DERIVATIVES UNIT PROBLEM SETS

DERIVATIVES UNIT PROBLEM SETS PROBLEM SET #1 – Rate of Change ***Calculators Not Allowed*** Find the average rate of change for each function betwee...
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DERIVATIVES UNIT PROBLEM SETS PROBLEM SET #1 – Rate of Change

***Calculators Not Allowed***

Find the average rate of change for each function between the given values:

1.

𝑦 = π‘₯ 2 + 4π‘₯ + 3 from π‘₯ = βˆ’2 to π‘₯ = 3

2.

𝑓 (π‘₯ ) = π‘₯ 3 βˆ’ π‘₯ 2 + 1 from π‘₯ = βˆ’2 to π‘₯ = 2

3.

𝑔(π‘₯) = 𝑠𝑖𝑛π‘₯ from π‘₯ = to π‘₯ = πœ‹

4.

β„Ž(π‘₯) = 2𝑠𝑖𝑛π‘₯π‘π‘œπ‘ π‘₯ from π‘₯ = to π‘₯ =

5.

π‘Ÿ(𝑑) = βˆšπ‘‘ 2 βˆ’ 9 from 𝑑 = βˆ’5 and 𝑑 = 3

6.

𝑦 = log10 𝑑 from 𝑑 = 10 to 𝑑 = 100

πœ‹ 2

πœ‹

3πœ‹

4

4

NJCTL.org

7.

𝑦 = 2 log 5 𝑑 from 𝑑 = 1 to 𝑑 = 25

8.

𝑔(𝑑) = 𝑒 𝑑 + 5 from 𝑑 = 0 and 𝑑 = 1

9.

𝑦 = |4 βˆ’ π‘₯ 2 | from π‘₯ = 0 to π‘₯ = 3

10.

𝑓(π‘₯) =

11.

Andrew is a physics student testing the rate of change of objects he can throw. Given his calculations, if he throws the baseball from the top of a hill, it follows the equation π‘₯(𝑑) = βˆ’4.9𝑑 2 + 14.7𝑑 + 25. He wants to know the average rate of change of the ball for each of the following time periods: (Calculator allowed)

π‘₯+2 π‘₯βˆ’2

from π‘₯ = 3 to π‘₯ = 8

a) 𝑑 = 0 π‘‘π‘œ 𝑑 = 1.5

b) 𝑑 = 1.5 π‘‘π‘œ 𝑑 = 3

c) 𝑑 = 1 π‘‘π‘œ 𝑑 = 3 NJCTL.org

PROBLEM SET #2 – Slope of a Curve

***Calculators Not Allowed***

For problems #1-8, find the limit of the function at the given point: 1.

a) Find the derivative of the function 𝑦 = 4π‘₯ + 1.

b) What is the value of the derivative at π‘₯ = 1 ?

2.

a) Find 𝑓 β€² (π‘₯) if the function is 𝑓(π‘₯) = 2π‘₯ 2 .

b) What is the value at π‘₯ = 2 ?

3.

a) Find the derivative of 𝑔(π‘₯) = 3π‘₯ 3 + 1.

b) What is the value of 𝑔′ (1) ?

4.

a) Find

𝑑𝑦 𝑑π‘₯

of the function 𝑦 = (π‘₯ + 1)2

b) What is the slope at π‘₯ = βˆ’1 ? NJCTL.org

5.

a) Find β„Žβ€² (π‘₯) if β„Ž(π‘₯) = 2π‘₯ 3 + 3π‘₯ 2 .

b) What is the slope at π‘₯ = βˆ’2 ?

6.

a) Find

𝑑𝑓 π‘‘π‘Ÿ

if 𝑓(π‘Ÿ) = πœ‹π‘Ÿ 2 .

b) What is the slope at π‘Ÿ = 3 ?

7.

a) Find the derivative of 𝑦 = √π‘₯.

b) What is the slope at π‘₯ = 2 ?

8.

Of the functions you have worked with, which type of functions have the same Δ𝑦 average rate of change as their instantaneous rate of change? (i.e. = 𝑦′) Try taking Ξ”π‘₯ the average rate of changes of the examples above. a) Linear b) Quadratic c) Cubic d) Square Root NJCTL.org

PROBLEM SET #3 – Derivative Rules 1.

Find the derivative of the function 𝑦 = 11.

2.

Find 𝑓 β€² (π‘₯) if the function is 𝑓(π‘₯) = 4π‘₯ 2 .

3.

a) Find the derivative of 𝑔(π‘₯) = 12π‘₯ 3 βˆ’ 4π‘₯ 2 + 2π‘₯.

***Calculators Not Allowed***

b) What is the value of 𝑔′ (βˆ’1) ?

𝑑𝑦

for the function 𝑦 = (π‘₯ + 2)2.

4.

Find

5.

a) Find β„Žβ€² (π‘₯) if β„Ž(π‘₯) = 2π‘₯ 3/4 .

𝑑π‘₯

b) What is the slope at π‘₯ = 16 ?

NJCTL.org

6.

a) Find

𝑑𝑓 𝑑𝑑

if 𝑓(𝑑) = 3βˆšπ‘‘ .

b) What is the slope at 𝑑 = 2 ?

7.

a) Find the derivative of 𝑦 = (2π‘₯ + 1)(2π‘₯ βˆ’ 1) .

b) What is the slope at π‘₯ = 2 ?

8.

a) Find 𝑔′ (π‘₯) if 𝑔(π‘₯) = (π‘₯ 2 + 4π‘₯ + 1)2 .

b) What is the slope at π‘₯ = 0 ?

9.

a) Find

𝑑𝑦 𝑑π‘₯

13

if 𝑦 = √π‘₯ . 2

b) What is the slope at π‘₯ = 8 ? NJCTL.org

PROBLEM SET #4 – Higher Order Derivatives

***Calculators Not Allowed***

1.

Find the 1st and 2nd derivative of the function 𝑦 = 20π‘₯ .

2.

Find 𝑓 β€² (π‘₯) and 𝑓 β€²β€² (π‘₯) of the function 𝑓(π‘₯) = 13π‘₯ 2 + 4π‘₯ .

3.

Find the 1st and 2nd derivative of 𝑔(π‘₯) = (π‘₯ + 4)3 .

4.

Find

5.

Find β„Žβ€²β€²β€² (π‘₯) if β„Ž(π‘₯) =

𝑑𝑦 𝑑π‘₯

and

𝑑2𝑦 𝑑π‘₯ 2

of the function 𝑦 = (π‘₯ βˆ’ 4)2 .

1 24

1

π‘₯4 βˆ’ π‘₯3 6

NJCTL.org

𝑑2𝑓

if 𝑓(𝑑) = βˆšπ‘‘ .

6.

Find

7.

Find the derivative of 𝑦 = √π‘₯ .

8.

Find the 1st and 2nd derivatives of 𝑓(π‘₯) = π‘₯ 𝑏

𝑑𝑑 2

3

a) where 𝑏 = 2

b) where 𝑏 = 3

c) generically for all 𝑏 > 3

9.

Looking back at problem 8c above, we begin to see a pattern with derivatives. As we take derivatives, our exponents are used as coefficients and multiplied together. If we were to continue taking derivatives until the exponent is zero, we can see that the last coefficient is equal to the factorial (!) of the original power (any derivative after that will equal zero). If we stopped at any other time along that path, we would have a permutation of the exponents equal to π‘Ž 𝑃𝑑 where β€œa” is the original exponent and β€œd” is the specific derivative. Using this method, find the following derivatives: (calculator OK)

a) 8th derivative of π‘₯ 9

b) 10th derivative of 2π‘₯ 10

NJCTL.org

PROBLEM SET #5 – Trigonometry Rules 1.

***Calculators Not Allowed***

a) Find the derivative of the function 𝑦 = 𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ .

πœ‹

b) What is the value of the derivative at π‘₯ = ? 2

πœ‹

c) What is the value of the derivative at π‘₯ = ? 6

2.

a) Find 𝑓 β€² (π‘₯) if the function is 𝑓(π‘₯) =

1 𝑠𝑖𝑛π‘₯

+

1 π‘π‘œπ‘ π‘₯

.

πœ‹

b) What is the value of the derivative at ? 4

πœ‹

c) What is the value of the derivative at ? 3

3.

a) Find the derivative of 𝑔(π‘₯) = 3π‘‘π‘Žπ‘›π‘₯ βˆ’ 4π‘π‘œπ‘‘π‘₯ .

πœ‹

b) What is the value of 𝑔′ ( ) ? 6

4.

a) Find

𝑑𝑦 𝑑π‘₯

of the function 𝑦 =

1 π‘π‘œπ‘‘π‘₯

+

2 π‘π‘œπ‘‘π‘₯

.

πœ‹

b) What is the derivative at π‘₯ = ? 4

NJCTL.org

5.

a) Find β„Žβ€² (π‘₯) if β„Ž(π‘₯) = βˆ’π‘π‘ π‘π‘₯ + 𝑠𝑒𝑐π‘₯ .

b) What is the slope at π‘₯ =

5πœ‹ 6

?

6.

Find the first four derivatives of 𝑦 = 4𝑠𝑖𝑛π‘₯ .

7.

Find the first four derivatives of 𝑔(π‘₯) = 5π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯

8.

Find

9.

a) Find the derivative of 𝑦 = 2 sinβˆ’1 π‘₯ .

π‘‘π‘š 𝑑π‘₯

if π‘š(π‘₯) = tanβˆ’1 π‘₯ + cot βˆ’1 π‘₯ .

b) What is the value of the derivative at π‘₯ = 0.

NJCTL.org

PROBLEM SET #6 – Product/Quotient Rule

***Calculators Not Allowed***

1.

Find 𝑓 β€² (π‘₯) if the function is 𝑓(π‘₯) = (π‘₯ + 2)(π‘₯ βˆ’ 2) .

2.

Find

3.

Find the derivative of 𝑔(π‘₯) = √π‘₯(π‘₯ 4 βˆ’ 3π‘₯ 2 βˆ’ 10π‘₯ + 1) .

4.

Find β„Žβ€² (π‘₯) if β„Ž(π‘₯) = √π‘₯ π‘‘π‘Žπ‘›π‘₯ .

5.

Find 𝑦′ if 𝑦 = π‘π‘œπ‘ π‘₯ βˆ™ 𝑠𝑖𝑛π‘₯ .

6.

Find 𝑓 β€² (π‘₯) if𝑓(π‘₯) = sin2 π‘₯ .

𝑑𝑦 𝑑π‘₯

of the function is 𝑦 = (π‘₯ 2 + 4π‘₯ βˆ’ 3)(3π‘₯ 2 βˆ’ 10) .

3

NJCTL.org

𝑠𝑖𝑛π‘₯

Find the derivative of 𝑦 =

8.

Find 𝑔′ (π‘₯) if 𝑔(π‘₯) =

9.

Find

10.

Using any prior rules, find the 1st and 2nd derivatives of 𝑦 = π‘‘π‘Žπ‘›π‘₯ .

11.

**Show using product rule that the derivative of sin3 π‘₯ is 3 sin2 π‘₯π‘π‘œπ‘ π‘₯ .

𝑑𝑦 𝑑𝑑

if 𝑦 =

𝑒 𝑑 +1 𝑑3

π‘₯ 2 +1 π‘₯ 2 βˆ’1

π‘π‘œπ‘ π‘₯

+

π‘π‘œπ‘ π‘₯

7.

𝑠𝑖𝑛π‘₯

(without using trig shortcuts).

.

.

NJCTL.org

PROBLEM SET #7 Derivatives Using Tables

1.

***Calculators Allowed***

x

f(x)

f’(x)

g(x)

g’(x)

-2

-19

16

-11

19

-1

-6

10

-2

4

0

1

4

-1

-1

1

2

-2

-2

6

2

-3

-8

1

7

Given β„Ž(π‘₯) = 𝑓(π‘₯) βˆ™ 𝑔(π‘₯)

find: β„Žβ€² (2)

find: β„Žβ€² (1)

2.

Given 𝑗(π‘₯) = 2π‘₯ 2 βˆ™ 𝑓(π‘₯)

find: 𝑗 β€² (0)

find: 𝑗 β€² (2)

3.

Given π‘˜(π‘₯) =

𝑓(π‘₯) 𝑔(π‘₯)

find: π‘˜ β€² (βˆ’1)

find: π‘˜ β€² (1)

4.

Given π‘š(π‘₯) = 𝑔(π‘₯)2

find: π‘šβ€² (0)

find: π‘šβ€² (βˆ’2)

NJCTL.org

5.

t

g(t)

g’(t)

h(t)

h’(t)

0

-10

6

-1

-9

1

-3

1

-3

5

2

4

14

-1

-1/4

3

2

-1.5

-2

5

4

0

8

1/2

1

Given 𝑓(𝑑) = 𝑔(𝑑) βˆ™ β„Ž(𝑑)

find: 𝑓 β€² (1)

find: 𝑓 β€² (2)

6.

Given π‘Ÿ(𝑑) =

𝑔(𝑑) β„Ž(𝑑)

find: π‘Ÿ β€² (0)

find: π‘Ÿ β€² (3)

7.

Given 𝑠(𝑑) = β„Ž(𝑑)2

find: 𝑠 β€² (0)

find: 𝑠 β€² (4)

8.

Given 𝑔(𝑑) = 2𝑑 3 β„Ž(𝑑)

find: 𝑔′ (1)

find: 𝑔′ (2)

NJCTL.org

PROBLEM SET #8 – Tangent/Normal Lines

***Calculators Not Allowed***

For each question find the equations of the tangent & normal lines at the given value.

1.

𝑓(π‘₯) = 4π‘₯ + 7 at π‘₯ = 2

2.

𝑓(π‘₯) = 3π‘₯ 2 βˆ’ 2π‘₯ + 1 at π‘₯ = βˆ’3

3.

𝑦 = 3π‘π‘œπ‘ π‘₯ + 1 at π‘₯ =

4.

𝑔(π‘₯) = βˆ’4√π‘₯ βˆ’ 2 at π‘₯ = 4

πœ‹ 2

NJCTL.org

5.

𝑦 = π‘π‘œπ‘ π‘₯ βˆ™ 𝑠𝑖𝑛π‘₯ at π‘₯ =

6.

𝑦 = |2π‘₯| at π‘₯ = βˆ’3

7.

β„Ž(π‘₯) =

8.

𝑓(π‘₯) = at π‘₯ = 2

π‘₯ 2 +1 π‘₯ 2 βˆ’1

πœ‹ 4

at π‘₯ = 0

1

π‘₯

NJCTL.org

PROBLEM SET #9 – Derivatives of Log & e

***Calculators Not Allowed***

1. Find the derivative of: 𝑓(π‘₯) = 5ln π‘₯ + 1

2. Find

dy dx

for 𝑦 = 23π‘₯ + 15π‘₯ + 6π‘₯

3. Differentiate: 𝑦 = 3 log 4 π‘₯ + 7 log 2 π‘₯

4. Find the derivative of: 𝑓(π‘₯) = π‘₯ 2 + 2π‘₯

5. Find

dy dx

for: 𝑦 = 5π‘₯ + 𝑒 π‘₯ + ln 𝑒

NJCTL.org

6. Find 𝑦′ for: 𝑦 = ln π‘₯ + π‘₯ 2 log 8 π‘₯ + π‘₯ 3

7. Find the derivative of: 𝑓(π‘₯) = 2𝑒 π‘₯ + 3π‘₯ 4 ln 2 + 2π‘₯

8. Find

dy dx

for: 𝑦 =

3π‘₯ log3 π‘₯ +3𝑒 π‘₯

9. Differentiate: 𝑓(π‘₯) = 5π‘₯ + log 7 π‘₯ + 9ln π‘₯

10. Find the derivative of : 𝑓(π‘₯) = 3π‘₯ ln π‘₯ + 3π‘₯ 2 ln π‘₯ + 6

NJCTL.org

PROBLEM SET #10 – Chain Rule

***Calculators Not Allowed***

1. Find 𝑦′ for : 𝑦 = (5π‘₯ 2 + 3π‘₯)3

2. Given: 𝑓(π‘₯) = 3π‘₯(2π‘₯ βˆ’ 17)3 + 15 Find 𝑓 β€² (π‘₯)

3. Given: 𝑓(π‘₯) = (4π‘₯ 3 + 2π‘₯ 2 +

1 2

π‘₯)5 Find 𝑓 β€² (π‘₯)

4. Find 𝑦′ for : 𝑦 = 3π‘π‘œπ‘  2 π‘₯

5. Differentiate: 𝑓(π‘₯) = (sin(2π‘₯ 3 βˆ’ 1))2

6. Given: 𝑓(π‘₯) = (3π‘₯ 4 + π‘₯)(10π‘₯ 2 βˆ’ π‘₯)5 Find 𝑓 β€² (π‘₯)

𝑑𝑦

7. Find 𝑑π‘₯ if : 𝑦 =

5π‘₯ 2 βˆ’3π‘₯ (3π‘₯ 7 +2π‘₯ 6 )4

3

8. Differentiate: 𝑓(π‘₯) = (15π‘₯ 3 βˆ’ 10π‘₯ 5 )2

NJCTL.org

9. Given: 𝑓(π‘₯) = (7π‘₯ + (3π‘₯ βˆ’ 9)6 )3 Find 𝑓 β€² (π‘₯)

10. Differentiate: 𝑓(π‘₯) = 3tan(8π‘₯)

𝑑𝑦

11. Find 𝑑π‘₯ if 𝑦 = (βˆ’3π‘₯ 3 + 2π‘₯ 2 + π‘₯)4

12. Find 𝑦′ for : 𝑦 = 2π‘₯ 8 (4π‘₯ 2 βˆ’ 3π‘₯ )2

13. Given: 𝑓(π‘₯) = (7π‘₯ 2 βˆ’ π‘₯ + 15)βˆ’3 Find 𝑓 β€² (π‘₯)

𝑑𝑦

14. Find 𝑑π‘₯ if 𝑦 = βˆ’2sin4 (3π‘₯ βˆ’ 5)

15. Differentiate: 𝑓(π‘₯) = (βˆ’4x 2 βˆ’ 6π‘₯)2 (π‘₯ 5 + 5π‘₯)3

NJCTL.org

PROBLEM SET #11 – Derivs. of Inv. Functions

***Calculators Not Allowed***

β€²

1. If 𝑓(1) = 4 and 𝑓 β€² (1) = 6 , find 𝑓 βˆ’1 (4)

2. Find the derivative of the inverse of 𝑓(π‘₯) = π‘₯ 4

β€²

3. If 𝑓(4) = 6 and 𝑓 β€² (4) = 5 find 𝑓 βˆ’1 (6)

4. If 𝑓(2) = 7 and 𝑓 β€² (π‘₯) = 3π‘₯ 2 + 5π‘₯ + 12 find 𝑓 βˆ’1 β€²(7)

5. If 𝑓(8) = 15 and 𝑓 β€² (π‘₯) = π‘₯ 3 + 2π‘₯ 2 + 3π‘₯ find 𝑓 βˆ’1 β€²(15)

NJCTL.org

6. Find the derivative of the inverse of 𝑓(π‘₯) = 𝑠𝑖𝑛π‘₯

7. If 𝑓 β€² (π‘₯) =

2π‘₯+3 3π‘₯+2

and 𝑓(3) = 6 find 𝑓 βˆ’1 β€²(6)

8. If 𝑓(2) = 23 and 𝑓 β€² (π‘₯) =

5π‘₯+8 5π‘₯ 2 +8

Find 𝑓 βˆ’1 β€²(23)

9. If 𝑓(π‘₯) = 5π‘₯ βˆ’ 7 Find 𝑓 βˆ’1 β€²(8)

10. If 𝑓(1) = 5 and 𝑓(π‘₯) = 2π‘₯ 4 + 2π‘₯ 3 + 2π‘₯ 2 + 2π‘₯ + 2 Find 𝑓 βˆ’1 β€²(5)

NJCTL.org

PROBLEM SET #12 – Contin. vs. Differentiability

***Calculators Not Allowed***

1. If a function is differentiable on a given interval, it is also continuous. a. True b. False 2. For a function to be differentiable, it: (choose all that apply) a. Must have no discontinuities b. Can have discontinuities c. Must have no vertical tangent lines d. Can have vertical tangent lines e. Must not have corners f. May have cusps

3. At which values of x is 𝑓(π‘₯) not differentiable?

4. At which values of x is 𝑓(π‘₯) not differentiable?

5. At which values of x is 𝑓(π‘₯) not differentiable?

NJCTL.org

PROBLEM SET #13 – Der. of Piecewise & Abs. Value

***Calcs. Not Allowed***

1. a. What is the equivalent piecewise function for the following? b. What is the derivative? 𝑦 = 2|π‘₯ + 3| βˆ’ 4

2. a. What is the equivalent piecewise function for the following? b. What is the derivative? 𝑓(π‘₯) = 4|2π‘₯ βˆ’ 2| + 5

3. Find the derivative of the following function: 3(βˆ’π‘₯ + 2) βˆ’ 1 π‘₯ ≀ 2 𝑦={ 3(π‘₯ βˆ’ 2) βˆ’ 1 π‘₯>2

4. Find the derivative of the following function: 2 𝑦 = {5π‘₯ βˆ’ 9π‘₯ π‘₯ < 1 𝑙𝑛π‘₯ π‘₯β‰₯1

5. What values of π‘˜ and π‘š will make the function differentiable over the interval (βˆ’1,10)? π‘šπ‘₯ + 4 βˆ’ 1 ≀ π‘₯ ≀ 2 𝑓(π‘₯) = { 2 π‘˜π‘₯ + 3 2 < π‘₯ ≀ 10

NJCTL.org

6. What values of π‘š and 𝑗 will make the function differentiable over the interval (βˆ’3,9)? π‘šπ‘₯ 2 + 5 βˆ’ 3 ≀ π‘₯ ≀ 3 𝑓(π‘₯) = { 15 + 𝑗π‘₯ 33 a. 𝑓(π‘₯) is continuous at π‘₯ = 3 b. 𝑓(π‘₯) is differentiable at π‘₯ = 3 c. 𝑓(π‘₯) is not continuous at π‘₯ = 3 d. 𝑓(π‘₯) is not differentiable at π‘₯ = 3

NJCTL.org

PROBLEM SET #14 – Implicit Differentiation 1. Find

2. Find

3. Find

4. Find

5. Find

6. Find

7. Find

𝑑𝑦 𝑑π‘₯

𝑑𝑦 𝑑𝑑

𝑑π‘₯ 𝑑𝑑

***Calculators Not Allowed***

: 𝑦 2 + π‘₯ 2 = 2π‘₯ βˆ’ 5𝑦 3

: 𝑦 = 3𝑑 + 2𝑑𝑦 + 𝑑 2

: 15𝑑 = 2π‘₯ 2 + 4π‘₯

𝑑𝑦 𝑑π‘₯

𝑑𝑉 𝑑𝑑

𝑑𝑦 𝑑π‘₯

𝑑𝑧 𝑑𝑦

:

3π‘₯ 2𝑦

+ 2π‘₯ 2 + 4𝑦 = 5π‘₯

4

: 𝑉 = πœ‹π‘Ÿ 3 3

: π‘₯ 4 + π‘₯𝑦 4 = 4π‘₯

: 3𝑦 3 + 3𝑧 2 = 5𝑧

NJCTL.org

8. Find

9. Find

𝑑𝑦 𝑑π‘₯

π‘‘πœƒ 𝑑𝑑

10. Find

11. Find

: sin π‘₯ = 2π‘₯ cos 𝑦 + 2

: 2 sin2 πœƒ + cos 2 πœƒ = 2𝑑

𝑑𝐴 𝑑𝑑

𝑑𝑦 𝑑π‘₯

: 𝐴 = 2πœ‹π‘Ÿβ„Ž + 2πœ‹π‘Ÿ 2

: sin 2π‘₯ + cos 𝑦 = 5

12. Find the slope of the tangent line at the point (1,2) for: 3𝑦 2 + 5π‘₯ 3 = 17

NJCTL.org

13. Find the slope of the tangent line at the point (16,-1) for: 2𝑦 2 + √π‘₯ βˆ’ 6 = 0

14. Find the slope of the tangent line at the point (1,1) for: 2π‘₯ 3 = 3π‘₯ 2 βˆ’ 3π‘₯𝑦 + 2

15. Find the equation of the tangent line at the point (4,8) for: 4x 2 βˆ’ 2𝑦 2 = βˆ’16π‘₯

16. Find the equation of the tangent line at the point (1,0) for:

1 π‘₯

+ 5 = 𝑦π‘₯ + 6

17. Find the equation of the tangent line at the point (-3,3) for: π‘₯𝑦 2 + 𝑦 = 8π‘₯

NJCTL.org

Limits and Continuity- Answer Keys Problem Set #1 – Rate of Change 1. 2. 3.

5 4 2 βˆ’

4. βˆ’ 5. 6. 7. 8. 9.

Problem Set #3 – Derivative Rules 1. 𝑦 β€² = 0 2. 𝑓 β€² (π‘₯) = 8π‘₯ 3. a. 𝑔′ (π‘₯) = 36π‘₯ 2 βˆ’ 8π‘₯ + 2 b. 𝑔′ (βˆ’1) = 46

πœ‹

4 πœ‹

βˆ’ 1

8. a. linear

1 2

4.

90 1

π‘’βˆ’1

b.

1

6. a.

2

10. βˆ’ 3 11. a. 7.35 b. βˆ’7.35 c. βˆ’4.9 Problem Set #2 – Slope of a Curve 1. a. 𝑦 β€² = 4 b. 4 2. a. 𝑓 β€² (π‘₯) = 4π‘₯ b. 𝑓 β€² (2) = 8 3. a. 𝑔′ (π‘₯) = 9π‘₯ 2 b. 𝑔′ (1) = 9 𝑑𝑦 4. a. = 2π‘₯ + 2 𝑑π‘₯ b. 0 5. a. β„Žβ€² (π‘₯) = 6π‘₯ 2 + 6π‘₯ b. β„Žβ€² (2) = 12 𝑑𝑓 6. a. = 2πœ‹π‘Ÿ π‘‘π‘Ÿ b. 6πœ‹ 1 7. a. 𝑦 β€² = 2 √π‘₯

b.

√2 4

𝑑π‘₯

5. a.

6

3

𝑑𝑦

= 2π‘₯ + 4 3 4

2 √π‘₯ 3 4 𝑑𝑓

=

𝑑𝑑 3 √2 4 β€²

3 2 βˆšπ‘‘

b. 7. a. 𝑦 = 8π‘₯ b. 16 8. a. 𝑔′ (π‘₯) = 4π‘₯ 3 + 24π‘₯ 2 + 36π‘₯ + 8 b. 𝑔′ (0) = 8 𝑑𝑦 1 9. a. = 3 2 b.

𝑑π‘₯ 1

6 √π‘₯

24

Problem Set #4 – Higher Order Derivatives 1. 𝑦 β€² = 20 𝑦 β€²β€² = 0 2. 𝑓 β€² (π‘₯) = 26π‘₯ + 4 𝑓 β€²β€² (π‘₯) = 26 3. 𝑔′ (π‘₯) = 3π‘₯ 2 + 24π‘₯ + 48 𝑔′′ (π‘₯) = 6π‘₯ + 24 𝑑𝑦 4. = 2π‘₯ βˆ’ 8 𝑑π‘₯ 𝑑2𝑦

5. 6.

=2

𝑑π‘₯ 2 β€²β€²β€² (π‘₯)

β„Ž

𝑑2𝑓 𝑑𝑑 2

=

=π‘₯βˆ’1 βˆ’1 4βˆšπ‘‘ 3

NJCTL.org

1

Problem Set #6– Product/Quotient Rule

7.

𝑦′ =

8.

a. 𝑓 β€² (π‘₯) = 2π‘₯ 𝑓 β€²β€² (π‘₯) = 2 b. 𝑓 β€² (π‘₯) = 3π‘₯ 2 𝑓 β€²β€² (π‘₯) = 6π‘₯ c. 𝑓 β€² (π‘₯) = 𝑏π‘₯ π‘₯βˆ’1 𝑓 β€²β€² (π‘₯) = (𝑏 βˆ’ 1)𝑏π‘₯ π‘βˆ’2 a. 362,880π‘₯ b. 7,257,600

9.

3

3 √π‘₯ 2

Problem Set #5 – Trigonometry Rules 1. a. 𝑦 β€² = π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ b.

πœ‹ 𝑦′ ( ) 2 β€² πœ‹

√3βˆ’1

4 β€² πœ‹ 𝑓 ( ) 3 β€² (π‘₯)

=

βˆ’2 3

+ 2 √3

= 3 sec 2 π‘₯ + 4 csc 2 π‘₯

πœ‹

b. 𝑔′ ( ) = 20 6 𝑑𝑦

4. a. = 3 sec 2 π‘₯ 𝑑π‘₯ b. 6 5. a. β„Žβ€² (π‘₯) = 𝑐𝑠𝑐π‘₯π‘π‘œπ‘‘π‘₯ + 𝑠𝑒𝑐π‘₯π‘‘π‘Žπ‘›π‘₯ 2 b. βˆ’2√3 + 3 6. 𝑦 β€² = 4π‘π‘œπ‘ π‘₯ 𝑦 β€²β€² = βˆ’4𝑠𝑖𝑛π‘₯ 𝑦 β€²β€²β€² = βˆ’4π‘π‘œπ‘ π‘₯ 𝑦 4 = 4𝑠𝑖𝑛π‘₯ 7. 𝑔′ (π‘₯) = βˆ’5𝑠𝑖𝑛π‘₯ βˆ’ π‘π‘œπ‘ π‘₯ 𝑔′′ (π‘₯) = βˆ’5π‘π‘œπ‘ π‘₯ + 𝑠𝑖𝑛π‘₯ 𝑔′′′ (π‘₯) = 5𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ 𝑔4 (π‘₯) = 5π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ π‘‘π‘š 8. =0 𝑑π‘₯

2

9. a. √1βˆ’π‘₯ 2 b. 2

15π‘₯

1/2

4. β„Žβ€² (π‘₯) 5.

= βˆ’1

𝑦 ( )= 6 2 β€² (π‘₯) 2. a. 𝑓 = βˆ’π‘π‘ π‘π‘₯π‘π‘œπ‘‘π‘₯ + 𝑠𝑒𝑐π‘₯π‘‘π‘Žπ‘›π‘₯ πœ‹ b. 𝑓 β€² ( ) = 0 3. a. 𝑔

1. 𝑓 β€² (π‘₯) = 2π‘₯ 𝑑𝑦 2. = 12π‘₯ 3 + 36π‘₯ 2 βˆ’ 38π‘₯ βˆ’ 𝑑π‘₯ 40 9 15 3. 𝑔′ (π‘₯) = π‘₯ 7/2 βˆ’ π‘₯ 3/2 βˆ’

6. 7. 8.

2 2 1 βˆ’1/2 + π‘₯ 2 1 βˆ’2/3 = π‘₯ π‘‘π‘Žπ‘›π‘₯ 3 2

+

π‘₯ 1/3 sec π‘₯ 𝑦 β€² = βˆ’ sin2 π‘₯ + cos 2 π‘₯ or 𝑦 β€² = π‘π‘œπ‘ 2π‘₯ 𝑑𝑓 = 2π‘π‘œπ‘ π‘₯𝑠𝑖𝑛π‘₯ or 𝑦 β€² = 𝑠𝑖𝑛2π‘₯ 𝑑π‘₯ 𝑦 β€² = sec 2 π‘₯ + csc 2 π‘₯ βˆ’4π‘₯ 𝑔′ (π‘₯) = 4 2 𝑑𝑦

π‘₯ βˆ’2π‘₯ +1 𝑑𝑒 𝑑 βˆ’3𝑒 𝑑 βˆ’3

9. = 𝑑𝑑 𝑑4 2 10. sec π‘₯ 2 sec 2 π‘₯π‘‘π‘Žπ‘›π‘₯ 11. must show work

Problem Set #7 – Derivative Tables 1. β„Žβ€² (2) = βˆ’29 β„Žβ€² (1) = 16 2. 𝑗 β€² (0) = 0 𝑗 β€² (2) = βˆ’88 3. π‘˜ β€² (βˆ’1) = 1 π‘˜ β€² (1) = βˆ’2 4. π‘šβ€² (0) = 2 π‘šβ€² (βˆ’2) = βˆ’418 5. 𝑓 β€² (1) = βˆ’18 𝑓 β€² (2) = βˆ’15 6. π‘Ÿ β€² (0) = βˆ’96 7 π‘Ÿ β€² (3) = βˆ’ 4 7. 𝑠 β€² (0) = 18 𝑠 β€² (4) = 1 8. 𝑔′ (1) = βˆ’8 𝑔′ (2) = βˆ’28 NJCTL.org

1

normal: 𝑦 βˆ’ = 4(π‘₯ βˆ’ 2) or

Problem Set #8 – Tangent/Normal Lines 1. tangent: 𝑦 βˆ’ 15 = 4(π‘₯ βˆ’ 2) or 𝑦 = 4π‘₯ + 7 1 normal: =𝑦 βˆ’ 15 = βˆ’ (π‘₯ βˆ’ 2) 1

𝑦 = 4π‘₯ βˆ’

1. 𝑓 β€² (π‘₯) =

or 𝑦 = βˆ’ π‘₯ + 4 2 2. tangent: 𝑦 βˆ’ 34 = βˆ’20(π‘₯ + 3) or 𝑦 = βˆ’20π‘₯ βˆ’ 26 1 normal: 𝑦 βˆ’ 34 = (π‘₯ + 3) or 𝑦=

1 20

π‘₯+

20

πœ‹

1

πœ‹

normal: 𝑦 βˆ’ 1 = (π‘₯ βˆ’ ) or 3 2 1

6βˆ’πœ‹

𝑦= π‘₯+ 3 6 4. tangent: 𝑦 + 10 = βˆ’(π‘₯ βˆ’ 4) or 𝑦 = βˆ’π‘₯ βˆ’ 6 normal 𝑦 + 10 = 1(π‘₯ βˆ’ 4) or 𝑦 = π‘₯ βˆ’ 14 1 5. tangent: 𝑦 =

1

2

15

𝑦= π‘₯+ 2 2 at π‘₯ = 3: tangent: 𝑦 = 2π‘₯ 1 normal: 𝑦 βˆ’ 6 = βˆ’ (π‘₯ βˆ’ 3) or 1

15

2

𝑦=βˆ’ π‘₯+ 2 2 7. tangent: 𝑦 = βˆ’1 normal: π‘₯ = 0 1 1 8. tangent: 𝑦 βˆ’ = βˆ’ (π‘₯ βˆ’ 2) or 1

2

𝑦 =βˆ’ π‘₯+1 4

4

𝑑π‘₯ π‘₯

π‘₯

1

𝑑π‘₯ 1 (3π‘₯ 𝑙𝑛3)(log3 π‘₯+3𝑒 π‘₯ )βˆ’(3π‘₯ )( +3𝑒 π‘₯ ) π‘₯𝑙𝑛3

9. 10.

(log3 π‘₯+3𝑒 π‘₯ )2 1 9 𝑓 β€² (π‘₯) = 5π‘₯ 𝑙𝑛5 + + π‘₯𝑙𝑛7 π‘₯ 3π‘₯ β€² (π‘₯) π‘₯ 𝑓 = 3 𝑙𝑛3𝑙𝑛π‘₯ + + π‘₯

6π‘₯𝑙𝑛π‘₯ + 3π‘₯

2 πœ‹

normal: π‘₯ = 4 6. at π‘₯ = βˆ’3: tangent: 𝑦 = βˆ’2π‘₯ 1 normal: 𝑦 βˆ’ 6 = (π‘₯ + 3) or

π‘₯ π‘₯

6. 𝑦 = 3π‘₯ 2 + 2π‘₯ log 8 π‘₯ + + 𝑙𝑛8 π‘₯ β€² (π‘₯) π‘₯ 3 7. 𝑓 = 2𝑒 + 24π‘₯ 𝑙𝑛2 + π‘₯ 2 𝑙𝑛2 𝑑𝑦 8. =

3πœ‹+2 2

𝑑𝑦

5

𝑑π‘₯ β€²

3. tangent: 𝑦 βˆ’ 1 = βˆ’3 (π‘₯ βˆ’ ) 2 or 𝑦 = βˆ’3π‘₯ +

2

= 23 𝑙𝑛23 + 15π‘₯ 𝑙𝑛15 + 6 𝑙𝑛6 3 7 3. 𝑦 β€² = + π‘₯𝑙𝑛4 π‘₯𝑙𝑛2 β€² (π‘₯) 4. 𝑓 = 2π‘₯ + 2π‘₯ 𝑙𝑛2 𝑑𝑦 5. = 5π‘₯ 𝑙𝑛5 + 𝑒 π‘₯ 2.

20

683

2

Problem Set #9 – Derivatives of Logs and e

4

31

15

Problem Set #10 – Chain Rule 1. 3(5x 2 + 3π‘₯)2 (10π‘₯ + 3) 2. 3(2π‘₯ βˆ’ 17)3 + 18π‘₯(2π‘₯ βˆ’ 17)2 4

1

1

3. 5 (4π‘₯ 3 + 2π‘₯ 2 + π‘₯) (12π‘₯ 2 + 4π‘₯ + ) 2

2

4. βˆ’6π‘π‘œπ‘ π‘₯ 𝑠𝑖𝑛π‘₯ 5. 12π‘₯ 2 sin(2π‘₯ 3 βˆ’ 1) cos(2π‘₯ 3 βˆ’ 1) 6. (12π‘₯ 3 + 1)(10π‘₯ 2 βˆ’ π‘₯)5 + 5(3π‘₯ 4 + π‘₯)(10π‘₯ 2 βˆ’ π‘₯)4 (20π‘₯ βˆ’ 1) 7.

8.

(10π‘₯βˆ’3)(3π‘₯ 7 +2π‘₯ 6 )4 βˆ’4(5π‘₯ 2 βˆ’3π‘₯)(3π‘₯ 7 +2π‘₯ 6 )3 (21π‘₯ 6 +12π‘₯ 5 ) (3π‘₯ 7 +2π‘₯ 6 )8

3 2 3

1

(15π‘₯ 3 βˆ’ 10π‘₯ 5 )2 (45π‘₯ 2 βˆ’ 50π‘₯ 4 ) or

(45x 2 βˆ’ 50π‘₯ 4 )√15π‘₯ 3 βˆ’ 10π‘₯ 5

2

NJCTL.org

9. 9(7π‘₯ + (3π‘₯ βˆ’ 9)6 )2 (7 + 6(3π‘₯ βˆ’ 9)5 ) 10. 24 𝑠𝑒𝑐 2 8π‘₯ 11. 4(βˆ’3π‘₯ 3 + 2π‘₯ 2 + π‘₯)3 (βˆ’9π‘₯ 2 + 4π‘₯ + 1) 12. 16π‘₯ 7 (4π‘₯ 2 βˆ’ 3π‘₯)2 + 4π‘₯ 8 (4π‘₯ 2 βˆ’ 3π‘₯)(8π‘₯ βˆ’ 3) βˆ’42π‘₯+3 13. 2 4 (7π‘₯ βˆ’π‘₯+15) 3 (3π‘₯

14. βˆ’24𝑠𝑖𝑛 βˆ’ 5) cos(3π‘₯ βˆ’ 5) 2 15. 2(βˆ’4x βˆ’ 6π‘₯)(βˆ’8π‘₯ βˆ’ 6)(π‘₯ 5 + 5π‘₯)3 + 3(βˆ’4π‘₯ 2 βˆ’ 6π‘₯)2 (π‘₯ 5 + 5π‘₯)2 (5π‘₯ 4 + 5) Problem Set #11 – Der. of Inverse Fns. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Problem Set #13 – Derivatives of Piecewise & Absolute Value Functions 1. βˆ’2π‘₯ βˆ’ 10 π‘₯ ≀ βˆ’3 2π‘₯ + 2 π‘₯ > βˆ’3 βˆ’2 π‘₯ < βˆ’3 b. 𝑦 β€² = { 2 π‘₯ > βˆ’3 a. 𝑦 = {

2. βˆ’8π‘₯ + 13 π‘₯ ≀ 1 8π‘₯ βˆ’ 3 π‘₯>1 βˆ’8 π‘₯ < 1 b. 𝑦 β€² = { 8 π‘₯>1 βˆ’3 π‘₯ < 2 β€² 3. 𝑦 = { 3 π‘₯>2 10π‘₯ βˆ’ 9 π‘₯ ≀ 1 4. 𝑦 β€² = {1 π‘₯>1 a. 𝑦 = {

π‘₯ 1

5. π‘˜ = βˆ’ , π‘š = βˆ’1

1 6 1

1

4

4 √π‘₯ 1

or π‘₯ 3 4

3 βˆ’ 4

5 1 34 1 664 1

6. 𝑗 = βˆ’

1.

9 1

2.

5 1

3.

20

4.

True a, c, e π‘₯ = βˆ’4, π‘₯ = βˆ’1 π‘₯ = βˆ’1, π‘₯ = 4, π‘₯ = 6 π‘₯ = βˆ’4, π‘₯ = βˆ’3, π‘₯ = βˆ’2, π‘₯ = 3

,π‘š = βˆ’

10 9

Problem Set #14 – Implicit Differentiation

9 14

1. 2. 3. 4. 5.

3

7. 𝑗 = βˆ’15, π‘˜ = βˆ’15 8. a, d 9. c, d 10. a, b

√1βˆ’π‘₯ 2 11

Problem Set #12 – Continuity vs. Diff.

4 20

5. 6. 7. 8.

dy dx dy dt 𝑑π‘₯ 𝑑𝑑 dy

= = = =

2βˆ’2π‘₯ 2𝑦+15𝑦 2 3+2𝑦+2𝑑 1βˆ’2𝑑 15 4π‘₯+4 10𝑦 2 βˆ’8π‘₯𝑦 2 βˆ’3𝑦

8𝑦 2 βˆ’3π‘₯ π‘‘π‘Ÿ = 4πœ‹π‘Ÿ 2 𝑑𝑑 𝑑𝑑 dy 4π‘₯ 3 +𝑦 4 βˆ’4 dx 𝑑𝑉

dx 𝑑𝑧 𝑑𝑦 dy dx

=

= =

βˆ’4π‘₯𝑦 3 9𝑦 2 5βˆ’6𝑧 cos π‘₯βˆ’2 cos 𝑦 βˆ’2π‘₯ sin 𝑦

NJCTL.org

9. 10. 11.

π‘‘πœƒ 𝑑𝑑 𝑑𝐴 𝑑𝑑 dy dx

=

1 sin πœƒ cos πœƒ π‘‘π‘Ÿ

= 2πœ‹β„Ž =

𝑑𝑑 2 cos 2π‘₯ sin 𝑦 5

12. π‘š = βˆ’

4

+ 2πœ‹π‘Ÿ

13. π‘š = π‘‘β„Ž 𝑑𝑑

+ 4πœ‹π‘Ÿ

π‘‘π‘Ÿ 𝑑𝑑

1 32

14. π‘š = βˆ’1 3

3

2

2

15. 𝑦 βˆ’ 8 = (π‘₯ βˆ’ 4) or 𝑦 = π‘₯ + 2 16. 𝑦 = βˆ’π‘₯ + 1 1 17. 𝑦 βˆ’ 3 = (π‘₯ + 3) π‘œπ‘Ÿ 𝑦 = 17

1 17

π‘₯+

54 17

NJCTL.org