example:
Find the second derivative of f(x) = 2x3 - 3x2 + 4x +5
Step 1: Find the first derivative f'(x)
Differentiating the given polynomial function using power rule
f'(x) = 6x2 - 6x + 4
Step 2: Differentiate f'(x) to get the second derivative
We again use the power formula to differentiate f'(x).
The second derivative f"(x) = 12x - 6.

Second Derivative Test to check the concavity of the graph of a function
Suppose f(x) is a twice differentiable function on an Interval I
  1. If f"(x) > 0 on I, then the graph of f(x) over the interval I is concave up.
  2. If f"(x) < 0 on I, then the graph of f(x) over the interval I is concave down.

The point where the concavity changes is called a point of inflection.

Second derivative Test to find the Local Extremum:
If f"(x) is continuous in an open interval that contain x = c,
  1. f'(c) = 0 and f"(c) < 0 ⇒ The function f has a local maximum at x = c
  2. f'(c) = 0 and f"(c) > 0 ⇒ The function f has a local minimum at x = c
  3. f'(c) = 0 and f"(c) = 0 ⇒ The test fails.
When the graph of the second derivative of a function is analyzed some information about the function itself are got.
  1. A function f(x) is concave up wherever the graph of the second derivative is above the x axis.
  2. f(x) is concave down where ever the graph of the second derivative is below the x axis.
  3. The points where the graph of the second derivative crosses the x axis are possible inflection points.
  4. The Inflection points may also occur where the graph of the second derivative display infinite jump, from positive to negative or vice-versa.

let us analyze two Second derivative graphs and infer the behavior of the original function.

  In the adjoining graph f"(x) is positive in the intervals
(-1,0) and (1, ∞) and negative in the intervals (-∞, -1) and
(0, 1).
Hence the function f(x) is concave up in (-1,0) and (1, ∞)
and concave down in ( -∞, -1) and (-1,0). f(x) and
also has possible points of inflection at x = -1, 0 and 1.
   
In the adjoining graph of second derivative we observe that
f"(x) alternate signs between successive intervals of
length π. Positive in (0, π) and negative in (-π, 0) and
(π, 2π).
Hence f(x) is concave up in (0, π) and concave down in
(-π,0) and (π, 2π).
The points where f"(x) is undefined and result in infinite
jumps of the graph from top to down and vice versa
like x = -π, 0, π and 2π are possible inflection points
 Second Derivatives

Thus if the acceleration function is given, then velocity and position functions can also be known roughly.

Functions are often defined using parameters, meaning both the variables x and y are defined as functions of a third variable say t.

If x = g(t) and y = f(t) the first derivative $\frac{dy}{dx}$ is given by the rule,
$\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ = $\frac{f'(t)}{g'(t)}$
The second derivative $\frac{d^{2}y}{dx^{2}}$ is got by using the formula,

$\frac{d^{2}y}{dx^{2}}$ = $\frac{d}{dx}(\frac{dy}{dx})$ = $\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$

It is wrong to use $\frac{d^{2}y}{dx^{2}}$ = $\frac{\frac{d^{2}y}{dt^{2}}}{\frac{d^{2}x}{dt^{2}}}$

Solved Example

Question: Find $\frac{d^{2}y}{dx^{2}}$ if x = t2 and y = t3 - 3t.
Solution:
 
Differentiating the two equations with respect to t we get

               $\frac{dx}{dt}$ = 2t and $\frac{dy}{dt}$ = 3t2 - 3

               $\frac{dy}{dx}$ = $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

                                       = $\frac{3t^{2}-3}{2t}$                                         First derivative.

               Now to find the second derivative we use the formula,
               $\frac{d^{2}y}{dx^{2}}$$\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$

                                                   = $\frac{\frac{2t(6t)-(3t^{2}-3)2}{4t^{2}}}{2t}$

                                                   = $\frac{9t^{2}+6}{8t^{3}}$   

 


Solved Examples

Question 1: Find the second derivative of y = x .cos x
Solution:
 
$\frac{dy}{dx}$ = - x sin x + cox                                                  First derivative found using the product rule for derivatives

    $\frac{d^{2}y}{dx^{2}}$ = - x cos x - sin x - sin x                                    Second derivative found by differentiating  $\frac{dy}{dx}$

            =  - x cos x - 2sin x
 

Question 2: Find the second derivative of the function f(x) = tan-1 (x2)
Solution:
 
f'(x) = $\frac{1}{1+(x^{2})^{2}}$ . 2x                            First derivative found using chain rule
                                     = $\frac{2x}{1+x^{4}}$

                              f"(x) = $\frac{(1+x^{4})2-2x(4x^{3})}{(1+x^{4})^{2}}$                  Second derivative found using quotient rule.

                                     = $\frac{-6x^{4}+2}{(1+x^{4})^{2}}$

Let us solve one problem where we find the second derivative using implicit differentiation.

 

Question 3: Find the second derivative x2 + y2 = 16
Solution:
 
Differentiating the equation implicit with respect to x,
                                         2x + 2y $\frac{dy}{dx}$ = 0

                                           x + y $\frac{dy}{dx}$ = 0

                                         $\frac{dy}{dx}$ = $\frac{-x}{y}$

   Differentiating again implicit with respect to x,

                      $\frac{d^{2}y}{dx^{2}}$ = $\frac{y(-1)-(-x)\frac{dy}{dx}}{y^{2}}$

                                                          = $\frac{-y+x(\frac{dy}{dx})}{y^{2}}$

                                                          = $\frac{-y+x(\frac{-x}{y})}{y^{2}}$

                                                          = $\frac{-(x^{2}+y^{2})}{y^{3}}$

                                                          = $\frac{-16}{y^{3}}$                            Second derivative simplified.