Second order derivatives

In a nutshell

Sometimes a function can be differentiated more than once. In particular, the second derivative represents the rate of change of the gradient. Recall that the gradient is the first derivative and represents the rate of change of the original function. Differentiating the first derivative is done using the same rules as differentiating a function.

Notation

The derivative of the expression $y=f(x)$ with respect to $x$ gives $\dfrac{\text{d}y}{\text{d}x}=f'(x)$ . You can think of the left-hand side as having applied $\dfrac{\text{d}}{\text{d}x}$ to $y$ . This makes it easier to see why when you differentiate this again, you get

$\boxed{\dfrac{\text{d}^2y}{\text{d}x^2}=f''(x)}$

Pay attention to the position of the "squared" parts. Also notice the double dash on the right-hand side. This indicates the function has been differentiated twice with respect to $x$ .

Example 1

Find the second order derivative of the equation below with respect to $x$ .

$y=5x^3-x^2+6x-3+\dfrac4{x}$ 

Start by expressing fractions with index notation:

$y=5x^3-x^2+6x-3+4x^{-1}$ 

The first derivative is

$\dfrac{\text{d}y}{\text{d}x}=15x^2-2x+6-4x^{-2}$ 

The second derivative if found by differentiating again:

$\underline{\dfrac{\text{d}^2y}{\text{d}x^2}=30x-2+8x^{-3}}$ 

This can also be given as

$\underline{\dfrac{\text{d}^2y}{\text{d}x^2}=30x-2+\dfrac8{x^{3}}}$ 

Displacement, velocity and acceleration

You may have a function of displacement in terms of time, $s(t)$ . If you differentiate this with respect to $t$ , you obtain a function for velocity in terms of time:

$\boxed{\dfrac{\text{d}s}{\text{d}t}=v(t)}$

If you differentiate again with respect to $t$ , you have a function for acceleration in terms of time:

$\boxed{\dfrac{\text{d}v}{\text{d}t}=a(t)}$

In other words, acceleration is the second order derivative of the function for displacement with respect to time:

$\boxed{\dfrac{\text{d}^2s}{\text{d}t^2}=a(t)}$

Note:  $s(t)$ , $v(t)$ and $a(t)$ means displacement, velocity and acceleration respectively are being represented by a function given in terms of time.

Example 2

The displacement of a projectile can be mapped with the function

$s=-t^2-5t+6$ 

where $s$ is the displacement in metres and $t$ is the time in seconds. Find an expression for the acceleration of the projectile for this journey.

To use a function for displacement to find a function for acceleration, you must differentiate twice with respect to time. Differentiating once gives a function for velocity:

$\dfrac{\text{d}s}{\text{d}t}=v=-2t-5$ 

Differentiating again gives a function for acceleration:

$\dfrac{\text{d}^2s}{\text{d}t^2}=\dfrac{\text{d}v}{\text{d}t}=\underline{a=-2}$ 

Therefore, the acceleration of this projectile is constant at $\underline{-2\ m/s^2}$ .

Note: Remember that acceleration and velocity are vector quantities so they have a magnitude and direction. Therefore, they can take on negative values.

Example 3

Find the acceleration after $3$ seconds of a vehicle whose displacement in metres from its start position is given by the function

$s(t)=t^3+2t^2$ 

where $t$  is the time in seconds since the start of the journey.

The function for displacement needs to be differentiated twice with respect to time:

$\dfrac{\text{d}s}{\text{d}t}=v(t)=3t^2+4t\\\space\\\dfrac{\text{d}^2s}{\text{d}t^2}=a(t)=6t+4$ 

Thus acceleration after $3$ seconds is found by inserting $t=3$ into the second derivative:

$\underline{a(3)=6(3)+4=22\text{ ms}^{-2}}$ 