Second Order Derivative

For understanding second order derivative, let us step back a bit and understand what a first derivative is. The first derivative  \( \frac {dy}{dx} \) represents the rate of the change in y with respect to x. Considering an example, if the distance covered by a car in 10 seconds is 60 meters, then the speed which is actually the first order derivative of the distance travelled with respect to time. Hence the speed in this case is given as \( \frac {60}{10} m/s \).

Now we know that speed also varies and does not remain constant forever. Now to measure this rate of change in speed we come across second derivative. The second order derivative is nothing but the derivative of the derivative of the given function. So, the variation in speed of the car can be found out by finding out the second derivative, i.e. the rate of change of speed with respect to time (second derivative of distance travelled with respect to the time).

Graphically the first derivative represents the slope of the function at a point and second derivative describes how the slope changes over the independent variable in the graph. For a function having a variable slope the second derivative explains the curvature of the given graph.

Second Order Derivative

In this graph the blue line indicates the slope i.e. the first derivative of the given function. And the second derivative is used to define the nature of the given function. For example, we use the second derivative test to determine the maximum, minimum or the point of inflexion.

Mathematically, if \( y = f(x) \)

Then \( \frac {dy}{dx}\) = f'(x)

Now if f'(x) is differentiable, then differentiating \( \frac {dy}{dx} \) again w.r.t. x  we get 2nd order derivative, i.e.

\( \frac {d}{dx} \left( \frac {dy}{dx} \right) \) = \( \frac {d^2y}{dx^2}\) = f”(x)

Similarly, higher order derivatives can also be defined in the same fashion like \( \frac {d^3y}{dx^3}\)  represents a third order derivative\( \frac {d^4y}{dx^4}\) ,  represents a fourth order derivative and so on.

Let us see an example to get acquainted with second order derivatives.

Example 1: Find \( \frac {d^2y}{dx^2}\) if y = \( e^{(x^3)} – 3x^4 \)

Solution 1: Given that y = \( e^{(x^3)} – 3x^4 \), then differentiating this equation w.r.t. x we get,

\(~~~~~~~~~~~~~~\)\( \frac {dy}{dx} = e^{(x^3)} ×3x^2 – 12x^3 \)

Now to find the 2nd order derivative of the given function, we differentiate the first derivative again w.r.t. x,

\(~~~~~~~~~~~~~~\)\( \frac {d^2y}{dx^2}\) = \(e^{(x^3)} × 3x^2 × 3x^2 + e^{(x^3)}  × 6x – 36x^2 \)

\(~~~~~~~~~~~~~~\)\(  \frac{d^2y}{dx^2} \) = \( xe^{(x^3)} × (9x^3 + 6 ) – 36x^2 \)

This is the required solution.

Example 2:Find \( \frac {d^2y}{dx^2}\)  if y = 4 \( sin^{-1}(x^2) \)

Solution 2:Given that y = 4 \( sin^{-1}(x^2) \) , then differentiating this equation w.r.t. x we get,

\( \frac {dy}{dx} \)=\( \frac {4}{\sqrt{1 – x^4}} × 2x \)

Now for finding the next higher order derivative of the given function, we need to differentiate the first derivative again w.r.t. x ,

\(~~~~~~~~~~~~~~\)\( \frac {d^2y}{dx^2} \) = \( 2x × \frac {d}{dx}\left( \frac {4}{\sqrt{1 – x^4}}\right) + \frac {4}{\sqrt{1 – x^4}} \frac{d(2x)}{dx} \)         (using  \( \frac {d(uv)}{dx} \) = \( u \frac{dv}{dx} + v \frac {du}{dx}\))

\(~~~~~~~~~~~~~~\)\( \frac {d^y}{dx^2} \) = \( \frac {-8(x^4 + 1)}{(x^4 – 1)\sqrt{1 – x^4}} \)<

We hope it is clear to you how to find out second order derivatives. To learn more about differentiation, download Byjus- the learning app


Practise This Question

d2xdy2 equals: