In this Module Project we will explore patterns that arise in repeated different
ID: 1721039 • Letter: I
Question
In this Module Project we will explore patterns that arise in repeated differentiation. Below you will find three functions. What we are going to do is repeatedly differentiate these functions and try to find a pattern we can use to describe the “nth” derivative. Here are a couple of useful tools for expressing your final results: The factorial of a positive integer n is: n!=n x (n-1) x (n-2) '.....' 1
So for example 2!=2 x 1,3!=3 x 2 x 1, 4! 4 x 3 x 2 x 1 As im sure you noticed it is clear n!= n x (n-1)!
An alternating sequence {-1, 1, -1, 1, -1, …} may be written as ( ) , where is the number of the term in the sequence starting with n = 1. (first term, second term, etc.)
For each of the following functions, find a formula for the “nth” derivative of the function. As a means to get started, do this for n = 1 (first derivative), 2 (second derivative) and 3 (third derivative), and maybe even n = 4, then look for a pattern. Leave negative exponents in each derivative in preparation for computing the next one.
A) f(x)= 1/x
B) f(x)= 1/1+x
C) f(x) squareroot x
D) f(x)= e^x
Answers please with how you did them so I can see the work Im still kind of confused with dervivatives.
Explanation / Answer
We will write Dn (x) for the nth derivative of f(x).
1) f(x) =1/x
Df(x) =-1/x2
D2 f(x) = 1.2/x3
D3 f(x) = -1.2.3/x4
D4 f(x) = 1.2.3.4/x5
The pattern is
Dn f(x) = (-1)n n!/x(n+1)
1) f(x) =(1+x)-1
Df(x) =-1(1+x)-2
D2 f(x) = 1.2(1+x)-3
D3 f(x) = -1.2.3 (1+x)-4
D4 f(x) = 1.2.3.4 (1+x)_5
The pattern is
Dn f(x) = (-1)n n!(1+x)_(n+1)
C) f(x) = x1/2
Df(x) = 1/2 x_1/2
D2 f(x) = (1/2)(-1/2) x-3/2
D3 f(x) = (1/2)(-1/2)(-3/2) x-5/2
More generally
Dn f(x) = (-1)n-1(1/2)n [1.3.5...(2n-3)/2] x-(2n-1)/2
D) Dn(ex) = ex for all n