Second Order Derivative

Introduction

In calculus, Derivatives are a basic tool in solving problems and differential equations. Derivatives are used as a tool by scientists to analyse dynamic systems to determine the change in variables, and use this to frame differential equations, and use integration to predict the future changes. To put simply, a second-order derivative is the two consequent derivatives of a function, or it is the derivative of the first order derivative of that function. In this tutorial, we will learn about derivatives, order of derivatives, second order derivatives, second order derivatives of parametric functions, and some solved examples.

Derivative

Derivative of a function which consists of real variables, measures the slope of that function or the changes in its values with respect to other variables. The output is referred to as function value whereas the input is referred to as argument. Derivative is denoted by $\mathrm{\frac{dy}{dx}}$ . Also written as $\mathrm{\frac{d}{dx}(y)}$ shows the change in y with respect to x. Derivative gives the slope of a function on a graph.

Order of Derivatives

gives the slope of the function, whereas the second order derivative gives the shape of for example, to measure the rate of change of distance over time, i.e., velocity, can be given by first order derivative of the function and the change in velocity, i.e., acceleration can be given by its second order derivative of that function. The first derivative of the function can be written as $\mathrm{\frac{dy}{dx}}$, while the second order derivative can be denoted by $\mathrm{\frac{d^{2}y}{dx^{2}}}$. The first order derivative the curve of the function.

Second Order Derivative

A special type of differential equation which is used to calculate derivatives of a function whose order is 2 is called second order derivative. It can be denoted as

$\mathrm{y"\:,\:d^{2}y\:/\:dx^{2}\:,\:y"(x )}$

A second order differential equation with variable coefficients can be written simply as $\mathrm{y"\:+\:p(x)y'\:+\:q(x)y\:=\:f(x)\:,\:Where\:p(x)\:,\:q(x)\:,\:and\:f(x)}$ are the functions of x in the equation. Whereas with constant coefficients, it can be given by $\mathrm{y"\:+\:py'\:+\:qy\:=\:0}$ .

Second Order Derivative of Parametric Functions

In some cases, it becomes complex to use functions to establish relations between independent and dependent variables. Therefore, a third variable is used to simplify the calculation. The third variable used is called a parameter, and the expression is called parametric equation.

Thus, now there are three variables x, y, and t. Since t is the parameter, it becomes the independent variable, and x and y become the dependent variables. Hence, x and y are represented as equations of t.

Therefore, the second order derivative of parametric function can be given by −

$$\mathrm{\frac{d^{2}y}{dx^{2}}\:=\:\frac{d}{dx}\:.\:\frac{dy}{dx}(t)}$$

But when we apply the derivative, the function is of the variable t and we are differentiating with x, by the chain rule the term $\mathrm{\frac{dt}{dx}}$ gets multiplied, to counter which we will multiply by $\mathrm{\frac{dx}{dt}}$.

$$\mathrm{\frac{d^{2}y}{dx^{2}}\:=\:(\frac{d^{2}y}{dx^{2}}\:.\:\frac{dt}{dx})\:.\:\frac{dx}{dt}}$$

Solved Examples

1. Evaluate the second order differential equation $\mathrm{b"\:-\:6b'\:+\:5b\:=\:0}$

Solution: Let us assume $\mathrm{b\:=\:e^{ra}}$ and find its first and second derivative − $\mathrm{b'\:=\:re^{ra}\:,\:b"\:=\:r^{2}e^{ra}}$

Next, substitute the values of $\mathrm{b,\:b'\:,\:and\:b"\:in\:b"\:-\:6b'\:+\:5b\:=\:0}$

$$\mathrm{r^{2}e^{ra}\:-\:6re^{ra}\:+\:5e^{ra}}$$

$$\mathrm{=\:e^{ra}(r^{2}\:-\:6r\:+\:5)}$$

$\mathrm{=\:r^{2}\:-\:6r\:+\:5}$

$\mathrm{\Longrightarrow\:(r\:-\:5)(r\:-\:1)\:=\:0}$

$\mathrm{\Longrightarrow\:r\:=\:1\:and\:5}$

The solution to given differential equation can be given by $\mathrm{b\:=\:Ae^{a}\:+\:Be^{5a}}$ because the roots of the differential equations are real.

2. Consider the given equation $\mathrm{q"\:-\:8q'\:+\:16q\:=\:0}$ and find the second derivative.

Solution − Let us suppose $\mathrm{q\:=\:e^{rp}}$ and first derivative is given by − $\mathrm{q'\:=\:re^{rp}}$ , and second derivative is given by − $\mathrm{q"\:=\:r^{2}e^{rp}}$

Next, substitute the values of $\mathrm{q\:,\:q'\:and\:q"\:in\:q"\:-\:8q'\:+\:16q\:=\:0}$ We have,

$$\mathrm{r^{2}e^{rp}\:-\:8re^{rp}\:+\:16e^{rp}\:=\:=}$$

$\mathrm{\Longrightarrow\:e^{rp}(r^{2}\:-\:8r\:+\:16)\:=\:0}$

$\mathrm{\Longrightarrow\:r^{2}\:-\:8r\:+\:16\:=\:0}$

$\mathrm{\Longrightarrow\:r^{2}\:-\:8r\:+\:16\:=\:0}$

$\mathrm{(r\:-\:4)(\:-\:4)\:=\:0}$

$\mathrm{r\:=\:4\:and\:4}$

The solution to given differential equation can be given by $\mathrm{q\:=\:Ae^{4p}\:+\:Bpe^{4p}}$ because the roots of the differential equations are identical and real.

Example 3: Consider the given equation $\mathrm{9q"\:+\:12q'\:+\:29q\:=\:0}$ , and find the second derivative.

Solution: Let us suppose $\mathrm{q\:=\:e^{rp}}$ and first derivative is given by − $\mathrm{q'\:=\:re^{rp}}$ , and second derivative is given by − $\mathrm{q"\:=\:r^{2}e^{rp}}$

Next, substitute the values of $\mathrm{q\:,\:q'\:,\:and\:q"\:in\:9q"\:+\:12q'\:+\:29q\:=\:0}$ We have,

$$\mathrm{9q"\:+\:12q'\:+\:29q\:=\:0}$$

$\mathrm{\Longrightarrow\:e^{rp}(9r^{2}\:+\:12qr\:+\:29\:=\:0)}$

$\mathrm{\Longrightarrow\:9r^{2}\:+\:12r\:+\:29\:=\:0\:\:\rightarrow\:Characteristic\:Equation}$

$\mathrm{\Longrightarrow\:r\:=\:[\frac{-12\:\pm\:\sqrt{12^{2}\:-\:4\:\times\:9\:\times\:29}}{2\:\times\:9}]}$

$\mathrm{\Longrightarrow\:r\:=\:(-2\:/\:3)\:\pm\:(5\:/\:3)i}$

The solution to given differential equation can be given by $\mathrm{q\:=\:e^{(-2\:3)x}[A\:\sin\:(5\:/\:3)x\:+\:B\:\cos\:(5\:/\:3)x]}$ , because the roots of the differential equations are complex numbers.

Conclusion

In this tutorial we learned about Differential Calculus, and Second Order Derivative of Functions and their significance. In calculus, Derivatives are a basic tool in solving problems and differential equations. Derivative of a function which consists of real variables measures the change of the output with respect to the input. The first derivative of the function can be written as $\mathrm{\frac{dy}{dx}}$, while the second order derivative can be denoted by $\mathrm{\frac{d^{2}y}{dx^{2}}}$. A special type of differential equation which is used to calculate derivatives of a function whose order is 2, is called a second order derivative.

FAQs

1. What does the first and second derivative tell us about the graph of a function?

The first order derivative gives the slope of the function, if the first order derivative is positive the graph of the function is increasing, negative means decreasing and neutral for 0. Whereas the second order derivative gives the shape of the curve of the function, if the second order derivative is positive then the graph is concave upward, and the point is local minima whereas if it is negative the graph is concave downward, and the point is local maxima and zero means no concavity and point is called inflection point.

2. Define a second order differential equation with constant coefficients?

A differential equation $\mathrm{y"\:+\:py'\:+\:qy\:=\:f(x)}$, where p, q denote constant coefficients is known as a second order differential equation with constant coefficients.

3. Give the equation for the second order derivative of parametric function.

$\mathrm{\frac{dy/dx}{dx/dt}}$ where dependent variables are “x” and “y”, whereas the independent variable variable is “t”

4. What is concave up and concave down in the second derivative of a function?

if the second derivative of a given function is greater than zero it is said to be concave up and if it is less than zero it is called concave down.

5. What is the point of inflection?

The second order derivative at this point is equal to zero. In other words, it is neither concave up or concave down at the point of inflection.