2.3 Product and Quotient Rules - Kennesaw State University

[Pages:8]2.3. PRODUCT AND QUOTIENT RULES

75

2.3 Product and Quotient Rules

2.3.1 Product rule

Suppose that f and g are two di?erentiable functions. Then (f (x) g (x))0 = f 0 (x) g (x) + f (x) g0 (x)

See 2.3.5 on page 77 for a proof. Example 124 Find x2ex 0

x2ex 0 = x2 0 ex + x2 (ex)0 (product rule) = 2xex + x2ex = xex (2 + x)

2.3.2 Quotient rule

Suppose that f and g are two di?erentiable functions. Then

f (x) 0 f 0 (x) g (x) f (x) g0 (x)

= g (x)

(g (x))2

See 2.3.5 on page 77 for a proof.

Example

125

Find

y0

for

y

=

x + ex x2 .

Since this function is a quotient, we ...rst apply the quotient rule.

y0 =

x + ex 0 x2

(x + ex)0 x2 (x + ex) x2 0

=

(x2)2

(1 + ex) x2 (x + ex) (2x)

=

x4

x2 + x2ex 2x2 2xex

=

x4

x2 + xex (x 2)

=

x4

x + ex (x 2)

=

x3

2.3.3 More Examples

All the applications we studied earlier which used the derivative can now be done more quickly by using the rules of di?erentiation. We look at some examples.

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CHAPTER 2. RULES OF DIFFERENTIATION

Example 126 Find (xex)0 and determine where xex is increasing. This function is a product of two functions, so we ...rst apply the product rule.

(xex)0 = (x)0 ex + x (ex)0 = (1) ex + xex = ex (1 + x)

A function is increasing where its derivative is positive. Since ex is always

positive, it follows that (1 + x) must also be positive, that is 1 + x > 0 or x > 1. So, xex is increasing on the interval ( 1; 1).

Example

127

Find

y0

for

y

=

x + ex x2 .

Since this function is a quotient, we ...rst apply the quotient rule.

y0 =

x + ex 0 x2

(x + ex)0 x2 (x + ex) x2 0

=

(x2)2

(1 + ex) x2 (x + ex) (2x)

=

x4

x2 + x2ex 2x2 2xex

=

x4

x2 + xex (x 2)

=

x4

x + ex (x 2)

=

x3

Example 128 Find f 00 (x) for f (x) = x3 By de...nition, f 00 (x) = (f 0 (x))0. So, we must ...rst ...nd f 0 (x).

f 0 (x) = 3x2 (power rule)

Therefore

f 00 (x) = 3x2 0 = 3 x2 0 (constant multiple rule) = 6x

Example 129 The position of an object is given by h (t) = 16t2 + 64t + 6 where h is in feet and t in seconds. Find the instantaneous velocity of the object for t = 1, t = 2. The instantaneous velocity of the object is h0 (t). We compute it using the rules of di? erentiation.

h0 (t) = 16t2 + 64t + 6 0

= 32t + 64

2.3. PRODUCT AND QUOTIENT RULES

77

Now, the instantaneous velocity at t = 1 is h0 (1) = 32 ft/s. The instantaneous velocity at t = 2 is h0 (2) = 0 ft/s.

2.3.4 Summary of the Rules Learned So Far

Let c and n denote any constants. Suppose that f and g are two di?erentiable functions. We have the following rules::

1. (c)0 = 0

2. (x)0 = 1

3. (cf (x))0 = c (f (x))0

4. (xn)0 = nxn 1

5. (f (x) g (x))0 = f 0 (x) g0 (x) (sum and di?erence rules)

6. (f (x) g (x))0 = f 0 (x) g (x) + f (x) g0 (x) (product rule)

0

f (x)

g (x) f 0 (x) f (x) g0 (x)

7.

=

g (x)

(g (x))2

(quotient rule)

8. (ex)0 = ex

Remark When computing derivatives, you should always try to simplify the function ...rst. You should also try to simplify your answer as much as possible. Chances are that after having computed a derivative, you will have to use it. You may need to ...nd where it is 0, positive, negative. If it has been simpli...ed, these tasks will be easier to perform.

2.3.5 Proofs of the Rules Learned So Far

1. Proof of the constant rule. Let f (x) = C. We want to prove that f 0 (x) = 0. By de...nition, we have

f 0 (x) = lim f (x + h)

h!0

h

CC = lim

h!0 h

= lim (0)

h!0

=0

f (x)

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CHAPTER 2. RULES OF DIFFERENTIATION

2. Proof of dx = 1 dx

Let f (x) = x. Then, by de...nition

f 0 (x) = lim f (x + h)

h!0

h

x+h x

= lim

h!0

h

h = lim

h!0 h

=1

f (x)

3. Proof of the power rule. Note: though the rule is true for any constant n, we only prove it in the case n is a positive integer. Let f (x) = xn, where n is a constant. Then, by de...nition

f 0 (x) = lim f (x + h)

h!0

h

(x + h)n

= lim

h!0

h

f (x) xn

This computation is a little bit more di? cult than the previous ones, we do it in several steps. First,

(x + h)n = xn + nxn 1h + n (n 1) xn 2h2 + ::: + nxhn 1 + hn 2

Therefore

(x + h)n xn = xn + nxn 1h + n (n 1) xn 2h2 + ::: + nxhn 1 + hn xn 2

= nxn 1h + n (n 1) xn 2h2 + ::: + nxhn 1 + hn 2

= h nxn 1 + n (n 1) xn 2h + ::: + nxhn 2 + hn 1 2

It follows that

f (x + h) h

h nxn 1 + n (n 1) xn 2h + ::: + nxhn 2 + hn 1

f (x) =

2

h

= nxn 1 + n (n 1) xn 2h + ::: + nxhn 2 + hn 1 2

Therefore

f (x + h)

lim

h!0

h

f (x) = lim

nxn 1 + n (n

1) xn 2h + ::: + nxhn 2 + hn 1

h!0

2

= nxn 1

Since all the other terms contain h as a factor.

2.3. PRODUCT AND QUOTIENT RULES

79

4. Proof of the constant multiple rule.

Let C be a constant and suppose that f is di?erentiable, prove that (Cf (x))0 = Cf 0 (x). Since we are assuming that f is di?erentiable, it

means that f 0 (x) exists and f 0 (x) = lim f (x + h)

f (x) . Let us de...ne

h!0

h

F (x) = Cf (x). We need to prove that F 0 (x) = Cf 0 (x). By de...nition,

F 0 (x) = lim F (x + h) F (x)

h!0

h

Cf (x + h) Cf (x)

= lim

(since F (x) = Cf (x))

h!0

h

C ((x + h) f (x))

= lim

(factor C)

h!0

h

f (x + h) f (x)

= C lim

(since C is a constant)

h!0

h

= Cf 0 (x) (de...nition of the derivative)

5. Proof of the sum rule

Suppose that f and g are two di?erentiable functions. Let F (x) = f (x) + g (x). We want to prove that F 0 (x) = f 0 (x) + g0 (x). By de...nition

F 0 (x) = lim F (x + h) F (x)

h!0

h

f (x + h) + g(x + h) f (x)

= lim

h!0

h

f (x + h) f (x) + g(x + h)

= lim

h!0

h

g(x) (since F (x) = f (x) + g (x) ) g(x)

f (x + h) f (x) g(x + h) g(x)

= lim

+

h!0

h

h

f (x + h) f (x)

g(x + h) g(x)

= lim

+ lim

(property of limits)

h!0

h

h!0

h

= f 0 (x) + g0 (x) (de...nition of the derivative)

6. Proof of the di?erence rule The proof is very similar to the proof of the sum rule. We leave it as an exercise.

7. Proof of the product rule Suppose that f and g are two di?erentiable functions. Let F (x) = f (x) g (x). We want to prove that F 0 (x) = f 0 (x) g (x) + f (x) g0 (x). By de...nition

F 0 (x) = lim F (x + h) F (x)

h!0

h

f (x + h) g (x + h)

= lim

h!0

h

f (x) g (x) (de...nition of F )

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CHAPTER 2. RULES OF DIFFERENTIATION

It helps when we know the result to prove. We are trying to obtain f 0 (x) g (x)+f (x) g0 (x). If we think in terms of the de...nition of the derivative, that means that in the numerator we must have g (x) (f (x + h) f (x))+ f (x) (g (x + h) g (x)) that is, if we expand it, we realize that we must have in the numerator g (x) f (x + h) g (x) f (x)+f (x) g (x + h) g (x) g (x). This is more terms than what we currently have in the numerator. This means that we need to insert new terms. Of course, we cannot simply insert what is missing, we would change the problem. We use the standard trick of adding something and subtracting it immediately. This way we haven't changed anything. We then rearrange the terms. We decide to insert the following term: g (x + h) f (x) + g (x + h) f (x) g (x + h) f (x). You can see that this term is simply 0, so we have not changed anything. Here is what we obtain after inserting this term:

F 0 (x) = lim f (x + h) g (x + h)

h!0

g (x + h) (f (x + h) = lim

h!0

g (x + h) f (x) + g (x + h) f (x) h

f (x)) + f (x) (g(x + h) g(x)) h

f (x) g (x)

g (x + h) (f (x + h) f (x)) f (x) (g(x + h) g(x))

= lim

+

h!0

h

h

g (x + h) (f (x + h) f (x))

f (x) (g(x + h) g(x))

= lim

+ lim

(property of limits)

h!0

h

h!0

h

Let us evaluate each limit separately.

g (x + h) (f (x + h) f (x))

f (x + h) f (x)

lim

= lim g (x + h) lim

h!0

h

h!0

h!0

h

Since g is di?erentiable, by a theorem studied in class, we also know that it is continuous. Therefore, lim g (x + h) = g (x). The second limit is, by

h!0

de...nition, f 0 (x). So,we see that

lim g (x + h) (f (x + h) f (x)) = f 0 (x) g (x)

h!0

h

Similarly, we have

f (x) (g(x + h) lim

g(x)) = f (x) g0 (x)

h!0

h

and therefore

F 0 (x) = f 0 (x) g (x) + f (x) g0 (x)

8. Proof of the quotient rule Suppose that f and g are two di?erentiable functions. Let F (x) = f (x) . g (x)

2.3. PRODUCT AND QUOTIENT RULES

81

We

want

to

prove

that

F 0 (x)

=

f 0 (x) g (x) f (x) g0 (x)

(g (x))2

.

By

de...nition

F 0 (x) = lim F (x + h)

h!0

h

f (x + h)

F (x) f (x)

= lim g (x + h) g (x)

h!0

h

f (x + h) g (x) f (x) g (x + h)

= lim

(reduced to the same denominator

h!0

hg (x + h) g (x)

and combined fractions)

As in the proof of the product rule, we realize that we are missing terms to obtain what we need. So, we insert some extra terms. This time, we insert f (x) g (x) + f (x) g (x). We obtain

F 0 (x) = lim f (x + h) g (x) f (x) g (x) + f (x) g (x) f (x) g (x + h)

h!0

hg (x + h) g (x)

g (x) (f (x + h) f (x)) f (x) (g(x + h) g(x))

= lim

h!0

hg (x + h) g (x)

g (x) (f (x + h) f (x))

f (x) (g(x + h) g(x))

= lim

lim

h!0 hg (x + h) g (x)

h!0 hg (x + h) g (x)

We evaluate each limit separately. As in the proof of the product rule, lim g (x + h) = g (x). So, we have

h!0

g (x) (f (x + h) f (x))

f (x + h) f (x)

lim

= g (x) lim

h!0 hg (x + h) g (x)

h!0 hg (x + h) g (x)

f (x + h) f (x)

1

= g (x) lim

lim

h!0

h

h!0 g (x + h) g (x)

f 0 (x) = g (x) (g (x))2

f 0 (x) g (x) = (g (x))2

Similarly, we obtain that

f (x) (g(x + h) g(x))

g(x + h) g(x)

lim

= f (x) lim

h!0 hg (x + h) g (x)

h!0 hg (x + h) g (x)

g(x + h) g(x)

1

= f (x) lim

lim

h!0

h

h!0 g (x + h) g (x)

=

f

(x)

g0

(x)

(g

1 (x))2

f (x) g0 (x) = (g (x))2

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CHAPTER 2. RULES OF DIFFERENTIATION

And therefore, combining the two limits (don't forget the minus sign in between), we obtain the desired result.

F0

(x)

=

f0

(x) g

(x) f (x) g0 (g (x))2

(x)

2.3.6 Problems

10 1. Find a rule for xn 2. Find a rule for ( pn x)0

3. Prove the di?erence rule.

4. Be able to do problems such as # 1, 3, 5, 7, 11, 13, 17, 21, 27, 37, 42, 44, 48 on pages 198-200.

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