2.01: Prerequisites to Numerical Differentiation
( \newcommand{\kernel}{\mathrm{null}\,}\)
Lesson 1: Prerequisites to Numerical Differentiation
Learning Objectives
After successful completion of this lesson, you should be able to:
1) recall the exact definition of differentiation
2) derive exact derivatives of a few functions
Introduction
In this lesson, we will review the necessary concepts of differential calculus to learn numerical differentiation. These include the concepts of finding exact derivatives of functions.
The derivative of a function represents the rate of change of a variable with respect to another variable. For example, the velocity of a body is defined as the rate of change of the location of the body with respect to time. The location is the dependent variable, while time is the independent variable. Now, if we measure the rate of change of velocity with respect to time, we get the acceleration of the body. In this case, velocity is the dependent variable, while time is the independent variable.
Whenever differentiation is introduced to a student, the two concepts of the secant line and tangent line (Figure 2.1.1.1) are revisited.
![Function curve with tangent and secant lines.](https://math.libretexts.org/@api/deki/files/97718/02.01.graph2.png?revision=1&size=bestfit&width=684&height=684)
Let P and Q be two points on the curve, as shown in Figure 1. The secant line is the straight line drawn through P and Q.
![Curve of a function f(x) with secant line drawn through points P and Q on the curve, which are located at x-values of a and a+h respectively.](https://math.libretexts.org/@api/deki/files/97719/Chp0201Calculationofthesecantline.png?revision=1)
The slope of the secant line (Figure 2.1.1.2) is then given as
mPQ,secant=f(a+h)−f(a)(a+h)−a=f(a+h)−f(a)h(2.1.1.1)
As Q moves closer and closer to P, the limiting portion is called the tangent line. The slope of the tangent line mPQ,tangent then is the limiting value of mPQ,secant from Equation (2.1.1.1) as h→0.
mPQ,tangent=limh→0f(a+h)−f(a)h(2.1.1.2)
Find the slope of the secant line of the curve y=4x2 between points (3,36) and (5,100).
Solution
The slope of the secant line between (3,36) and (5,100) from Equation (2.1.1.1) is
m=f(5)−f(3)5−3=100−365−3=32
Find the slope of the tangent line of the curve y=4x2 at point (3,36).
Solution
The slope of the tangent line at (3,36) from Equation (2.1.1.2) is m=limh→0f(3+h)−f(3)h=limh→04(3+h)2−4(3)2h=limh→04(9+h2+6h)−36h=limh→036+4h2+24h−36h=limh→0h(4h+24)h=limh→0(4h+24)=24
![Secant line for the function y=4x^2, drawn through the points (3, 36) and (5, 100).](https://math.libretexts.org/@api/deki/files/97720/Chapter02.01Lesson100Figure3.png?revision=1&size=bestfit&width=965&height=526)
Derivative of a Function
Recall from calculus, the derivative of a function f(x) at x=a is defined as the slope of the tangent line at x=a
f′(a)=limh→0f(a+h)−f(a)h(2.1.1.3)
Find f′(3) if f(x)=4x2.
Solution
f′(x)=4(2x)=8x=24
![Graph of the function y = 4x^2 and the tangent line to the function at x=3.](https://math.libretexts.org/@api/deki/files/97721/Chapter02.01Lesson100Figure4.png?revision=1&size=bestfit&width=616&height=336)
Second Definition of Derivatives
There is another form of the definition of the derivative of a function. The derivative of the function f(x) at x=a is defined as
f′(a)=limx→af(x)−f(a)x−a(2.1.1.4)
As x→a, the definition is nothing but the slope of the tangent line at P.
![Graph showing the second definition of the derivative.](https://math.libretexts.org/@api/deki/files/97722/Chp0201Graphshowingtheseconddefinitionofthederivative.png?revision=1)
Find f′(3) if f(x)=4x2 by using the form
f′(a)=limx→af(x)−f(a)x−a
of the definition of a derivative.
Solution
f′(3)=limx→3f(x)−f(3)x−3=limx→34x2−4(3)2x−3=limx→34x2−36x−3=limx→34(x2−9)x−3=limx→34(x−3)(x+3)x−3=limx→34(x+3)=4(3+3)=24
Higher-Order Derivatives
So far, we have limited our discussion to calculating the first derivative, f′(x) of a function f(x). What if we are asked to calculate higher-order derivatives of f(x). A simple example of this is finding the acceleration of a body from a function that gives the location of the body as a function of time. The derivative of the location with respect to time is the velocity of the body, followed by the derivative of velocity with respect to time being the acceleration. Hence, the second derivative of the location function gives the acceleration function of the body.
Given f(x)=3x3−2x−7, find the second derivative, f′′(x), and the third derivative, f′′′(x). Also, find the value of f′′(2) and f′′′(2).
Solution
Given
f(x)=3x3−2x−7
we have
f′(x)=3(3x2)−2=9x2−2
f′′(x)=ddx(f′(x))=ddx(9x2−2)=9(2x)=18x
f(2)=18×2=36
f′′′(x)=ddx(f′′(x))=ddx(18x)=18
f′′′(2)=18
Other Notations of Derivatives
Derivatives can be denoted in several ways. For the first derivative, the notations include
f′(x), ddxf(x), y′,˙y, and dydx
For the second derivative, the notations include
f′′(x), d2dx2f(x), y′′, ¨y, and d2ydx2
For the nth derivative, the notations include
f(n)(x), dndxnf(x), y(n), and dnydxn
Lesson 2: Equation of a Tangent Line
Learning Objectives
After successful completion of this lesson, you should be able to:
1) find the equation of the tangent line to a function.
Finding equations of a tangent line
One of the numerical methods used to solve a nonlinear equation is called the Newton-Raphson method. This method is based on the prerequisite knowledge of finding the tangent line to a curve at a point. Let us look at an example to illustrate finding the equation of the tangent line to a curve.
Find the equation of the line tangent to the function
f(x)=x3−0.165x+3.993×10−4
at x=0.05.
Solution
The line tangent is a straight line of the form
y=mx+c
To find the equation of the tangent line, let us first find the slope m of the straight line.
f′(x)=3x2−0.165
At x=0.05f′(0.05)=3(0.05)2−0.165=−0.1575
Hencem=−0.1575
To find the value of the y-intercept c of the straight line, we first find the value of the function at x=0.05.
f(0.05)=(0.05)3−0.165(0.05)+3.993×10−4=−0.0077257
The tangent line passes through the point (0.05,−0.0077257), so
−0.0077257=m(0.05)+c
−0.0077257=−0.1575(0.05)+c
c=0.0001493
![Graph of function f (x) and the tangent line at x = 0.05.](https://math.libretexts.org/@api/deki/files/97723/Chapter02.01Lesson200Figure1.png?revision=1&size=bestfit&width=842&height=459)
Hence,
y=mx+c=−0.1575x+0.0001493
is the equation of the tangent line.
Multiple Choice Test
(1). The definition of the first derivative of a function f(x) is
(A) f′(x)=f(x+Δx)+f(x)Δx
(B) f′(x)=f(x+Δx)−f(x)Δx
(C) f′(x)=limΔx→0f(x+Δx)+f(x)Δx
(D) f′(x)=limΔx→0f(x+Δx)−f(x)Δx
(2). Given y=5e3x+sinx, dydx is
(A) 5e3x+cosx
(B) 15e3x+cosx
(C) 15e3x−cosx
(D) 2.666e3x−cosx
(3). Given y=sin2x, dydx at x=3 is most nearly
(A) 0.9600
(B) 0.9945
(C) 1.920
(D) 1.989
(4). Given y=x3lnx, dydx is
(A) 3x2lnx
(B) 3x2lnx+x2
(C) x2
(D) 3x
(5). The velocity of a body as a function of time is given as v(t)=5e−2t+4, where t is in seconds, and v is in m/s. The acceleration in m/s2 at t=0.6 s is
(A) −3.012
(B) 5.506
(C) 4.147
(D) −10.00
(6). If x2+2xy=y2, then dydx is
(A) x+yy−x
(B) 2x+2y
(C) x+1y
(D) −x
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/02dif/quiz_02dif_background_solution.pdf