section 2.3 product & quotient rules and higher-order derivatives

Section 2.3Product & Quotient Rules and Higher-Order Derivatives

Unit 2 – Differentiation

Objectives:

1. Find the derivative of a function using the Product Rule.2. Find the derivative of a function using the Quotient Rule.3. Find the derivative of a trigonometric function.4. Find a higher-order derivative of a function.

Theorem 2.7The Product Rule

» The product of two differentiable functions and is itself differentiable. Moreover, the derivative of is the derivative of the first function times the second, plus the first times the derivative of the second function.

» This can be extended to more functions as so:

𝑑𝑑𝑥 [ 𝑓 (𝑥 )𝑔 (𝑥)]= 𝑓 ′ (𝑥 )𝑔 (𝑥 )+ 𝑓 (𝑥 )𝑔 ′ (𝑥)

𝑑𝑑𝑥 [ 𝑓 (𝑥 )𝑔 (𝑥 )h(𝑥)]= 𝑓 ′ (𝑥)𝑔 (𝑥 )h (𝑥 )+ 𝑓 (𝑥 )𝑔 ′ (𝑥)h (𝑥 )+ 𝑓 (𝑥 )𝑔 (𝑥 )h′ (𝑥)

Example 1

» Find the derivative of the following:

𝑓 (𝑥) 𝑔 (𝑥)

Example 1 continued…

» Find the following derivative:

h ′ (𝑥 )=−24 𝑥2−24 𝑥+10Since the two function pieces were polynomials, we can also find the derivative without using the product rule.

Example 2

» Find the derivative of the following :

𝑓 (𝑥)𝑔 (𝑥)

𝑑𝑦𝑑𝑥 =

𝑑𝑑𝑥 [ 4 𝑥2 ]sin (𝑥 )+4 𝑥2 𝑑𝑑𝑥 [sin (𝑥)]−3 𝑑

𝑑𝑥 [cos (𝑥)]

Example 3

» Find the derivative of the following :

Here it is necessary to use the several rules together.¿ { ¿¿Product Rule Constant Multiple Rule

¿ Difference Rule

¿8 𝑥 sin (𝑥 )+4 𝑥2cos (𝑥 )+3sin (𝑥)

This answer can be represented multiple ways, & when compared to an answer (like from the back of the book), it could be given differently.


» Alternative way to write the answer:

Factor GCF of

Theorem 2.8The Quotient Rule

» The quotient of two differentiable functions and is itself differentiable at all values of for which . Moreover, the derivative of is given by the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

Example 4

» Find the derivative of the following:𝑑𝑑𝑥 [ 𝑓 (𝑥 )

𝑔(𝑥 ) ]= 𝑓 ′ (𝑥 )𝑔 (𝑥 )− 𝑓 (𝑥 )𝑔 ′ (𝑥)

[𝑔 (𝑥)]2,𝑔 (𝑥)≠0

Example 5

» Find an equation of the tangent line to the graph of at .

𝑓 ′ (1 )=

4(1 )2− 21 −2

(1−4 )2=4−2−2

(−3 )2=09=0



𝑓 ′ (1 )=−2 (1 )2−2 (1 )+4

[ (1 )2−4 (1 ) ]2=−2−2+4

(−3 )2=09=0

𝑓 (𝑥 )=𝑥 (2+ 1𝑥 )𝑥 (𝑥−4 )

=2 𝑥+ 𝑥𝑥𝑥2−4 𝑥

= 2𝑥+1𝑥2−4 𝑥

Alternatively, the function is multiplied by which clears the fraction on top.



𝑓 ′ (1 )=0 𝑠𝑙𝑜𝑝𝑒⇒

𝑚=0

𝑓 (1 )=2+ 111−4

=3−3

=−1𝑝𝑜𝑖𝑛𝑡⇒

(1 ,−1 )

So, is the equation of the line tangent to the function at .

Example 6

» Find the derivative of the following using both the Quotient Rule & the Constant Multiple Rule:

First with the Quotient Rule.


» Find the derivative of the following using both the Quotient Rule & the Constant Multiple Rule:

Now with the Constant Multiple Rule.

Theorem 2.9Derivatives of Trigonometric Functions

To aid in memorizing these, remember the derivatives of the cofunctions (cosine, cotangent, and cosecant) all require negatives in front.

Example 7

» Prove the trigonometric derivatives using the Quotient Rule.

It will always be to your advantage to memorize these rules for speed and ease in working problems. However, it is important to remember that even though you may forget a rule, they can easily be derived using the Quotient Rule & a few trigonometric identities.

𝑓 (𝑥 )=tan (𝑥 )= sin (𝑥)cos (𝑥)

)


» Prove the trigonometric derivatives using the Quotient Rule.

𝑓 (𝑥 )=csc (𝑥 )= 1sin(𝑥)

The proofs of the other two derivatives are similar.

Example 8

» Find the derivatives of the following:

𝑦 ′=−sin (𝑥 )+sec2(𝑥)

𝑦 ′=−3 𝑥 csc (𝑥 )cot (𝑥)+3 csc (𝑥)

Example 9

» Differentiate both forms of the following:


» Differentiate both forms of the following:

Now to differentiate it in its other form:

Higher-Order Derivatives

» In section 2.2 we discussed the position function, We also discussed the derivative of this function, which is the velocity function. Now, the velocity function, just like any other function has a derivative as well, and this is called the acceleration function, denoted by .

Position functionVelocity functionAcceleration function

Higher-Order Derivatives

The acceleration function is called the second derivative of the position function and is an exampled of a higher-order derivative. Below are how higher-order derivatives are denoted:

Example 10

» Because the moon has no atmosphere, a falling object on the moon encounters no air resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall at the same rate on the moon. The position function for each of these falling objects is given by where is the height in meters and is the time in seconds. What is the ratio of Earth’s gravitational force to the moon’s?

Moon:

Earth:

Eart h ′ s gravitational forceMoo n ′ s gravitational force

=−9.8−1.62

≈6.05

Interpreting a Derivative from a Graph

» Once again, one of the ways the AP exam will test the concept of a derivative rather than just your memorization of rules will be in how it asks you to interpret derivatives from a graph. The following example will illustrate one way on how this is achieved.

Example 11

» The graphs of functions and are shown below. If and find the following:

f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

A. and B. require no more than finding the slopes of the functions at & as that is what the derivative means.

¿12

¿−2



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

C. and D. are much trickier. At first glance one might think to find C. you would multiply the slopes of & at , and similarly, divide them for D. at Unfortunately, this would not be right.



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

For C. we will first differentiate the function as defined above using the Product Rule:

This gives us a formula to use.



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

Now for , is the slope of , is the y-value of , is the y-value of , and is the slope of .



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

For D. we will differentiate the function as defined above using the Quotient Rule:



f

g1 2 3 4 5 6–1–2–3–4–5–6 x

1

2

3

4

5

6

–1

–2

–3

–4

–5

–6

y

Example 12A

A. If and , which is closest to

B. 0.016C. 1.0D. 5.0E. 8.0F. 32.0

Here is an example of a particular multiple choice AP test question. This question would be on the non-calculator portion of the test. What it really wants us to find is an approximation using the average rate of change (the slope of the secant line).

Example 12B

B. A differentiable function has the values shown. Estimate

Again we want to find just the average rate of change, but we will find it between the x-values of 1.4 and 1.6.

𝑓 ′ (1 .5 )≈ 𝑓 (1.6 )− 𝑓 (1.4 )1.6−1.4

=22−140.2

= 80.2

=4 0

1.0 1.2 1.4 1.6

8 10 14 22

section 2.3 product & quotient rules and higher-order derivatives

Documents

continued example

following example

product quotient rules

higherorder derivativesunit

product ruleexample

quotient ruleexample

function pieces

multiple ways