Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 4: More On The Derivative – Section 4.1: Higher-Order Derivatives

 

4.1
Higher-Order Derivatives

 

 

Return To Contents
Go To Problems & Solutions

 

1. Mathematical Induction

 

Principle Of Mathematical Induction

 

 

Putting To Practice

 

 

Example 1.1

 

 

Solution

EOS

 

Go To Problems & Solutions     Return To Top Of Page

 

2. Higher-Order Derivatives

 

Let y = f(x) be a differentiable function. If the derivative y' = f '(x) is itself differentiable, its derivative is called the
second derivative of y = f(x) and is denoted y'' or f ''(x) or d 2y/dx2 or (d 2/dx2) f(x):

 

 

 

Go To Problems & Solutions     Return To Top Of Page

 

3. Factorials

 

Let n be a positive integer. The factorial of n is defined as the product of all integers from 1 thru to n and is denoted by
n!, read “n factorial”, so that n! = 1 Χ 2 Χ 3 Χ ... Χ n. It's a product of n factors, hence the name factorial. The factorial of
0 is defined to be 1: 0! = 1 (there's a reason for this definition; it's not needed here). We have:

 

 

The factorial expansion is also written in decreasing order of the factors:

 

n! = n Χ (n – 1) Χ (n – 2) Χ … Χ 2 Χ 1.

 

Go To Problems & Solutions     Return To Top Of Page

 

4. Derivatives Of All Orders And Mathematical Induction

 

Example 4.1

 

Let f(x) = 1/x. Find enough derivatives of different orders of f to enable you to guess the general formula for f (n)(x),
where n is a positive integer. Then use mathematical induction to prove your guess.

 

Solution
We have:

 

 


EOS

 

Note the use of (–1)n to specify the sign. If n is even then we get the “ +” sign; if n is odd then we get the “ – ” sign. For
(–1)n+1, if n is even then n + 1 is odd, hence we get the “ – ” sign; if n is odd then n + 1 is even, hence we get the “ +”
sign.

 

Return To Top Of Page

 

Problems & Solutions

 

1.  Prove the following formula by using mathematical induction:

 

   

 

     where n is any positive integer. Evaluate:

 

    1 + 2 + ... + 5,000.

 

Solution

 

For n = 1 we have:

 

 

Return To Top Of Page

 

 

2.  Let f and g be twice-differentiable functions. Prove that ( fg)'' = f ''g + 2 f 'g' + fg''.

 

Solution

 

( fg)''   =   (( fg)')'   =   ( f 'g + fg' )'   =   ( f 'g)' + ( fg' )'   =   ( f ''g + f 'g' ) + ( f 'g' + fg'' )   =   f ''g + 2 f 'g' + fg''.

 

Return To Top Of Page

 

 

3.  Let f(x) = 1/(x + 2). Calculate enough derivatives of different orders of f to enable you to guess the general formula
     for f (n)(x), where n is any positive integer. Then use mathematical induction to prove your guess.

 

Solution

 

We have f(x) = (x + 2)–1. Then:

 

 

Return To Top Of Page

 

 

 

Solution

 

 

 

Return To Top Of Page

 

 

5.  Let f(t) = t2/3. Calculate enough derivatives of different orders of f to enable you to guess the general formula for
      f (n)(t), where n is any positive integer. Then use mathematical induction to prove your guess.

 

Solution

 

We have:

 

 

 

Return To Top Of Page     Return To Contents