Second Order Derivatives

What is a second order derivative?

The second order derivative is the derivative of the derivative. If we differentiate once, we get $\frac{dy}{dx}$.  Differentiating again gives $\frac{d^2y}{dx^2}$, which is known as the second derivative.  It can also be written as $f''(x)$. The second derivative gives the rate of change of the derivative, or the gradient of the curve, with respect to $x$.   Important: The second derivative is less than zero at a local maximum. The second derivative is greater than zero at a local minimum. The second derivative is zero at a point of inflection.  

Displacement, velocity and acceleration

Imagine a car is travelling along a straight road.  At any given time, its distance from its starting point can be assigned to the value $x$.  Therefore, the rate with which this distance changes with respect to time is $\frac{dx}{dt}$ - the velocity. The rate at which their velocity changes with respect to time - the acceleration - can then be expressed as the second order derivative,  $\frac{d^2x}{dt^2}$. This is demonstrated in the diagram below.

Displacement, Velocity And Accceleration (binary/octet-stream)

In this example, the Displacement (Distance from starting position) vs Time graph is a quadratic function.   Differentiating this function should give us a linear function.  This is because when we differentiate a function containing an $x^2$ term, this will be come an $x$ term. This means what was previously a quadratic equation now becomes an equation of a line.  The second graph shows $\frac{dx}{dt}$, the velocity, as a function of time. The slope of the velocity vs time graph is $\frac{d^2x}{dt^2}$, the acceleration. As you may already know, the slope of a straight line is constant.  This is because when we differentiate the $x$ term, we get a constant.  Therefore, in this case, $\frac{d^2x}{dt^2}$ is a constant. As a result, the graph of $\frac{d^2x}{dt^2}$ as a function of time is simply a constant.

Example 1

Part 1: Without detailed calculation, draw a rough sketch of the graph of the slope of the function

$$f(x) = 2x^3 - 3x^2 -36x +4$$ Part 2: Use the differentiation rules you have learnt to find and plot the first and second derivatives of the above function. Answer: Part 1: $f(x)$ is a cubic function.  Therefore, the graph of its slope will be a quadratic curves.   Part 2: Use the rules you have learnt to differentiate and simplify the equation: $$f'(x) = (2 \times 3)x^{3-1} -  (2 \times 3)x^{2-1} - (6 \times 6)x^{1-1} = 6(x^2 - x - 6)$$ Remember that $x^{0}=1$ and that differentiating a constant always gives zero.  This expression can be factorised to give $$6(x-3)(x+2)$$ which is easily graphed as shown below

Quadratic Graph 2 (binary/octet-stream)

  Simply differentiate again to find the second derivative.  

$$f''(x) = (6)(2x^{2-1} - x^{1-1}) = 6 (2x-1)$$ which is graphed as shown below.    

Linear Small (binary/octet-stream)

Example 2

Question:A cyclist starts cycling at $t=0$ from $x=0$.  For the first ten seconds, her journey can be described by the equation

$$x=4\sqrt{5t} + 6t^2$$   i) Find the cyclist's velocity after 5 seconds. ii) Find her acceleration after   5 seconds. Answer: i) The cyclist's velocity is $\frac{dx}{dt}$, which can be found by differentiating the expression for $x$ found above. First, change the surd to an exponent. $$x= 4\sqrt{5} t^\frac{1}{2} + 6t^2$$ Now differentiate x with respect to t to find  the general equation for $\frac{dx}{dt}$. $$\frac{dx}{dt}=4\sqrt {5} (\frac{1}{2})t^{-\frac{1}{2}} + (6)(2)t = 2\sqrt {\frac{5}{t}} + 12t$$ The velocity after five seconds is found by subbing 5 in for $t$ in the above expression. $$\frac{dx}{dt}(5)= 2\sqrt {\frac{5}{5}} + 12(5)= 2+60 = 62ms^{-1}$$ Make sure to include the units!   ii) The acceleration can be found by differentiating the general expression for $\frac{dx}{dt}$. If $$\frac{dx}{dt}=2\sqrt {5}t^{-\frac{1}{2}} + 12t$$ then  $$\frac{d^2x}{dt^2}= 2\sqrt {5} (-\frac{1}{2})t^{-\frac{3}{2}} + 12 = -\sqrt {\frac{5}{t^3}}  + 12$$  The acceleration after 5 seconds can be found by subbing 5 into the general expression: $$\frac{d^2x}{dt^2}= -\sqrt {\frac{5}{5^3}}  + 12=-\sqrt {\frac{5}{125}}  + 12=- {\frac{1}{5}}  + 12 = \frac{59}{5}ms^{-2}$$

Bike 1877387 960 720 (binary/octet-stream)