Differential Calculus.docx

CLASSIFICATION OF FUNCTIONS

functions

algebraic

trencendendal

rational

rational

integral

fractional

elementary

trigometric inverse trigo

higher

exponential logarithmic

Relation Is any set of one or more ordered pair.

Function Is a rule of correspondence between empty set such that two each element of the 1st set, there is corresponds one and only one element of the 2nd set.

Notation The function notation (f) (x) means the value of the function f at number x.

1

Vertical Line Test A graph of a relation is a function if any vertical line drawn passing through the graph intersects it at exactly one point. 𝑓(𝑥) = 2𝑥 2 + 5𝑥 − 3

It at exactly one point:

𝑓(2) = 2(2)2 + 5(2) − 3 = 8 + 10 − 3 𝑓(2) = 15

𝑦 = 2𝑥 2 + 5𝑥 − 3 𝑦 = 2𝑥 2 + 5𝑥 − 3 5 𝑦 + 3 = 2 (𝑥 2 + 𝑥) 2 25 5 25 + 𝑦 + 3 = 2 (𝑥 2 + 𝑥 + ) 8 2 16 49 5 2 𝑦+ = 2 (𝑥 + ) 8 4 𝑦 49 5 2 √ √ + = (𝑥 + ) 2 16 4 𝑦 49 5 𝑥=√ + − 2 6 4

2

Kinds of Functions A. Constant function A constant function C is a function the range of which contain a single real number K Example: The graph shows a constant function y = 12. The value of y is 12 for any x.

For all real number in its domain C(x) =k

Domain: all real number

Y=k

Range: 12

3

B.

Identity function: 𝐟(𝐱) = 𝐱 Evaluating any value for x will result in that same value.

For example, 𝑓(0) = 0 and 𝑓(2) = 2. The identity function is linear, 𝑓(𝑥) = 1𝑥 + 0, with slope m=1 and y-intercept (0, 0).

𝑓(𝑥) = 𝑥

x

𝑓(𝑥)

0 2

0 2

The domain and range both consist of all real numbers.

4

C.

Linear function Is a function that can be written in the form of 𝑓(𝑥) = 𝑚𝑥 + 𝑏 where m and

b are real numbers and m ≠ 0. Example: Graph the function f(x) = 2x – 1.

The graph is always a line with either positive or negative slope. The domain and range both consist of all real numbers.

5

D. Absolute function The absolute value parent function, written as 𝑓(𝑥) = |𝑥|, is defined as To graph an absolute value function, choose several values of x and find some ordered pairs.

X

𝑦 = |𝑥|

-2 -1 0 1 2

2 1 0 1 2

Plot the points on a coordinate plane and connect them.

Domain: All real Numbers Range: All real numbers greater than or equal to 0 (𝑦 ≥ 0)

6

E.

Square Root Function: 𝑓(𝑥) = √𝑥

Its domain is 𝑥| 𝑥 ≥ 0 & its range is {𝑦|𝑦 > 0| The parent function of the functions of the form f(x) = √x − a + b is f(x) = √x.

Note that the domain of 𝑓(𝑥) = √𝑥 is 𝑥 ≥ 0and the range is𝑦 ≥ 0. The graph of 𝑓(𝑥) = √𝑥 − 𝑎 + 𝑏 can be obtained by translating the graph of 𝑓(𝑥) = √𝑥 to a units to the right and then b units up.

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𝑁(𝑥)

F. Rational Function: 𝑓(𝑥) = 𝐷(𝑥)

N(x) & D(x) are polynomial functioning 𝐷(𝑥) ≠ 0 Asymptotes: Is an imaginary line being approached but does not reached or intersected by a graph as it. Example: 𝑓(𝑥) = 1⁄𝑥 . X -4 -2 -1 -0.1 -0.01 0.01 0.1 1 2 4

f(x) -0.25 -0.5 -1 -10 -100 100 10 1 0.5 0.25

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G. Greatest Integer Function: 𝐺(𝑥) = [𝑥] Domain of 𝐺(𝑥) = [𝑥] where [ ] denotes greatest integer function. Example: 𝒇(𝒙) =

𝟏 √[𝒙] − 𝒙

Domain: All real numbers Range: All real numbers

H. Piece wise Function These functions are defined compositely using several expressions on different interval domains. Example:

9

𝑥 − 2, 𝑥 ≤ 0 𝑓(𝑥) = { 𝑥 + 3, 𝑥 > 0

Domain: All real numbers Range: y > 3 & y <=-2 I. Signum Function The signum function, denoted sgn, is defined as follows: 1, 𝑠𝑔𝑛(𝑥) = {−1, 0,

𝑥>0 𝑥<0 𝑥=0

Graph:

10

Domain: All real numbers Range: y = 1, -1, 0 J. Unit Step Function The Unit-step function is defined by 𝑢. (𝑡) = {0 𝑖𝑓 𝑜 ≤ 𝑡 < 𝑎} That is, u is a function of time t, and u has value zero when time is negative and value one when time is positive.

Domain: All real numbers Range: 1 & 0

K. Transcendental Function Is a function that does not satisfy a polynomial equation whose coefficients are themselves roots of polynomials, in contrast to an algebraic function, which does satisfy such an equation. Example: The following functions are transcendental: 𝑓1 (𝑥) = 𝑥 𝜋 𝑓2 (𝑥) = 𝑐 𝑥 , 𝑐 ≠ 0,1 𝑓3 (𝑥) = 𝑥 𝑥 = 2𝑥 11

1

𝑓4 = 𝑥 2 𝑓5 (𝑥) = log 𝑐 𝑥. 𝑐 ≠ 0,1 𝑓6 (𝑥) = sin 𝑥

Note that in particular for ƒ2 if we set c equal to e, the base of the natural logarithm, then we get that ex is a transcendental function. Similarly, if we set

c equal to e in ƒ5, then we get that ln(x), the natural logarithm, is a transcendental function. For more information on the second notation of ƒ3, see tetration.

𝐸𝑥. 𝐼𝑓 𝑓(0) = 𝑥 2 − 𝑥 + 3 Find 𝑎. 𝑓(0) = 02 − 0 + 3 = 3 𝑏. 𝑓(2) = 22 − 2 + 3 = 4 − 2 + 3 = 5 𝑐. 𝑓(−4) = (−4)2 − (−4) + 3 = 16 + 4 + 3 = 23 𝑑. 𝑓(−2𝑥) = (−2𝑥)2 − (−2𝑥) + 3 = 4𝑥 2 + 2𝑥 + 3

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Name: Course/Year & Section:

ASSIGNMENT no.1 One Value & Many Valued Functions 1. 𝐼𝑓𝐹(𝑦) = 𝑦(𝑦 − 3)2 , Find: 𝐹(𝑐) = 𝑐(𝑐 − 3)2 = 𝑐(𝑐 2 − 6𝑐 + 9) = 𝒄𝟑 − 𝟔𝒄𝟐 + 𝟗𝒄

𝐹(0) = 0(0 − 3)2 = 0(9) =𝟎 𝐹(3) = 3(3 − 3)2 = 3(0)2 =𝟎 𝐹(−1) = −1(−1 − 3)2 = 1(4)2 = 1(16) = 𝟏𝟔 𝐹(𝑥 + 3) = (𝑥 + 3)(x + 3 − 3)2 = (𝑥 + 3)(x 2 ) = 𝒙𝟐 (𝒙 + 𝟑) 2. 𝐼𝑓𝑔(𝑥) = 4𝑥 4 − 3𝑥 3 + 2𝑥 − 2, Find: 𝑎. 𝑔(2) = 4(2)4 − 3(2)3 + 2(2) − 2 = 4(16) 3(8) + 4  2 = 64  24 + 4  2 = 𝟒𝟐 𝑏. 𝑔(−2) = 4(−2)4 − 3(−2)3 + 2(−2) − 2 13

= 4(16) − 3(−8) − 4 − 2 = 64 + 24  4  2 = 𝟖𝟐

1 1 4 1 3 1 𝑐. 𝑔 ( ) = 4 ( ) − 3 ( ) + 2 ( ) − 2 2 2 2 2 1 1 1 3 −𝟗 = 4( ) 3( ) + 1  2 =  − 1 = 16 8 4 8 𝟖

𝑑. 𝑔(−𝑥) = 4(−𝑥)4 − 3(−𝑥)3 + 2(−𝑥) − 2 = 4(𝑥 4 ) 3(−𝑥 3 ) + (−2𝑥)  2 = 𝟒𝒙𝟒 + 𝟑𝐱 𝟑 − 𝟐𝐱 − 𝟐

3. 𝐼𝑓𝜑(𝑥) = 𝑐𝑜𝑠 𝑥, Find: 𝑎. 𝜑(0) = 𝑐𝑜𝑠(0) = 𝟏

1 1 𝑏. 𝜑 ( 𝜋) = 𝑐𝑜𝑠 𝜋 2 2 = 𝑐𝑜𝑠 90 = 𝟎

𝑐. 𝜑(𝜋) = 𝑐𝑜𝑠𝜋 = 𝑐𝑜𝑠 180 = 𝟏

𝑑. 𝜑(−𝑥) = 𝑐𝑜𝑠(−𝑥) = 𝒄𝒐𝒔 𝒙 e. 𝜑(𝜋 − 𝑦) = 𝑐𝑜𝑠(  𝑦) = 𝑐𝑜𝑠(180  𝑦) =  𝒄𝒐𝒔 𝒚

4. 𝐼𝑓 𝜓(𝑥) = 𝑡𝑎𝑛 𝑥, 14

Find: 𝜋 𝜋 𝑎. 𝜓 ( ) = 𝑡𝑎𝑛 6 6 = 𝑡𝑎𝑛 30 √𝟑 = 𝟑 1 1 𝑏. 𝜓 (𝑥 − 𝜋) = 𝑡𝑎𝑛 (𝑥 − π) 2 2 = 𝑡𝑎𝑛(𝑥 − 90) 𝑡𝑎𝑛 𝑥  𝑡𝑎𝑛 90 = 1 + 𝑡𝑎𝑛𝑥𝑡𝑎𝑛90 = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅

𝑐. 𝜓(−𝑥) = 𝑡𝑎𝑛(−x) = − 𝒕𝒂𝒏 𝒙 Express 𝜓(2𝑥) as a function of 𝜓(𝑥). (2𝑥) = 𝑡𝑎𝑛 2𝑥 = 2𝑡𝑎𝑛 𝑥 = 𝟏 𝒕𝒂𝒏𝟐 𝐱

15

Examples: 1.) The amount of $1 @ 4% simple interest, as a function of time. 𝐼 = 𝑃𝑟𝑡 𝐼 = 1(0.04)𝑡 𝑰 = 𝟎. 𝟎𝟒 𝒕

2.) The volume of a sphere as a function of the radius. 4𝜋 3 𝑟 3 4 𝑉 = 𝜋 𝑟3 3 𝑉=

3.) The radius of the sphere as a function of the volume. 3

𝑟 =

√3𝑣 4𝜋

4.)The length, e of an edge of a cube as a function of the surface area, A of a cube. e A = 6e2 𝐴 = 6𝑒 2

e

𝐴

e = √6

e

e

Definition of a Limit Let 𝑓(𝑥) be a function of x & let “a” be constant. If there is a number L such that, in order to make a value of 𝑓(𝑥) as close to L as many be desired, it is sufficient to choose x close enough to a , but different from a ,then we say that the limit of 𝑓(𝑥) as a x approaches a , is L.

𝐿𝑖𝑚𝑓(𝑥) = 𝐿 16

𝑥→𝑎 𝐿𝑖𝑚(2𝑥 + 1) = 7 𝑥→3 (2𝑥 + 1)7 ∈ |2𝑥 − 6| ∈ ∈ 2 ∈ 𝐿𝑒𝑡𝑓 = 2 |𝑥 − 3|

(2 − 3) <

∈ 𝑥≠3 2

It follows that |2𝑥 − 6| <∈ From which 1(2𝑥 + 1) − 7 <∈ |𝑥 − 3| < ∈⁄2 − 3 < ∈ 𝑥⁄2 ≠ 3 It follows that |2𝑥 − 6 | <∈ From which 𝑙(2𝑥 + 1) − 7 <∈ 𝐿𝑖𝑚𝑖𝑡 (𝑥 2 + 1) = 5 𝑥2 |(𝑥2 + 1) 5| <∈ |𝑥 2  4| <∈ |𝑥  2|  |𝑥 + 2| <∈ |𝐴 + 𝐵| ≤ |𝐴| + |𝐵| Since 𝑥 + 2 = 𝑥  2 + 4, if follows that |𝑥 − 2| ≤ |𝑥 − 2| + 4 Thus if we choose |𝑥 − 2| < 8 |𝑥 + 2| < 8 + 4

Theorem 1: The limit of the sum of two (or more) functions is equal to the sum of their limits.

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𝐿𝑖𝑚 [𝑢(𝑥) + 𝑣(𝑥)] = 𝐿𝑖𝑚 𝑢(𝑥) + 𝐿𝑖𝑚 𝑣(𝑥) 𝑥𝑎 𝑥𝑎 𝑥𝑎

Theorem 2: The limit of the product of two or more functions is equal to the product of their limits. 𝐿𝑖𝑚 [𝑢(𝑥) 𝑢(𝑥)] = [𝐿𝑖𝑚 𝑢 (𝑥)] [𝐿𝑖𝑚 𝑢(𝑥)] 𝑥𝑎 𝑥𝑎 𝑥𝑎 Theorem 3: The limit of the quotient of two functions is equal to the quotient of the limits, provided the limit of the denominator is not zero.

𝐿𝑖𝑚

𝑢(𝑥) 𝐿𝑖𝑚 𝑢(𝑥) 𝑥𝑎 = 𝑣(𝑥) 𝐿𝑖𝑚 𝑣(𝑥) 𝑥𝑎

𝐼𝑓 𝐿𝑖𝑚 𝑣(𝑥) ≠ 0 𝑥 𝑎

Examples: Theorem 1 Evaluate 𝐿𝑖𝑚 (𝑥 3 + 4𝑥) = 𝐿𝑖𝑚 𝑥 3 + 𝐿𝑖𝑚 4𝑥

𝑥 3

𝑥3 = 33 + 4(3) = 27 + 12 = 𝟑𝟗

𝑥3

Theorem 2 𝐿𝑖𝑚 (𝑥 3 + 4𝑥) = [𝐿𝑖𝑚 𝑥][𝐿𝑖𝑚 𝑥][𝐿𝑖𝑚 𝑥] + [𝐿𝑖𝑚 4][𝐿𝑖𝑚 𝑥] 𝑥 3

𝑥3

𝑥3

𝑥3

𝑥→3

𝑥3

= (3)(3)(3) + 4(3) = 27 + 12 = 𝟑𝟗

Lim

x3 −9x+10 x2 −4

18

Lim (x3  9x + 10) = 0 Lim (x2  4) = 0 𝑥2

𝑥2

Lim (x2  4) = 0 𝑥2 𝐿𝑖𝑚

[𝑥 2 + 2𝑥 − 5][𝑥 − 2] (𝑥 2 + 2𝑥 − 5) = 𝐿𝑖𝑚 (𝑥 + 2)(𝑥 − 2) 𝑥+2 𝑥2

𝑥2 22 + 2(2) − 5 2+2 4+4−5 = 4 𝟑 = 𝟒 =

Right hand and Left hand limits 𝐿𝑖𝑚 𝑓(𝑥) = 𝐿 𝑥 𝑎+ & mean by 𝑥  𝑎+ that each x involved is greater than a. Right hand limit - the independent variable x approaches “a” from the right. Left hand limit – the independent variable x approaches “a” from the left. It means that x  a- that each x involved is lesser than “a”. #12: Limit of 𝑠𝑖𝑛⁄ as  approaches zero Theorem 4: If  is measured in ratios 𝐿𝑖𝑚

𝑠𝑖𝑛



= 1

Area RB T 19

1 𝐵𝑇 . 𝑅𝑇 2 1 = 𝑟 𝑐𝑜𝑠 ∝ 𝑟 𝑠𝑖𝑛 ∝ 2

𝐴𝑅𝑇𝐵

=

𝑟 2 𝑠𝑖𝑛 𝑐𝑜𝑠 = 2 𝐵𝑇 𝑅𝑇

𝐶𝑜𝑠 =

𝐵𝑇 = 𝑟 𝑐𝑜𝑠  𝑅𝑇 = 𝑟 𝑠𝑖𝑛 ∝ 𝑅𝑇 = 𝑟𝑠𝑖𝑛 Area of the sector



𝐴𝑠𝑒𝑐 = =

2𝑟

(𝑟2)



𝑟 2 2

Area of SUB = 1⁄2 (𝐵𝑉)(𝑆𝑉) 1 = 𝑟 (𝑟 𝑇𝑎𝑛 ) 2 𝑟𝑡𝑎𝑛 = 2 𝑠𝑣 𝑟 𝑟 𝑠𝑖𝑛 𝑐𝑜𝑠

𝑇𝑎𝑛  =

[

2

𝑆𝑉

𝑆𝑉 = 𝑟 𝑇𝑎𝑛

𝑟2 <

2

𝑟2 𝑡𝑎𝑛  <

[𝑠𝑖𝑛 𝑐𝑜𝑠 <  < 𝑡𝑎𝑛]

2

𝐵𝑉

]

2 1−2

1 𝑠𝑖𝑛

 1 < 𝑠𝑖𝑛 𝑐𝑜𝑠 1  > > 𝑐𝑜𝑠  𝑐𝑜𝑠  𝑠𝑖𝑛  𝐶𝑜𝑠 <

1 > 𝑢𝑛𝑑 > 1 20

𝐶𝑜𝑠  >

𝑠𝑖𝑛



> 𝑐𝑜𝑠 − 

1 > 1 > 1 𝐿𝑖𝑚

𝑠𝑖𝑛 



= 1

EXAMPLES: 1. 𝐿𝑖𝑚 (𝑥 2 + 3𝑥 – 5) 𝑥4 = 42 + 3 (4) – 5 = 16 + 12 – 5 = 23 2. 𝐿𝑖𝑚 (𝑦 3 – 2𝑦 + 7) 𝑦 3 = 33 – 2(3) + 7 = 27  6 + 7 = 𝟐𝟖

3 + 2 − 8 − 12 3. 𝐿𝑖𝑚 3  − 42 − 3 + 18 3 33 + 32 − 8(3) − 12 = 3 3 − 4(3)2 − 3(3) + 18 =

27+9−24−12 27−36−9+18

= 𝑵𝒐 𝑳𝒊𝒎𝒊𝒕

2 + 4 + 4 = 0 ( + 2)2 ( − 3) ( + 2)2 ( − 3) = ( + 2)( − 3)2

21

=

 +2 3+2 𝟓 = = 𝑵𝒐 𝑳𝒊𝒎𝒊𝒕  −3 3−3 𝟎

4. 𝐿𝑖𝑚

3 + 7 − 6 23 − 112 + 12 + 9

3 33 − 7(3) − 6 = 2(3)3 − 11 (3)2 + 12 (3) + 9 27 − 21 − 6 𝟎 = 𝑼𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 54 − 99 + 36 + 9 𝟎 ( + 1)( + 2)( − 3) = (2 + 1)( − 3)2 =

=

( + 1)( + 2) (2 + 1)( − 3)

3 =

(3 + 1)(3 + 2) (2(3) + 1)(3 − 3)

=

𝟐𝟎 𝑵𝒐 𝑳𝒊𝒎𝒊𝒕 𝟎

5. 𝐿𝑖𝑚 𝑥

=

=

=

=

4𝑥 3 − 3𝑥 + 1 12𝑥 3 − 8𝑥 2 − 𝑥 + 1

1 2

1 3 1 4 (2) − 3 (2) + 1 1 3 1 2 1 12 (2) − 8 (2) − (2) + 1 1 1 4 (8) − 3 (2) + 1 1 1 1 12 (8) − 8 (2) − 2 + 1 0 3 1 2 −2 + 2 0 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 0 22

(x + 1)(2x − 1)2 = (12x + 4)(2x − 1)2 =

(x + 1) (12x + 4)

1 3⁄ (2 + 1) 𝟑 = = 2= 1 10 𝟐𝟎 (12 (2) + 4)

6. 𝐿𝑖𝑚

𝑇𝑎𝑛𝜃 𝑠𝑖𝑛2 𝜃

 = 90 sin 𝜃 sin 𝜃 cos 𝜃 = = 2 𝑠𝑖𝑛 𝜃 cos 𝜃𝑠𝑖𝑛2 𝜃 sin 90 1 = 2 cos 90𝑠𝑖𝑛 90 0(1)2 =

𝟎 𝒖𝒏𝒅 𝟎

7. 𝐿𝑖𝑚

𝐶𝑜𝑠2 𝑇𝑎𝑛 

π 2 𝑐𝑜𝑠 180 = 𝑡𝑎𝑛 90 −𝟏 = 𝒖𝒏𝒅 →

8. 𝐿𝑖𝑚

𝑐𝑜𝑠 2  − 𝑠𝑖𝑛2  𝑠𝑖𝑛 𝑐𝑜𝑠

  90 𝑐𝑜𝑠 𝑐𝑜𝑠 2  − 𝑠𝑖𝑛2  𝑠𝑖𝑛 𝐿𝑖𝑚 𝑥 𝑠𝑖𝑛 𝑐𝑜𝑠 𝑐𝑜𝑠 𝑠𝑖𝑛

23

𝑐𝑜𝑠  (𝑐𝑜𝑠 2  − 𝑠𝑖𝑛2 ) = 𝑠𝑖𝑛 =

0 (0 − 1) 0 = 1 1

= 𝟎

√𝑥 − 2

9. 𝐿𝑖𝑚

√𝑥 2 − 4

𝑥2 𝐿𝑖𝑚 √

(𝑥 − 2)1 1 1 𝟏 = √ = √ = (𝑥 − 2)(𝑥 + 2) 𝑥+2 4 𝟐

10. 𝐿𝑖𝑚

𝑥−3 √𝑥 2 − 9

= 0

𝑥 3 2

(𝑥 − 3)2

𝐿𝑖𝑚

1

(𝑥 2 − 9)2 (𝑥 − 3)2 √ = (𝑥 − 3)(𝑥 + 3) 3−3 = √ 3+3

= √

0 = 𝟎 6

1

11. 𝐿𝑖𝑚

(𝑥 4 − 4𝑥 + 5𝑥 2 − 4𝑥 + 4)2 1

(𝑥 2 − 3𝑥 + 2)2 𝑥2 1

1

(𝑥 − 2)2 (𝑥 2 + 1) 2 (𝑥 − 2)(𝑥 2 + 1) 2 𝐿𝑖𝑚 [ ] = 𝐿𝑖𝑚 [ ] (𝑥 − 2)(𝑥 − 1) (𝑥 − 1) 1

(2 − 2)(22 + 1) 2 =( ) 1

1

0 2 = ( ) 1

= 𝟎 24

1

12. 𝐿𝑖𝑚

(𝑥 2 + 4𝑥 − 5)2 1

(𝑥 2 − 4𝑥 + 3)2 𝑥 1 1

1

(𝑥 + 5)(𝑥 − 1) 2 (𝑥 + 5) 2 𝐿𝑖𝑚 [ ] = 𝐿𝑖𝑚 [ ] (𝑥 − 3)(𝑥 − 1) (𝑥 − 3) 1

1

1+5 2 6 2 = ( ) = ( ) = √−𝟑 1−3 −2

13. 𝐿𝑖𝑚

4𝑥 3 − 3𝑥 + 1 12 𝑥 3 − 8 𝑥 2 − 𝑥 + 1

1 2 (2𝑥 − 1)2 (4 𝑥 + 4) = (2𝑥 − 1)2 (12𝑥 + 4) 𝑥

=

4(𝑥 + 1) 4(3𝑥 + 1)

=

(𝑥 + 1) (3𝑥 + 1)

1 +1 = 2 1 3 (2) + 1

3 =2 5 2

=

𝟑 𝟓

25

Name: Course/Year & Section:

ASSIGNMENT no.2 Limits 1. Lim(2𝑥 2 + 𝑥 + 4) 𝑥→ −1

= [2(1)2 + (1) + 4] = [2(1)  1 + 4] = 𝟓 2. Lim(𝑦 2 + 5 − 1) 𝑦→ −2

= [(2)2 + 5  1] = [4 + 5  1] = 𝟖 2𝑡 2 + 1 3. Lim 3 𝑡→0 𝑡 + 3𝑡 − 4 2(0)2 + 1 = (0)3 + 3(0) − 4 = −

4. lim 𝑡→1

𝟏 𝟒

(𝑡 + 1)2 2(𝑡 2 + 3) (1+1)2 = 2[(1)2 + 3] =

(2)2 2(4)

=

𝟏 𝟐

26

3𝑤 2 − 4𝑤 + 2 5. lim 𝑤→2 𝑤3 − 5 3(2)2 − 4(2) + 2 = (2)3 − 5 =

12 − 8 + 2 8−5

= 𝟐 3𝑤 3 − 2𝑤 + 7 𝑤→−1 𝑤2 + 1

6. lim

=

3(−1)3 − 2(−1) + 7 (−1)2 + 1

=

−3 + 2 + 7 1+1

= 𝟑

𝑠𝑖𝑛2 𝜃 7. lim 3 1 𝜃→ 𝜋 𝑡𝑎𝑛 𝜃 2

𝑠𝑖𝑛2 𝜃 = 𝑠𝑖𝑛3 𝜃 𝑐𝑜𝑠 3 𝜃 𝑐𝑜𝑠 3 𝜃 = 𝑠𝑖𝑛 𝜃 . 𝑠𝑖𝑛3 𝜃 0 = 1 2

= 𝟎

8. lim𝜋 𝜃→

6

sin 2𝜃 sin 𝜃 tan 𝜃 =

2 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃 2 𝑐𝑜𝑠 𝜃 == 𝑠𝑖𝑛 𝜃 𝑡𝑎𝑛 𝜃 𝑡𝑎𝑛 𝜃

2√3 = 2 1 √3

=

√3 1 √3

=𝟑

27

𝑥2 − 1 9. lim 2 𝑥→1 𝑥 + 3𝑥 − 4 (𝑥 − 1)(𝑥 + 1) = (𝑥 + 4)(𝑥 − 1) =

(𝑥 + 1) (𝑥 + 4)

1+1 1+4 𝟐 = 𝟓 =

𝑥 2 + 𝑥 − 12 10. lim 2 𝑥→3 2𝑥 − 7𝑥 + 3 (𝑥 + 4)(𝑥 − 3) = (2𝑥 − 1)(𝑥 − 3) 𝑥+4 2𝑥 − 1 3+4 = 2(3) − 1 =

=

𝟕 𝟓

2𝑥 2 − 𝑥 − 3 𝑥→−1 3𝑥 3 + 5𝑥 + 2

11. lim

=

(𝑥 + 1)(2𝑥 − 3) 3(−1)3 + 5(−1) + 2

(−1 + 1)[2(−1) − 3] −3 − 5 + 2 0 = −6 =

= 𝟎

28

2𝑥 3 − 7𝑥 − 4 12. lim 2 𝑥→4 𝑥 − 𝑥 − 12 2(4)3 − 7(4) − 4 = (𝑥 − 4)(𝑥 + 3) =

128 − 28 − 4 (4 − 4)(4 − 3)

= 𝑼𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝑦3 − 𝑦2 − 𝑦 − 2 13. lim 3 𝑦→2 2𝑦 −5𝑦 2 + 5𝑦 − 6

𝑦 3 − 13𝑦 + 12 𝑦→3 𝑦 3 − 14𝑦 + 15

14. lim

2𝑎3 − 5𝑎2 − 4𝑎 + 12 15. lim 𝑎→2 𝑎3 − 12𝑎 + 16

29

𝑥 4 + 5𝑥 + 6 16. lim 4 𝑥→−2 𝑥 + 5𝑥 − 6

2𝑥 4 − 2𝑥 3 − 𝑥 2 + 1 𝑥→1 𝑥 4 − 𝑥 2 − 2𝑥 + 2

17. lim

1 − cos 𝑦 𝑦→0 𝑠𝑖𝑛2 𝑦

18. lim

𝑠𝑖𝑛2 𝑦 19. lim 𝑦→𝜋 1 + cos 𝑦

sin 𝑎 sin 2𝑎 𝑎→0 1 − cos 𝑎

20. lim

𝑠𝑖𝑛3 𝑎 𝑎→0 sin 𝑎 − tan 𝑎

21. lim

30

sin 𝑎2 𝑎→0 𝑎 2

22. lim

sin 𝑎2 23. lim 𝑎→0 𝑎

𝜃2 𝜃→0 sin 𝜃

24 lim

𝜃 θ→0 𝑠𝑖𝑛2 𝜃

25. lim

sin 𝑘𝑥 𝑥→0 𝑥

26 lim

31

27. lim 𝑥 csc 3𝑥 𝑥→0

𝑥 sin 𝑠𝑖𝑛3𝑥 3𝑥 = 3 sin 3𝑥 1 = 3 =

sin 𝑎𝑥 𝑥→0 tan 𝑏𝑥

28. lim

1 − cos 4𝑥 𝑥→0 1 − cos 2𝑥

29. lim

2𝑦 − 𝜋 𝜋 cos 𝑦

30. lim 1 𝑦→ 2

31. lim 1

𝑦→ 𝜋 2

1 (𝑦 − 2 𝜋)2 1 − sin 𝑦

32

1

32. lim

𝑥→1

33. lim

𝑥→1

(1 − 𝑥²)2 1

(1 − 𝑥³)2

√1 − 𝑥 3 √1 − 𝑥 2

33

CONTINUITY A function 𝑓(𝑥) is said to be continuous at 𝑥 = 𝑎 𝐼𝑓 𝑓(𝑎) = 𝑒𝑥𝑖𝑠𝑡

𝐹(𝑥) = √𝑥

𝑥= 0

Lim f(x) exist

𝐹 (0) = √0 = 0

𝑥𝑎

Lim √𝑥 = 0 does not exist 𝑥0

Lim f(x) = f (a) 𝑥𝑎

𝑓 (𝑥) = √𝑥 right hand

continuity Ex. At x = 2; f (x) = x2 + 1 𝑓 (2) = 22 + 1 Lim x2 + 1 = 22 + 1 X2

=5

Missing Point Discontinuities

Consider a function f(x) which is not defined when x = a but such that Lim f (x) exist Lim f(x) = L 𝑥𝑎 6 (𝑥) = 𝑓 (𝑥)

𝑥 = 𝑎

6 (𝑥) = 1

𝑥 = 𝑎

𝐿𝑖𝑚

𝑥 3 − 9𝑥 + 10 𝑥−2

𝑥2 (𝑥 − 2)(𝑥 2 + 2𝑥 − 5) = (𝑥 − 2)

𝐿𝑖𝑚 𝑥2 + 2𝑥 − 5 𝑥2

= 22 + 2(2) – 5 = 4 + 4– 5 = 𝟑

34

Finite Jump

Infinite Discontinuity

𝐿𝑖𝑚𝑓 (𝑥) = 𝐿1

𝑦 =

𝑥𝑎−

𝑥 = − 2| − 1 | 0 | 1 | 2 | 3 | 4

𝐿𝑖𝑚 𝑓 (𝑥) = 𝐿2

𝑦 =

1 (𝑥 − 1)2 1 1 1 1 | | 1 |𝑢𝑛𝑑|1| | 9 4 4 9

𝑥𝑎+ Function with argument approaching infinity 𝐿𝑖𝑚

𝑥2 − 1 𝑥2 + 1

𝑥

2 − 2  = 2 = =   + 1 

Rational Algebraic Functions

Theorem 6: a polynomial is continuous for all values of x.

Theorem 7: a rational algebraic function is continuous except for those values of x which the denominator vanishes. 𝑥2 + 3 1. ) 2  𝑥 2 − 16 = 0  √𝑥 2 = √16  𝒙 = ± 𝟒 𝑥 − 16

2. )

3𝑥 + 2  𝑥2 – 6𝑥 + 9 = 0  (𝑥 – 3)2 = 0  𝒙 = 𝟑 𝑥 2 − 6𝑥 + 9

𝑥 2 − 3𝑥 3. ) 2  𝑥2 + 9  𝑥 = √−9  3√−1  𝟑 𝑥 + 9

35

4. )

𝑥+3  𝑥 − 𝑥+2

4𝑥 2

=

−𝑏 ± √𝑏 2 − 4𝑎𝑐 1 ± √12 − 4(4)(7) 𝟏 ± √−𝟑𝟏   2𝑎 2(4) 𝟖

𝑥 2 − 3𝑥 5. ) 3  𝑥3 + 2𝑥2 + 5𝑥 = 0  𝑥(𝑥 2 + 2𝑥 + 5) 𝑥 + 2𝑥 2 + 5𝑥 = 0 

−2 ± √22 − 4(1)(5) 2



−𝟐 ± √−𝟏𝟔 𝟐

Intermediate Value Function Theorem

The function f (x) is said to be continuous over the closed interval a ≤ 𝑥 ≥ b if f(a) is continuous at every pt a ≤ 𝑥 ≥ b and f(x) has right hand continuity x = a left hand continuity at x = b.

LEMMA 1: If f(x) is continuous over the closed interval a ≤ 𝑥 ≥ b; if f(a) < 0 and f(b) > 0, there exist a number “c” in the open interval a < c < b for which f(c) = 0

Theorem 8: if the single value function(x) is continuous over the closed interval a ≤ 𝒙 ≥ b then in that interval f(x) takes on every value between f(a) and f(b).

Theorem 9: if f(x) is continuous over the closed interval a ≤ 𝒙 ≥ b if (x) takes on a greatest value and the least value in the closed interval

36

DERIVATIVE  rate of change

 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 “𝑦” 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 “𝑥” 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜

∆𝑦 𝑤ℎ𝑒𝑛 ∆𝑥 ∆𝑥

𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ𝑒𝑠 𝑧𝑒𝑟𝑜 𝑑𝑦 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) = 𝐿𝑖𝑚 𝑑𝑥 ∆𝑥

Steps in to determine the derivative of a function 1. Replace x by (𝑥 + ∆𝑥) 𝑎𝑛𝑑 𝑦 𝑏𝑦 (𝑦 + ∆𝑦) 2. By subtraction, eliminate y b/n, thus obtaining a formula for ∆𝑦 in terms of x and ∆x 3. By some suitable transformation throw the right member into a form which contains ∆𝑥 explicity as factor 4. Divide through by ∆x 5. Determine the limit as ∆x approaches zero

Examples: 1. 𝑦 = 𝑥 3 – 2𝑥 𝑦 + ∆𝑦 = (𝑥 + ∆𝑥)3 − 2 (𝑥 + ∆𝑥) ∆𝑦 = (𝑥 + ∆𝑥)3 – 2 (𝑥 + ∆𝑥) – 𝑦 ∆𝑦 = 𝑥 3 + 3𝑥 2 (∆𝑥) + 3𝑥 (∆𝑥)2 + (∆𝑥)3 – 2𝑥  2(∆𝑥) – (𝑥 2 – 2𝑥) ∆𝑦 = 3𝑥 2 (∆𝑥) + 3𝑥 (∆𝑥)2 + (∆𝑥)3  2(∆𝑥) ∆𝑦 = ∆𝑥 [3𝑥 2 + 3𝑥 (∆𝑥) + (∆𝑥)3  2] ∆𝑦 = 3𝑥 2 + 3𝑥 (∆𝑥) + (∆𝑥) 2  2 ∆𝑥 ∆𝑦 = 3𝑥 2  2 ∆𝑥 2. 𝑥 =

1 𝑡

37

𝑥 + ∆𝑥 =

1 𝑡 + ∆𝑡

1 − 𝑥 𝑡 + ∆𝑡 1 1 ∆𝑥 = − 𝑡 + ∆𝑡 𝑡 𝑡 − (𝑡 + ∆𝑡) ∆𝑥 = (𝑡 − ∆𝑡) 𝑡 ∆𝑥 =

∆𝑥 = ∆𝑥 =

𝑡 − 𝑡 − ∆𝑡 (𝑡)2 + (∆𝑡)𝑡 𝑡2

∆𝑡 + 𝑡 (∆𝑡)

∆𝑥 −1 = 2 ∆𝑡 𝑡 + 𝑡 (∆𝑡) ∆𝑥 −1 = 2 ∆𝑡 𝑡 3.

𝑦 = √𝑥

𝑦 + ∆𝑦 = √𝑥 + ∆𝑥 −𝑦 + (𝑦 + ∆𝑦) = √𝑥 + ∆𝑦 − 𝑦 ∆𝑦 = √𝑥 + ∆𝑥 − √𝑥 ∆𝑦 = (√𝑥 + ∆𝑥 − √𝑥 ) . √𝑥 + ∆𝑥 + √𝑥 √𝑥 + ∆𝑥 + √𝑥 ∆𝑦 =

∆𝑥 √𝑥 + ∆𝑥 + √𝑥 1

∆𝑦 = ∆𝑥 √𝑥 + ∆𝑥 + √𝑥 ∆𝑥 → 0 ∆𝑦 1 = ∆𝑥 2√𝑥

38

4. 𝑦 = 𝑠𝑖𝑛 𝑥 𝑦 + ∆𝑦 = 𝑠𝑖𝑛 (𝑥 + ∆𝑥) −𝑦 + (𝑦 + ∆𝑦) = 𝑠𝑖𝑛 (𝑥 + ∆𝑥) – 𝑦 ∆𝑦 = 𝑠𝑖𝑛 (𝑥 + ∆𝑥) − 𝑠𝑖𝑛 𝑥 ∆𝑦 = 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 ∆𝑥 + 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥 – 𝑠𝑖𝑛 𝑥 ∆𝑦 = 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥 – 𝑠𝑖𝑛 𝑥 (1 – 𝑐𝑜𝑠 ∆𝑥) ∆𝑥 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥 − 2𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛2 2 ∆𝑦 = ∆𝑥 ∆𝑥 ∆𝑦 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 ∆𝑥 ∆𝑥 = – 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛2 ∆𝑥 ∆𝑥 2 ∆𝑥 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛2 2 ∆𝑦 = 𝑐𝑜𝑠 𝑥 – ∆𝑥 ∆𝑥 2 ∆𝑦 ∆𝑥 = 𝑐𝑜𝑠 𝑥 – 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥 2 ∆𝑥 → 0 ∆𝑦 = 𝑐𝑜𝑠 𝑥 ∆𝑥 5. 𝑦 = 𝑐𝑜𝑠 𝑥 𝑦 + ∆𝑦 = 𝑐𝑜𝑠 (𝑥 + ∆𝑥) −𝑦 + (𝑦 + ∆𝑦) = 𝑐𝑜𝑠 (𝑥 + ∆𝑥) – 𝑦 ∆𝑦 = 𝑐𝑜𝑠 (𝑥 + ∆𝑥) – 𝑐𝑜𝑠𝑥

∆𝑥 − 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛2 2 ∆𝑦 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥 = – ∆𝑥 ∆𝑥 ∆𝑥 2

∆𝑦

∆𝑦 ∆𝑥 = − 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 – 𝑠𝑖𝑛 𝑥 ∆𝑥 2

= 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 ∆𝑥 – 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥 – 𝑐𝑜𝑠 𝑥

∆𝑥  0

∆𝑦

∆𝑦 = − 𝑐𝑜𝑠𝑥 (0) – 𝑠𝑖𝑛 𝑥 ∆𝑥 ∆𝑦 = − 𝑠𝑖𝑛 𝑥 ∆𝑥

= – 𝑐𝑜𝑠 𝑥 (1 – 𝑐𝑜𝑠 ∆𝑥) – 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥

∆𝑦 − 𝑐𝑜𝑠 𝑥 (2 𝑠𝑖𝑛2 ∆𝑥) − 𝑠𝑖𝑛 𝑥 𝑠𝑖𝑛 ∆𝑥 = ∆𝑥 2 −𝑐𝑜𝑠𝑥(1 − 𝑐𝑜𝑠∆𝑥) − 𝑠𝑖𝑛𝑥𝑠𝑖𝑛∆𝑥 ∆𝑦 = ∆𝑥

39

Name: Course/Year & Section:

ASSIGNMENT no.3: DERIVATIVES 1. ) 𝑥 = 𝑦 4 − 2𝑦 3

1 2. ) 𝑦 = (3𝑥 2 + 1)² 2

40

3. )𝑦 =

1 𝑥+7

4. ) 𝑦 =

2𝑥 𝑥−1

41

5. ) 𝑦 = 2 − 3𝑥 −

6. ) 𝑥 =

1 𝑥

1 𝑡2

42

7. ) 𝑦 = √2 − 3𝑥

8. ) 𝑦 =

1 √𝑥

43

9. ) 𝑦 =

1 √𝑥 − 2

10. ) 𝑦 = 𝑥√𝑥 − 1

44

Tangent to Plane Curves 1. ) 𝑦 = 2− 𝑥 2 (3, −7) 𝑦 + 𝑦 = 2 – (𝑥 + 𝑥)2

𝑦 = 2 − (𝑥 2 + 2𝑥𝑥 + (𝑥) 2) (2 – 𝑥2) 𝑦 = 2 − 𝑥 2 − 2 𝑥𝑥 − (𝑥)2 − 2 + 𝑥 2 𝑦 = 𝑥(−2𝑥 − 𝑥) 𝑦 = −2𝑥 − 𝑥 𝑥 𝑥 = 0 𝑑𝑦 = −2𝑥 𝑑𝑥 𝑚 = −6 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦  ( 7) = −6(𝑥 − 3) 𝑦 + 7 = −6𝑥 + 18 𝑦 = −6𝑥 + 11  𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑡

2. ) 𝑦 2 = 3𝑥 + 1 (

−1 , 0) 3

(𝑦 + 𝑦)2 = 3(𝑥 + 𝑥) + 1 𝑦 2 + 2𝑦𝑦 + (𝑦)2 = 3𝑥 + 3𝑥 + 1 2𝑦𝑦 + (𝑥)2 = 3𝑥 + 3𝑥 + 1 – (3𝑥 + 1) 2𝑦𝑦 + (𝑦)2 = 3𝑥

𝑦(2𝑦 + 𝑦) = 3𝑥 𝑦 2 3 −1 = = ( , 0) 𝑥 2𝑦 + 𝑦 2𝑦 3 𝑦 0

𝑦 3 = 𝑥 2(0) 𝑦 = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝑥 45

3.) Find how fast a.) the circumference b.) the area of a circle increases when the radius increases. 𝑎. )𝑐 = 2𝑟

𝑏. ) 𝐴 = 𝑟2

𝑐 + 𝑐 = 2(𝑟 + 𝑟)

𝑎 + 𝑎 = (𝑟 + 𝑟)2

𝑐 = 2𝑟 + 2𝑟 – 𝐶

𝑎

= [𝑟2 + 2𝑟𝑟 + (𝑟)2] − 𝐴

𝑐 = 2𝑟 + 2𝑟 − 2𝑟

𝑎

= 𝑟2 + 2𝑟𝑟 + (𝑟)2 − 𝑟2

𝑐 = 2𝑟

𝑎 = 2𝑟 + 𝑟 𝑟

𝑎 = 2𝑟 𝑟

𝑟 0 𝑐 = 2 𝑟

4.) The dimension of a box , 𝑏, 𝑏 + 1, 𝑏 + 4.Find how fast the total surface area A increases as b increases. 𝐴 = 2(𝑏) (𝑏 + 1) + 2𝑏 (𝑏 + 4) + 2 (𝑏 + 1) (𝑏 + 4) 𝐴 = 2𝑏2 + 2𝑏 + 2𝑏2 + 8𝑏 + 2𝑏 + 10𝑏 + 8

b

b+4 b+1

𝐴 = 6𝑏2 + 20𝑏 + 8 𝐴 + 𝐴 = 6(𝑏 + 𝑏)2 + 20 (𝑏 + 𝑏) + 8

𝐴 = 6(𝑏2 + 2𝑏𝑏 + (𝑏)2 + 20𝑏 + 20𝑏 + 8 – (6𝑏2 + 20𝑏 + 8) 𝐴 = (6𝑏2 + 12𝑏𝑏 + 6(𝑏) 2 + 20𝑏 + 20𝑏 + 8 – 6𝑏2 – 20𝑏 – 8 𝐴 = 12𝑏𝑏 + 6(𝑏) 2 + 20𝑏 𝐴 = 𝑏(12𝑏 + 6𝑏 + 20) 𝐴 = 12𝑏 + 6𝑏 + 20 𝐵 𝑏0 𝐴 = 12𝑏 + 20 𝐵

46

5. ) 𝑎. )𝐴 = 6𝑠 2 𝐴 + 𝐴 = 6(𝑠 + 𝑠)2

𝐴 = 6[𝑠2 + 2𝑠(𝑠) + (𝑠)2]  𝐴 𝐴 = 6𝑠2 + 12𝑠(𝑠) + 6(𝑠)2  6𝑠2 𝐴 = 𝑠[12𝑠 + 6(𝑠)] 𝑠 = 0 𝐴 = 12𝑠 𝑠 𝐴 𝑠 = 12𝑠 𝑡 𝑡

= 12(6)(2)

𝑚 𝑚2 = 144 𝑠 𝑠

𝑏) 𝑣 = 𝑠 3 𝑑𝑣 𝑑𝑠 = 3𝑠 2 𝑑𝑡 𝑑𝑡 𝑑𝑣 = 3(6)2 (2) 𝑑𝑡 𝑑𝑣 = 63 𝑑𝑡 𝑑𝑣 𝑚3 = 216 𝑑𝑡 𝑠 6. ) 𝑎. )𝐴 = 4𝜋𝑟 2 𝐴 + 𝐴 = 4(𝑟 + 𝑟)2

𝐴 = 4[𝑟 2 + 2𝑟(𝑟) + (𝑟)2 ]  𝐴 𝐴 = 4𝑟 2 + 8𝑟(𝑟) + 4(𝑟)2 ]  4𝑟 2 𝐴 = 𝑟[8𝑟 + (𝑟)] 𝑟 = 0 𝐴 = 8𝜋𝑟 𝑟

47

4𝜋𝑟 3 𝑏)𝑣 = 3 4𝜋 𝑣 + 𝑣 = (𝑟 + 𝑟)3 3 4𝜋 3 𝑣 = (𝑟 + 3(𝑟)𝑟 2 + 3𝑟(𝑟)2 + (𝑟)3 ) − 𝑣 3 4𝜋𝑟 3 4𝜋(𝑟)3 2 2 𝑣 = + 4𝜋(𝑟)𝑟 + 4𝜋𝑟(𝑟) + 3 3 𝑣 𝑟 4𝜋 = [4𝜋𝑟 2 + 4𝜋(𝑟)𝑟 + (𝑟)2 ] 𝑡 𝑡 3

𝑟 = 0 𝑣 𝑟 [4𝜋𝑟 2 ] = 𝑡 𝑡 𝑣 𝑚 = (2 )(4𝜋)(6𝑚)2 𝑡 𝑠 𝑣 𝑚3 = 288𝜋 𝑡 𝑠 𝑐. ) 𝑦 = 3 + 4𝑥𝑥 2 𝑦 + 𝑦 = 3 + 4(𝑥 + 𝑥) − (𝑥 + 𝑥)2 𝑦 = 3 + 4𝑥 + 4𝑥 − (𝑥)2 − 2𝑥𝑥 − 𝑥 2 − 𝑦 𝑦 = 3 + 4𝑥 + 4𝑥 − (𝑥)2 − 2𝑥𝑥 − 𝑥 2 − (3 + 4𝑥 − 𝑥2) 𝑦 = 4𝑥 − (𝑥)2 − 2𝑥𝑥 𝑦 = 𝑥(4𝑥 − 2𝑥) 𝑦 = 4 − 𝑥 − 2𝑥 𝑥 𝑥0 𝒁=

𝒚 = 𝟒 − 𝟐𝒙 𝑓𝑖𝑟𝑠𝑡 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝒙

𝑍 = 4 − 2𝑥

𝑧 + 𝑧 = 4 − 2(𝑥 + 𝑥) 𝑧 = 4 − 2𝑥 − 2𝑥 − 𝑧 𝑧 = 4 − 2𝑥 − 2𝑥 − (4 − 2𝑥) 𝑧 = −2𝑥 𝒛 = −𝟐 𝒙

𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 48

𝑦 = 3𝑥 2 − 2𝑥(2,8) 𝑦 + 𝑦 = 3(𝑥 + 𝑥)2 − 2(𝑥 + 𝑥)

𝑦 = 3((𝑥)2 + 2𝑥𝑥 + 𝑥 2 ) − 2𝑥 − 2𝑥 − 𝑦 𝑦 = 3(𝑥)2 + 6𝑥𝑥 + 3𝑥 2 − 2𝑥 − 2𝑥 − (3𝑥2 − 2𝑥) 𝑦 = 3(𝑥)2 + 6𝑥𝑥 − 2𝑥 𝑦 = 𝑥(3𝑥 + 6𝑥 − 2)

𝑦 = 3𝑥 + 6𝑥 − 2 𝑥 𝑥0

𝑦 = 6(2) − 2 (2,8) 𝑥 𝑦 = 12 − 2 𝑥 𝑦 = 10 𝑥 7.) Find the vertex of the parabola 4𝑥 2 + 2𝑥 − 6𝑦 + 100 = 0 4(𝑥 + 𝑥)2 + 2𝑥𝑥 + 𝑥2 + 2𝑥 + 2𝑥 − 6𝑦 − 6𝑦 + 100 = 0 𝑦 =

4𝑥 2 + 2𝑥 + 100 6

2𝑥 2 𝑥 50 + + 3 3 3 2 1 50 𝑦 + 𝑦 = (𝑥 + 𝑥)2 + (𝑥 + 𝑥) + 3 3 3 2 𝑥 𝑥 50 𝑦 = ((𝑥)2 + 2𝑥𝑥 + 𝑥2) + + + −𝑦 3 3 3 3 𝑦 =

𝑦 =

2(𝑥)2 4𝑥𝑥 2𝑥 2 𝑥 𝑥 50 2𝑥 2 𝑥 50 + + + + + −( + + ) 3 3 3 3 3 3 3 3 3

2(𝑥)2 4𝑥𝑥 𝑥 + + 3 3 3 2𝑥 4𝑥 1 = 𝑥 [ + + ] 3 3 3 =

49

𝛥𝑦 2𝑥 4𝑥 1 = + + 𝛥𝑥 3 3 3

𝑥 = 0 4𝑥 1 + 3 3 4𝑥 −1 = 3 3 0=

12𝑥 = −3

𝑥 = −

1 4

−1 2 −1 4 ( ) + 2 ( ) – 6𝑦 = – 100 4 4 4 1 − – 6𝑦 = – 100 16 2 1 1 − – 6𝑦 = – 100 4 2 −400 + 1 – 6𝑦 = 4(−6) 𝑦=

−399 4(+6)

𝑉 (−

1 133 ) 4, 8

50

8.) Find the points where the tangent is parallel to the x-axis 𝑦 = 12𝑥 2 + 6𝑥 − 18 𝑦 − 𝑦 = 12(𝑥 + 𝑥)2 + 6(𝑥 + 𝑥) − 18

𝑦 = 12((𝑥)2 + 2𝑥𝑥 + 𝑥 2 ) + 6𝑥 + 6𝑥 − 18 − 𝑦 𝑦 = 12(𝑥)2 + 24𝑥𝑥 + 12𝑥 2 + 6𝑥 + 6𝑥 − 18 − 𝑦 𝑦 = 𝑥(12𝑥 + 24𝑥 + 6) 𝑥 = 0 𝑦 = 24𝑥 + 6 𝑥 24𝑥 + 6 = 0 −6 24 −1 𝑥= 4 𝑥 =

−1 2 −1 𝑦 = 12 ( ) + 6 ( ) − 18 4 4 12 3 = − – 18 6 2 12 − 24 − 18 (16) = 16 −12 − 288 = 16 −300 = 16 −150 = 8 −75 = 4 −1 −75 , 4 4

51

Derivative of Constant 𝑦 = 7𝑥 0 7(0)𝑥 −1 0 General Power Formula 𝑦 = 𝑐𝑥 𝑛 𝐝𝐲 = 𝐲 ′ = 𝐜𝐧(𝐱)𝐧−𝟏 𝐝𝐱 Derivative of Sum 𝒅(𝒖 + 𝒗) =

𝒅𝒖 𝒅𝒗 + 𝒅𝒙 𝒅𝒙

1. 𝑦 = 2𝑥 3 − 7𝑥 + 1 𝒅𝒚 = (𝟔𝒙𝟐 − 𝟕) 𝒅𝒙 2. 𝑦 = 3𝑥 −1 − 4𝑥 −2 𝑑𝑦 = 3(−1)x −2 − 4(−2)𝑥 −3 𝑑𝑥 𝐝𝐲 −𝟑 𝟖 = 𝟐 + 𝟑 𝐝𝐱 𝒙 𝒙

3. 𝑦 =

3 5 − 𝑥 𝑥2

𝑑𝑦 = 3𝑥 −1 − 5𝑥 −2 𝑑𝑥 𝑑𝑦 = 3(−1)x −2 − 5(−2)x −3 𝑑𝑥 𝒅𝒚 −𝟑 𝟏𝟎 = + 𝒅𝒙 𝒙𝟐 𝒙𝟑 1

1

4. 𝑦 = 𝑥 2 − 4𝑥 −2

52

1 1 1−1 1 x 2 − 4 (− ) x −2−1 2 2 𝟑 𝟏 𝟏 𝟏 𝟐 𝒚′ = 𝒙−𝟐 + 𝟐𝒙−𝟐 = + 𝟑/𝟐 𝟏/𝟐 𝟐 𝟐𝒙 𝒙

𝑦′ =

5. 𝑥 = √𝑡 −

1 √𝑡

= [𝑡 −

1 −1 2 𝑡2]

1 1 1 + 3/2 −1 2−1 1 1 −1−1 2t 𝑥’ = [𝑡 − 𝑡 2 ] [1 + 𝑡 2 ] = 1 2 2 1 2 2 [𝑡 − 1 ] 𝑡2 1

1 2

2t 3/2 + 1 [𝑡 − 1 ] 𝑡2 2t 3/2 𝑥’ = 1∙ 1 1 2 1 2 2 [𝑡 − 1 ] [𝑡 − 1 ] 𝑡2 𝑡2 𝟏/𝟐

𝒙’ =

𝟐𝐭 𝟑/𝟐 + 𝟏 𝐭 𝟑/𝟐 − 𝟏 ∙ ( 𝟏/𝟐 ) 𝟐𝐭 𝟑/𝟐 𝒕 𝟐 [𝒕 −

𝟏

𝟏] 𝒕𝟐

Derivative of Product 𝒅(𝒖𝒗) =

1.

𝒅𝒖 𝒅𝒗 𝒗+𝒖 𝒅𝒙 𝒅𝒙

𝑦 = (1 + 𝑥 2 )(3 − 2𝑥)

𝑦’ = (2𝑥)(3 − 2𝑥) + (1 + 𝑥 2 )(−2) 𝑦 ′ = 6𝑥 − 4𝑥 2 − 2 − 2𝑥 2 𝒚′ = −𝟔𝒙𝟐 + 𝟔𝒙 − 𝟐 = −(𝟔𝒙𝟐 − 𝟔𝐱 + 𝟐) 2. 𝑦 = (𝑣 2 + 3)( 𝑣 2 + 𝑣 + 1) y ′ = (2𝑣) (𝑣 2 + 𝑣 + 1) + (𝑣 2 + 3) (2𝑣 + 1) y ′ = 2𝑣 3 + 2𝑣 2 + 2𝑣 + 2𝑣 3 + 𝑣 2 + 6𝑣 + 3 53

𝒚′ = 𝟒𝒗𝟑 + 𝟑𝒗𝟐 + 𝟖𝒗 + 𝟑

Derivative of a Quotient (𝐝𝐮) 𝐯 − 𝐮 (𝐝𝐯) 𝐮 𝐝 ( )= 𝐯 𝐯𝟐 1. 𝑦 = 𝑦′ =

𝑥2

𝑥 −1

1(𝑥 2 − 1) – 𝑥 (2𝑥) (𝑥 2 − 1)2

𝑥 2 − 1 − 2𝑥 2 y = (𝑥 2 − 1)2 ′

y′ =

−𝑥 2 − 1 (𝑥 2 − 1)2

−𝟏(𝒙𝟐 + 𝟏) 𝒚 = (𝒙𝟐 − 𝟏)𝟐 ′

𝑥2 2. 𝑦 = 2 𝑥 −1 2𝑥(𝑥 2 − 1) − 𝑥 2 (2𝑥) y = (𝑥 2 − 1)2 ′

y′ = 𝒚′ =

2𝑥 3 − 2𝑥 − 2𝑥 3 (𝑥 2 − 1)2 −𝟐𝒙 − 𝟏)𝟐

(𝒙𝟐

1 3𝑥 − 2 0(3𝑥 − 2) − (3)(1) 𝑦′ = (3𝑥 − 2)2

3. 𝑦 =

𝐲′ =

−𝟑 (𝟑𝒙 − 𝟐)𝟐

y′ =

−𝑑(𝑣)𝑐 𝑖𝑓 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣2 54

𝑦′ =

3𝑥 − 2 4

𝑦′ =

3(4) − (3𝑥 − 2)0 42

=

12 𝟑 = 16 𝟒

MORE PROBLEMS 1. 𝑦 = 𝑥 2 𝒚′ = 𝟐𝒙 2. 𝑦 = (𝑥 2 + 1) (𝑥 2 + 1) 𝑦 ′ = (2𝑥) (𝑥 2 + 1) + (𝑥 2 + 1) (2𝑥) 𝑦 ′ = 2𝑥 3 + 2𝑥 + 2𝑥 3 + 2𝑥 𝑦 ′ = 4𝑥 3 + 4𝑥 𝒚′ = 𝟒𝒙(𝒙𝟐 + 𝟏)

3. 𝑦 = 4𝑥(𝑥 2 + 1) 𝑦′ =

𝑑𝑦 = 4(𝑥 2 + 1) + 4𝑥(2𝑥) 𝑑𝑥

𝑦 ′ = 4𝑥 2 + 4 + 8𝑥 2 𝑦 ′ = 12𝑥 2 + 4 𝒚 = 𝟒(𝟑𝒙𝟐 + 𝟏) 4. 𝑦 = (3𝑥 2 + 4𝑥 − 1)3 𝑦 ′ = 3(3𝑥 2 + 4𝑥 − 1)2 (6𝑥 + 4) 𝒚′ = 𝟔(𝟑𝒙 + 𝟐) (𝟑𝒙𝟐 + 𝟒𝒙 − 𝟏)𝟐 D((𝐮)𝐧 ) = n(𝐮𝐧−𝟏 ) du

1. 𝑦 = 6√4 + 𝑥 1 1 1 𝑦 = 6(4 + 𝑥)2 = 6 ( ) (4 + 𝑥)2−1 2

55

1

𝑦 ′ = 3(4 + 𝑥)−2 (1) 𝒚′ =

𝟑 √𝟒 + 𝒙

Name: Course/Year & Section: Assignment no. 4 Derive

1. )𝑦 = 7𝑥 4 − 3𝑥 3 − 5𝑥 + 1000 𝒚′ = 𝟐𝟖𝒙𝟑 − 𝟗𝒙𝟐 − 𝟓 2. )2𝑦 = 8𝑥 3 − 10𝑥 + 4 𝑦 = 4𝑥 3 − 5𝑥 + 2 𝒚′ = 𝟏𝟐𝒙𝟐 − 𝟓 3. )𝑦 = (4𝑥 2 + 3𝑥 + 1) (2𝑥 2 + 7) 𝑦 ′ = (8𝑥 + 3)( 2𝑥 2 + 7) + (4𝑥 2 + 3𝑥 + 1)(4𝑥) 𝑦 ′ = 16𝑥 3 + 56𝑥 + 6𝑥 2 + 21 + 16𝑥 3 + 12𝑥 2 + 4𝑥 𝒚′ = 𝟑𝟐𝒙𝟑 + 𝟏𝟖𝒙𝟐 + 𝟔𝟎𝒙 + 𝟐𝟏 4. )𝑦 = 𝑥(𝑥 − 1) (𝑥 + 1) 𝑦’ = 𝑥(𝑥 2 − 1) 56

𝑦 ′ = 1(𝑥 2 − 1) + 𝑥(2𝑥) 𝒚’ = 𝒙𝟐 − 𝟏 + 𝟐𝒙𝟐 = 𝟑𝒙𝟐 − 𝟏 5. )𝑦 =

3𝑥 2 − 2𝑥 + 5 2𝑥 − 5

(6𝑥 − 2)(2𝑥 − 5) − (2)(3𝑥 2 − 2𝑥 + 5) (2𝑥 − 5)2 2 12𝑥 − 30𝑥 − 4𝑥 + 10 − 6𝑥 2 + 4𝑥 − 10 ′ 𝑦 = (2𝑥 − 5)2 2 6𝑥 − 30𝑥 𝑦′ = (2𝑥 − 5)2 𝟔𝒙(𝒙 − 𝟓) 𝒚′ = (𝟐𝒙 − 𝟓)𝟐 𝑦′ =

𝑥2 − 4 6. )𝑦 = 2 𝑥 + 4𝑥 + 4 2𝑥(𝑥 2 + 4𝑥 + 4) − (𝑥 2 − 4)(2𝑥 + 4) (𝑥 2 + 4𝑥 + 4)2 3 2 2𝑥 + 8𝑥 + 8𝑥 − 2𝑥 3 − 4𝑥 2 + 8𝑥 + 16 𝑦’ = (𝑥 2 + 4𝑥 + 4)2 4𝑥 2 + 16𝑥 + 16 4(𝑥 + 2)2 𝑦′ = 2 = (𝑥 + 4𝑥 + 4)2 (𝑥 + 2)4 𝟒 𝒚’ = (𝒙 + 𝟐)𝟐 𝑦’ =

7. ) 𝑦 = √4𝑥 2 + 9 (2𝑥 + 3)6 1

𝑦 = (4𝑥 2 + 9)2 (2𝑥 + 3)6 𝑦′ = 𝑦′ =

1 1 (4𝑥 2 + 9)−2 (8𝑥) (2𝑥 + 3)6 + 6(2𝑥 + 3)5 (4𝑥 2 + 9)1/2 (2) 2

8𝑥(2𝑥 + 3)6 1

1

+ 12(4𝑥 2 + 9 )2 (2𝑥 + 3)5

2(4𝑥 2 + 9)2 𝑦 ′ = (2𝑥 + 3)5 [

5[

𝑦’ = (2𝑥 + 3)

4𝑥(2𝑥 + 3)

1

2 2 1 + 12(4𝑥 + 9) ] 2 (4𝑥 + 9 )2

8𝑥 2 + 12𝑥 + 12 (4𝑥 2 +9) ] 1 (4𝑥 2 + 9)2

57

=

(2𝑥 + 3)5 (4𝑥 2

8. ) 𝑦 =

𝑦′ =

1 9)2

+

(56𝑥 2

+ 12𝑥 + 108) =

(𝟒𝒙𝟐 +

4𝑥 2 + 9

𝑦=

√2𝑥 − 3

(8𝑥)(2𝑥 −

1 3)2

4𝑥 2 + 9 1

−1

4𝑥 2 + 9 1

(2𝑥 − 3)2 (2𝑥 − 3)

𝑦′ =

8𝑥(2𝑥 − 3) − (4𝑥 2 + 9) 3

=

16𝑥 2 − 24𝑥 − 4𝑥 2 − 9

(2𝑥 − 3)2 𝑦′ =

(𝟏𝟒𝒙𝟐 + 𝟑𝒙 + 𝟐𝟕)

(4𝑥 2 + 9 )(2𝑥 − 3) 2 (2) − 2 (2𝑥 − 3)

1

𝑦 =

𝟏 𝟗)𝟐

(2𝑥 − 3)2

8𝑥(2𝑥 − 3 )2 −

′

𝟒(𝟐𝒙 + 𝟑)𝟓

3

(2𝑥 − 3)2

12𝑥 2 − 24𝑥 − 9 3

(2𝑥 − 3)2 ′

𝒚 =

𝟑(𝟒𝒙𝟐 − 𝟖𝒙 − 𝟑) 𝟑

(𝟐𝒙 − 𝟑)𝟐

9. ) 𝑦 =

−100 √5𝑥 2 − 3

1

𝑦′ = −100 (5𝑥 2 − 3)−2 3 −1 𝑦′ = −100 ( ) (5𝑥 2 − 3 )−2 (10𝑥) 2 𝟓𝟎𝟎𝒙 𝒚′ = 𝟑 (𝟓𝒙𝟐 − 𝟑)𝟐

3

10. ) 𝑦 = √7𝑥 2 + 5 √5𝑥 − 3 𝑦 = (7𝑥 2 + 5)1/3 (5𝑥 − 3)1/2

58

𝑦′ = 𝑦′ =

2 1 1 1 (7𝑥 2 + 5)−3 (14𝑥) (5𝑥 − 3)2 + (7𝑥 2 + 5)1/3 ( ) (5𝑥 − 3)−1/2 (5) 3 2

14𝑥(5𝑥 − 3)1/2 2

+

3(7𝑥 2 + 5)3

5(7𝑥 2 + 5)1/3 2(5𝑥 − 3)1/2

28𝑥(5𝑥 – 3) + 15(7𝑥 2 + 5) 𝑦′ = 6(7𝑥 2 + 5)2/3 (5𝑥 − 3)1/2 𝑦′ =

140𝑥 2 − 84𝑥 + 105𝑥 2 + 75 2

1

6(7𝑥 2 + 5 )3 (5𝑥 − 3)2 𝒚’ =

𝟐𝟒𝟓𝒙𝟐 − 𝟖𝟒𝒙 + 𝟕𝟓 𝟔(𝟕𝒙𝟐 + 𝟓)𝟐/𝟑 (𝟓𝒙 − 𝟑)𝟏/𝟐

Examples: 1. y = 2 − x 2 (3, − 7) y ′ = −2x = −2(3) = −𝟔

Tangent

2. y = 3x 2 − 2x(2, 8) y ′ = 6x − 2 = 6(2) − 2 = 𝟏𝟎

59

4 3. 𝑣 = 𝜋𝑟 3 3 𝑑𝑣 4𝜋 𝑑𝑟 (3)𝑟 2 = 𝑑𝑡 3 𝑑𝑡 𝑑𝑣 6𝑓𝑡 = 4𝜋𝑟 2 ( ) 𝑑𝑡 𝑠

𝑑𝑟 𝑓𝑡 𝑟 =6 = 𝑑𝑡 𝑠 𝑡 6=

𝑟 ; 0.25

𝑡 = 0.25

𝑟 = 1.5

𝑑𝑣 = 24𝜋𝑟 2 = 24𝜋(1.5)2 = 𝟓𝟒𝝅 𝑑𝑡 𝑆=

d t

𝐯 𝐨𝐟 𝐬𝐩𝐡𝐞𝐫𝐞 P of circle rectangle dr r =6 = 1 dt 4 1 6( ) = 𝑟 4 1.5=r 𝑉 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟 2 ℎ;

𝑖𝑓 ℎ = 6

𝑣 = 6𝜋𝑟 2 𝑑𝑣 𝑑𝑟 = 12𝜋𝑟 𝑑𝑡 𝑑𝑡 𝑑𝑣 = 12𝜋(1.5)(6) 𝑑𝑡 𝑑𝑣 = 𝟏𝟎𝟖𝛑 𝑑𝑡

SLOPE 1. 𝑦 2 =

𝑥3 @ (𝑎, 𝑎) 2𝑎 − 𝑥

3𝑥 2 𝑑𝑥 (2𝑎 − 𝑥 ) − 𝑥 3 (−1)𝑑𝑥 2𝑦𝑑𝑦 = (2𝑎 − 𝑥)2 60

6𝑎𝑥 2 − 3𝑥 3 + 𝑥 3 2𝑦𝑦’ = (2𝑎 − 𝑥)2 𝑦’ =

6𝑎𝑥 2 − 2𝑥 3 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 2𝑦(2𝑎 − 𝑥)2

𝑦′ =

6aa2 − 2a3 2a(2a − a)2

4a3 𝑦′ = 3 2a 𝐲’ = 𝟐

𝑦3 =

2.

2𝑎𝑥 3 @ (𝑎, 𝑎) 𝑥+𝑎

6ax 2 (x + a) − 2ax 3 (1) (x + a)2 3 6ax + 6a2 x 2 − 2ax 3 ′ y = 3y 2 (x + a)2 3 4𝑎𝑥 + 6𝑎2 𝑥 2 ′ 𝑦 = 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 3𝑦 2 (𝑥 + 𝑎)2

3y 2 y ′ =

y′ =

4aa3 + 6a2 a2 3a2 (a + a)2

y′ =

10a4 12a4

𝒚′ =

𝟓 𝟔

POLYNOMIAL OF A TERM 1. 𝑦 = 3𝑥 2 − 2𝑥 = 1 @ (1 ,2) y ′ = 6x − 2 Derivative y ′ = 6(1) – 2 𝐲 ′ = 𝟒 Slope 𝑚=

𝑦 − 𝑦1 𝑥 − 𝑥1

𝑺𝒍𝒐𝒑𝒆 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 61

4=

y−2 x−1

4x − 4 = y – 2 𝟒𝒙 − 𝒚 = 𝟐

𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝑳𝒊𝒏𝒆

−1 y − y1 = 4 x − x1 −1 y − 2 = 4 x−1 −x + 1 = 4y − 8 𝐱 + 𝟒𝐲 = 𝟗

2.

Normal Line

y = x 2 − 2x @ its pt. of intersection w/ the line 𝑦 = 3

y + 1 = x 2 − 2x + 1 (-1, 3)

(3, 3)

y + 1 = (x − 1)2 V (1, -1) up P.I (-1, 3) (3, 3) y = x 2 − 2x 3 = x 2 − 2x 0 = x 2 − 2x − 3

FACTOR

(x – 3) (x + 1) = 0 x = 3 x = −1 y = x 2 − 2x 𝐲 ′ = 𝟐𝐱 − 𝟐 @ (-1, 3) y’ = 2 (−1) − 2 = −𝟒 62

y’ = 2 (3) – 2 = 𝟒

−4 =

y − 3 x + 1

−4x – 4 = y – 3 𝟒𝐱 + 𝐲 = − 𝟏 4 =

y− 3 x − 3

𝟒𝐱 – 𝟏𝟐 = 𝐲 – 𝟑

𝟒𝒙 – 𝒚 = 𝟗 Parabola x 2 = y + c Circle

x 2 + y 2 = a2

Ellipse ax 2 + by 2 = c Hyperbola ax 2 − by 2 = c

1. To the ellipse x 2 + 4y 2 = 8 || to the line 𝑥 + 2𝑦 = 6 x 2 + 4y 2 = 8 8 x2 y2 + =1 8 2 x2 y2 + =1 a2 b2 x + 2y = 6 2y = −x + 6 2 y = −

1x + 3; 2

m=−

1 2

slope of line 63

x 2 + 4y 2 = 8 2x + 8yy’ = 0 y’ =

−2x −x 1 = = − 8y 4y 2

x =1 2y x =2y x 2 + 4y 2 = 8 (2y)2 + 4y 2 = 8 4y 2 + 4y 2 = 8 8y 2 = 8 8 y2 = 1 𝐲 =±𝟏 y=1

y = -1

x= 2 x = - 2 (-2 , -1)

−1 y+1 = 2 x+2 -x -2 = 2y + 2 x + 2y = -4 −1 y−1 = 2 x−2 −x + 2 = 2y – 2 x + 2y = 4

2. To the parabola y 2 = 6x − 3 ∟ to the line x + 3y = 7 𝑦 2 = 6(𝑥 − 1⁄2) 64

𝑣(1⁄2 , 0 ) 𝑥 𝑖𝑛𝑡 𝑥 – 𝑖𝑛𝑡

1 2 3 4 5 6 7 8

𝑥 = 0 𝑦 = 0 𝑛𝑜𝑛𝑒

(1/2 , 0)

(2𝑦𝑦′ = 6) 2𝑦 𝑦’ =

3 𝑦

3 =

3 𝑦

y=1 3y = −x + 7 y =

−1 7 x+ 3 3

m =

−1 3

m ⥕= 3 𝑦 2 = 6𝑥 − 3 12 = 6𝑥 − 3 4 2 =𝑥= 6 3 2 ( , 1) 3 3=

𝑦−1 2 𝑥−3

𝟑𝒙 − 𝟐 = 𝒚 − 𝟏 𝟑𝒙 − 𝒚 = 𝟏

65

3. To the cubic y = x 3 − 2x + 3 || to 10𝑥 – 𝑦 = 3 𝑦 = 10𝑥 − 3 𝑚 = 10 y’ = 3x 2 − 2 = 10 y′ =

3x 2 = 12 3

x2 = 4 𝐱 𝟏 = +𝟐 𝐱 𝟐 = −𝟐 If x = + 2 y = 23 − 2(2) + 3 = 8 -4 + 3 y=7

(2, 7)

If x = - 2 y = −23 − 2(−2) + 3 = -8 + 4 + 3 y= -1 (-2, -1) 10 =

y−7 x−2

10x -20 = y -7 10x –y = 13 10 −

y +1 x+2

10x + 20x = y + 1 10x – y = -19

4. Make the parabola y = ax 2 + bx + c pass thru (2, 1) to be tangent to the line y = 2x + 4 @ (1, 6) @ (2, 1) 66

1 = a(2)2 + b(2) + c 1 = 4a + 2b +c

1 equation

@ (1 , 6) 6 = a(1)2 + b(1) +c 6= a + b + c

2 equation

y = ax 2 + bx + c y’ = 2ax + b 2 = 2a (1) + b y = 2x + 4 y’= 2

2 = 2a + b

3equation

1equation - 2equation 1 = 4a + 2b + c -(6 = a + b + c)

-5 = 3a + b

4equation

3equation to 4equation -5 = 3a + b -(2 = 2a + b) -7 = a a = - 7 5 equation

5equation in 4equation -5 = 3 (-7) + b -5+21 = b = 16 b = 16

6= a + b + c 6 = -7 + 16 + c C = -3 67

y = −7x 2 + 16x − 3

5. Make the cube y = ax 3 + bx 2 + cx + d be tangent to y = 12x + 13 @ (-1 , 1) to have a horizontal tangent line @ (1 , 5). @ (-1 , 1) 1 = -a + b – c + d

equation 1

@(1 , 5) 5= a + b + c + d

equation 2

y ′ = 3ax 2 + 2bx + c 12 = 3a(−1)2 + 2b(1) + c 12 = 3a – 2b + c

3equation

0 = 3a(1)2 + 2b(1) + c 0 = 3a + 2b + c

4 equation

1equation to 2equation 1=-a+b–c+d -(5 = a + b +c +d) -4 = -2a – 2c 2=a+c

5equation

y = 12x + 13 y’ = 12 3equation to 4equation 12 = 3a – 2b + c 0 = 3a + 2b + c 12 = 6a + 2c 2

68

6 = 3a+c

6 equation

6=3a+c -(2 = a + c) 4=2a a=2

7equation

c=0

8 equation

7 equation to 8 equation in 4 equation 0=3a+2b+c 0=3(2)+2b -6=2b b=-3 9 equation 7equation, 8 equation to 9 equation in 4 equation 1=-a+b-c+d 1=-2-3+d d=6 y = ax 3 + bx 2 + cx + d y=𝟐𝒙𝟑 − 𝟑𝒙𝟐 + 𝟔 MAXIMA AND MINIMA 1. y = 2 + 12x − x 3 y’ = 12 − 3x 2 = 0 =

12 = 3x 2 3

4 = x2 x = ±2 y = 2 + 12(2) – (2)3 = 2 + 24 – 8 =18 y = 2 + 12(−2) – (−2)3 69

= 2 – 24 + 8 = -14

(2, 18) (-2 , -14) (2, 18) max

(-2, -14) min

x= 1

2 3 -1 -2 -3

y = 13 18 11 -9 -14 -7

  

y = 0 if y changes from + to – as x- increase y is a maximum if y’ changes from – to + y is minimum y’ + to – y is maximum y’ does not change sign y ℵ neither maximum or minimum if y’ > 0 y increases y’ < 0 y decreases Name: Course/Year & Section: Assignment no.5: maxima and minima Answers:

1. y = − 6x + x 2 𝑦’ = 0 = − 6 + 2𝑥 6 = 2𝑥 𝒙 = 𝟑 y = − 6 (3) + (3)2 𝑦 = − 18 + 9 𝒚 = −𝟗 C. P (3 , -9) ,minimum @ (2 𝑦’ = −6 + 2𝑥 𝑦′ = − 6 + 2(2) 70

𝒚′ = −𝟐 @ (4) 𝑦’ = − 6 + (2)(4) y′ = +2

2. y = 4x 2 + 16x + 9 𝑦’ = 8𝑥 + 16 = 0 𝒙= −𝟐 y = 4(−2)2 + 16(−2) + 9 𝑦 = 16 – 32 + 9 𝒚 = −𝟕 C.P (-2 , -7) minimum 𝐼𝑓 𝑥 = −3 𝑦’ = 8 (−3) + 16 𝑦′ = −24 + 16 𝒚′ = −𝟖 @ 𝑥 = −1 𝑦’ = 8(−1) + 16 𝑦′ = −8 + 16 𝒚′ = 𝟖 3. y = (2x − 1)2 = 4(2x - 1) = 0 x=½ 1 y = (2 ( ) − 1)2 2 =0 C.P (1/2 , 0) minima y ′ = 4(2(0) − 1) y=4 y’ = 4(2(1) − 1) = 4(2-1) =4 4. y = −4(x + 2)2 y’ − 8(x + 2)1 (1) 𝑦’ = − 8(𝑥 + 2) = 0 𝑥 + 2 = 0⁄8 71

𝑥 + 2 = 0 𝒙 = −𝟐 y = −4(−2 + 2)2 𝑦= 0 C.P (−2 , 0) max 𝑦’ = −8 (−3 + 2) 𝒚′ = +𝟖 𝑥 = −1 𝑦’ = −8 (−1 + 2) 𝒚′ = −𝟖



y” is + y’ is increasing as x- increases the tangent turns in a ccw sense to the curve is concave upward.

x

CCW

CW

y” < 0 y is maximum y” > 0 y is minimum y“ = 0 test fails

E.G. y =

1 3 1 2 x − x − 2x + 2 3 2

𝑦′ = 𝑥2 − 𝑥 − 2 = 0 (𝑥 − 2)(𝑥 + 1) = 0 72

𝑥=2

𝑥 = −1

If 𝑥 = 2 1 1 (2)3 − (2)2 − 2(2) + 2 3 2 8 𝑦 = −2−4+2 3 𝑦=

𝑦=

8 4 − 3 1

𝑦 =

8 − 12 3

−4 3 −4 C. P(2 ) 3 𝑦=

If x = 1 1 1 (−1)3 − (−1)2 − 2(−1) + 2 3 2 𝑦 = − 1⁄3 − 1⁄2 + 2 + 2 4 𝑦 = − 1⁄3 − 1⁄2 + 1 𝑦 = −2 − 2 + 24⁄6 = 19⁄6 𝑦=

𝐌𝐚𝐱𝐢𝐦𝐚

19

C.P (−1, 6 ) Minima (2, -4/3) 𝑥 = −2 𝑦′ = 𝑥2 − 𝑥 − 2 𝑦’ = (−2)2 − (−2) − 2 𝑦’ = 4 + 2 − 2 𝒚’ = 𝟒 𝑥 = 0 𝑦 ′ = 12 − 1 − 2 𝑦’ = 1 − 1 − 2 𝒚 = −𝟐 𝑥 = 3 𝑦’ = 32 − 3 − 2 𝑦’ = 9 − 5 𝒚 = 𝟒 𝑦′ = 𝑥2 − 𝑥 − 2 = 0 𝑦” = 2𝑥 − 1 𝑦’’ = 2(−1) − 1 73

𝒚” = −𝟑 𝑦” = 2(2) − 1 𝑦” = 4 − 1 𝒚” = 𝟑 𝑦” = 2𝑥 − 1 = 0 1 𝑥= 2 1 1 3 1 1 2 1 𝑦 = ( ) − ( ) − 2( ) + 2 3 2 2 2 2 1 1 1 1 1 1 − 3 + 24 𝑦 = ( )− −1+2 𝑦= − +1= 3 8 8 24 8 24

𝑦=

22 11 = 24 12

Point of Inflection A point @ which the curve changes from concave upward to concave downward or vice versa. Sketching Polynomial Curve 1. Find the points of intersection with the axes. 2. Determine the behavior of y for large values of x. 3. Locate points where 𝑦 ′ = 0 to determine max or min. 4. Locate points where 𝑦 " = 0. 5. Plot additional points. 𝑦 = 𝑥 3 − 3𝑥 y = x(𝑥 2 − 3) Intercepts y=0 0 = x (𝑥 2 − 3) x = 0 + √3√3 (0 , 0) (√3, 0)(−√3, 0) x=0 y=0 𝑦 = 𝑥 3 − 3𝑥 𝑦 ′ = 3𝑥 2 − 3 = 0 𝑥2 = 1 𝑦 = (1)3 − 3(1) y= 1 – 3 y= -2 𝑦 = (−1)3 − 3(−1) 74

𝑦 = −1 + 3 𝑦= 2 𝐶𝑃 (1 , −2) (−1 , 2) 𝑦 " = 6𝑥 𝑦” = 6(1) 𝑦” = 6 𝑦 " = 6(−1) = −6 𝑦 " = 6𝑥 = 0 𝑥= 0 𝑦= 0

Name: Course/Year & Section:

Assignment no.6: Point of Inflection 1. )y = x³ − 3x² 𝑦’ = 3𝑥 2 − 6𝑥 0 = 3𝑥 (𝑥 − 2) 𝑥 = 0

𝑥 = 2

𝑦 = 03 − 3(0)2 = 0 𝑦 = 23 − 3(2)2 = 8 − 12 = −4 0 = 𝑥 2 (𝑥 − 3) 𝑥 =0

𝑥 =3

C.P Maxima(0,0) Minima(2, −4) 𝑦” = 6𝑥 − 6 = 0 𝑥 = 1 𝑦 = 13 − 3(1)2 𝑦= 1− 3 75

𝑦 = −2 𝑃. 1 (1, −2) 𝑦” = 6(0) − 6 = −6 𝑦” = 6(2) − 6 𝑦" = 12 − 6 𝑦" = 6 2. )y = x 3 + 3x 2 + 3x 𝑦 ′ = 3(𝑥 2 + 2𝑥 + 1) 𝑦’ = 3𝑥 2 + 6𝑥 + 3 = 0 0 = (3𝑥 + 3)(𝑥 + 1) 𝑥 = −1 𝑦 = (−1) 3 + 3(−1)2 + 3(−1) 𝑦 = −1 + 3 − 3 𝑦 = −1 C.P (−1, −1) 𝑦” = 6𝑥 + 6 𝑦” = 6(−1) + 6 𝑦” = 0 0 = 6𝑥 + 6 𝑥 = −1 P.I (−1, −1) 𝑥 = 0; 𝑦 = 0 𝑦 = 0 0 = 𝑥(𝑥 2 + 3𝑥 + 3)

3. )y = 36 + 12x − x³ 𝑦’ = 12 − 3𝑥 2 = 0 0 = −3(−4 + 𝑥 2 ) 𝑥 = ±2 𝑦 = 36 + 12(2) – (2)3 𝑦 = 36 + 24 − 8 76

𝑦 = 20 C.P. (2,52) (−2,20) 𝑦” = 0 = −6𝑥 𝑥 = 0𝑦 = 36 P.I (0, 36) Interceps (0,36)

4. )y = 2x 3 − 9x 2 + 12x − 2 𝑦’ = 6𝑥 2 – 18𝑥 + 12 0 = 6(𝑥 2 − 3𝑥 + 2) 0 = (𝑥 – 2) (𝑥 – 1) 𝑥 = 2𝑥 = 1 𝑦 = 2(2)3 − 9(2) 2 + 12(2) – 2 𝑦 = 16 – 36 + 24 – 2 𝑦 = 2 𝑦 = 3

C.P. (2,2) (1,3) 𝑦” = 12𝑥 – 18 𝑦” = 12(2) – 18 = 6 minima 𝑦” = 12(1) – 18 = −6 maxima 𝑦” = 0 = 6(2𝑥 − 3) 𝑥 = 1.5 𝑦 = 2.5 P.I (1.5, 2.5) 𝑥 = 0 ; 𝑦 = −2 77

𝑦 = 0; 𝑥 = 5. )y = 28 − 15x + 6x² − x³ y’ = -15 + 12x – 3x2 y’ = -3(5 - 4x + x2) no. critical points y” = 12 – 6x = 0 x=2 y = 14 P.I = (2,14)

x y

-1 50

0 28

1 18

2 14

3 10

4 0

5 -22

APPLICATION OF DERIVATIVES 1. A box is to be made of a piece of cardboard 16x10in by cutting equal squares out the corners and turning up the sides. Find the volume of the largest box that can be made in this way. 𝑎𝑣 = 𝑥(10 − 2𝑥)(16 − 2𝑥) 𝑣 = 2(10 − 2𝑥) (16 – 2𝑥) 2) 𝑣 = 𝑥(160 − 20𝑥 − 32𝑥 + 4𝑥 𝑣 = 2(10 – 4) (16 – 4) 𝑣 = 160𝑥 − 52𝑥² + 4𝑥 3 𝑣 = 2(6) (12) 𝑣 ′ = 160 − 104𝑥 + 12𝑥² = 0 𝑣 = 144 𝑖𝑛3 ′ 2 𝑣 = 12𝑥 − 104𝑥 + 160 = 0 𝑣 ′ = 6𝑥 2 − 52𝑥 + 80 = 0 𝑣 ′ = 3𝑥 2 − 26𝑥 + 40 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 26 ± √262 − 4(3)(40) 𝑥= 2(3) 26 ± √676 − 480 𝑥= 6 26 ± √196 𝑥= 6 26 ± 14 𝑥= 6 40 20 𝑥1 = = = 6.67 6 3 𝑥2 = 2

78

2. Find the area of the largest rectangle that can be inscribed in a given circle. A = 4xy x² + y² = r² y² = r² − x² y = √r 2 − x 2 A = 4x√r 2 − x 2 1 𝐴′ = 4√𝑟 2 − 𝑥 2 + 4𝑥 ( ) (𝑟 2 − 𝑥²)−1/2 (−2𝑥) 2 4𝑥 2 𝐴’ = 4(𝑟 2 – 𝑥 2 )1/2 − √𝑟 2−𝑥2 2 2) 2 4(𝑟 − 𝑥 − 4𝑥 𝐴′ = =0 √𝑟 2 + 𝑥 2 4𝑟 2 8𝑥 2 = 8 8 𝑟2 = 𝑥² 2 r √2r x= = 2 √2 2 y = √r + x 2 y = √r 2 − r

r2 2

√2r =x 2 √2 2x = 2y = r√2 A = 4xy r√2 r√2 A = 4( )( ) = 2r² 2 2 y=

=

3. Find the altitude of the largest circular cylinder that can be inscribed in a cone of radius r and height h. v = πx 2 y 𝑣𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑥 2 𝑦 x h−y = r h r(h − y) x= h 𝜋𝑟 2 (ℎ − 𝑦)2 𝑦 𝑣𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = ℎ2 2 πr v′ = [(h − y)² + 2(h − y)(−1) y] = 0 h 2 (h − y) − 2y(h − y) = 0 (h − y)2 = 2y(h − y) 79

h − y = 2y h = 3y h y= 3

x=

h r (h − 3)

h 2 rh x=3 h 2 x= 3r 2r 2 h v = π( ) ( ) 3 3 4πr 2 h v= 27

USE OF AUXILLARY VARIABLE 1. Find the shape of the largest rectangle that can be inscribed in a given circle. 𝐴 = 4𝑥𝑦 𝑥² + 𝑦² = 𝑟² 𝐴′ = 4𝑥𝑦 ′ + 4𝑦 𝐴 = 4(𝑥𝑦 ′ + 𝑦) = 0 𝑥𝑦 ′ + 𝑦 = 0 𝑦 𝑦′ = − 𝑥 2𝑥 + 2𝑦𝑦 ′ = 0 𝑥 𝑦′ = − 𝑦 ′ ′ 𝑦 =𝑦 𝑦 𝑥 − =− 𝑥 𝑦 2 √𝑦 − √𝑥 2 𝑟√2 𝑥=𝑦= 2 𝐴 = 2𝑟²

2. What odd numbers added to its reciprocal gives the minimum sum? 1 𝑠=𝑥+ 𝑥 80

𝑠 ′ = 1 + (− 1 0= 1− 2 𝑥 1 =1 𝑥2

1 ) 𝑥2

√1 = 𝑥 2 𝑥 = ±1

3. What number exceeds its square by the maximum amount? 𝑥 𝑥² 𝑦 = 𝑥 − 𝑥² 𝑦 ′ = 1 − 2𝑥 = 0 1 = 2𝑥 𝟏 𝒙= 𝟐 4. The sum of two numbers is k. find the minimum value of the sum of their squares. 𝑥+𝑦 =𝑘 𝑥 =𝑘−𝑦 𝑟 = 𝑥2 + 𝑦2 𝑟 = (𝑘 − 𝑦)2 + 𝑦² 𝑟 ′ = 2(𝑘 − 𝑦)(−1) + 2𝑦 = 0 𝑦 =𝑘−𝑦 2𝑦 = 𝑘 𝑘 𝑦= 2 𝑟 = 𝑥2 + 𝑦2 𝑘 2 𝑟 = (𝑘 − 𝑦) + ( ) 2 2

𝑟 = 𝑘 2 − 2𝑘𝑦 + 𝑦 2 +

𝑘2 4

𝑘 𝑘 2 𝑘2 𝑟 = 𝑘 − 2𝑘 ( ) + ( ) + 2 2 4 2 2 𝑘 𝑘 𝑟 = 𝑘2 − 𝑘2 + + 4 4 𝒌𝟐 𝒓= 𝟐 2

81

5. The sum of two numbers is k. find the minimum value of their cubes. 𝑥+𝑦 =𝑘 𝑥 =𝑘−𝑦 𝑟 = 𝑥3 + 𝑦3 𝑟 = (𝑘 − 𝑦)3 + 𝑦³ 𝑟 ′ = 3(𝑘 − 𝑦)2 (−1) + 3𝑦 2 = 0 √𝑦 2 = √(𝑘 − 𝑦)2 𝑦 =𝑘−𝑦 2𝑦 = 𝑘 𝑘 𝑦= 2 𝑟 = 𝑥3 + 𝑦3 𝑟 = (𝑘 − 𝑦)3 + 𝑦 3 𝑘 3 𝑘 3 𝑟 = (𝑘 − ) + ( ) 2 2 𝑘 3 𝑘 3 𝑘3 𝑘3 𝑟 =( ) +( ) 𝑟= + 2 2 8 8

𝑘3 𝑟= 4

6. The sum of two + number is 2. Find the sum of the cube of one number and the square of the other. 𝑥 + 𝑦 = 2 𝑥 = 2– 𝑦 𝑟 = 𝑥3 + 𝑦2 𝑟 = (2 − 𝑦)3 + 𝑦 3 𝑟 ′ = 3(2 − 𝑦)2 (−1) + 2𝑦 = 0 2𝑦 = 3(4 − 4𝑦 + 𝑦 2 ) 𝑟 ′ = 12 − 12𝑦 + 3𝑦 2 − 2𝑦 𝑟’ = 12 – 14𝑦 + 3𝑦 2 = 0 3𝑦 2 − 14𝑦 + 12 = 0 14 ± √142 − 4(3)(12) 𝑦= 2(3) 14 ± √196 − 144 𝑦= 6 14 ± √52 𝑦= 6 14 ± 2√13 𝑦= 6 7 + √13 𝑦1 = = 3.54 3 7 − √13 𝑦2 = = 1.13 3 𝑥 = 2 – 1.13 = 0.87 𝑟 = 0.873 + 1.132 𝑟 = 1.94 82

Name: Course/Year & Section: Assignment no.7: AUXILLARY VARIABLE 1. The sum of two numbers is 4. Find the sum of the cube of one number and the square of the other.

2. A balloon leaving the ground 60ft from an observer, rises vertically at the rate of 10ft/s. how fast the balloon receding from the observer, after 8 seconds? d= vt s 2 = 602 + (10t)2 s² = 3600 + 100t² 83

s = √3600 + 100t² s = √100(36 + t²) s = 10√36 + t² s = 10(36 + t²)1/2 ds dt ds dt ds dt ds

=

10 2

(36 + t 2 )-1/2(2t)

= 10t/(36 + t² )1/2 = 10(8)/(36 + 8²)1/2

= 80/(100)1/2 ds 80 = dt 10 ds = 8 ft /s dt dt

2. As a man walks across a bridge at the rate of 5ft/s , a boat passes directly beneath him at 10ft/s. if the bridge is 30ft after water, how fast are the man and the boat separating 3s later? d = vts2 = x 2 + y 2 + z 2 s² = (5t)2 + (30)2 + (10t)2 s² = 25t² + 900 + 100t² √s2 = √125t 2 + 900 s = √125t 2 + 900 ds 1 = 2 (125t 2 + 900)-1/2(250t) dt ds 125t = dt 125t 2 + 900 ds 125(3) = dt √(125(3)2 + 900 ds 375 = 1 dt (1125 + 900)2 ds = 375/(2025)1/2 dt ds 375 = dt 45 ds ft = 8.33 dt s

84

4. A man on the wharf 20ft above the water pulls a rope to which the boat is attached, at the rate of 4ft/s. at what rate is the boat approaching the wharf when there is 25ft of rope out? 2 2 2 s =x +y (4t)2 = 202 + x 2 16t 2 = 400 + x 2 dx 32t= 2x dt r2 = x2 + y2 √x 2 = √r 2 − y 2 x = √r 2 − y 2 =√r 2 − 202 =√r 2 − 400 1 dv 1 dr = (r 2 − 400)−2 (2r ) dt 2 dt 𝑟 𝑑𝑟/𝑑𝑡 = (𝑟2 − 400)1/2 4(4𝑓𝑡/𝑠) = 1 (𝑟 2 − 400)2 25(4) 100 = = 2 (25 − 400) (625 − 400)1/2 100 = (225)1/2 𝑑𝑥 100 20 = = = 6.667𝑓𝑡𝑠 𝑑𝑡 5 3

5. Water is flowing into a coined reservoir 20ft deep to 10ft across the top, at the rate of 15cu ft/min. Find how fast the surface is rising when the water is 8ft deep. 𝜋𝑟 2 ℎ 𝑉= 3 𝜋ℎ ℎ2 𝑉= ( )2 3 4 𝜋ℎ3 𝑉= 48 5 𝑟 ℎ = , 𝑟= 204 ℎ 4 𝑑𝑣 3𝜋ℎ2 𝑑ℎ = 𝑑𝑡 48 𝑑𝑡 85

𝜋(8)2 𝑑ℎ 15 = 16 𝑑𝑡 𝑑ℎ 15(16) 240 60 15 = = = = = 1.194 𝑓𝑡/𝑚𝑖𝑛 𝑑𝑡 𝜋(64) 𝜋(64) 𝜋(16) 4𝜋

Name: Course/Year & Section: Assignment no.8: 1. Water is flowing into a vertical cylinder tank at the rate of 24 cu. ft/min. if the radius of the tank is 4ft., how fast is the surface rising? v = πr 2 h v = π(4)2 h v = 16πh dv 16πdh = dt dt 16πdh 24 = dt 24ft 3 dh = min 2 dt 16πft dh = 0.48ft/min dt

2. Water flows into a vertical cylindrical tank at 12ft³/min., the surface rises 6in/min., find the radius of the tank.

86

dh 6in 1ft = ∗ = 0.5ft/min dt min 12in v = πr²h dv πr 2 dh = dt dt 12ft 0.5ft = πr 2 ( ) min min 24 √ = r² π r = 2.76ft

3. A rectangular through is 10ft long and 3ft wide. Find how fast the surface rises, if the water flows in at the rate of 12ft³/min. v = lwh v = 10(3)(h) dv 30dh = dt dt 12ft 3 30ft 2 dh = min dt 12ft 3 dh = min2 dt 30ft dh 2ft = dt 5min dh 0.4ft = dt min 4. A triangular through is 10ft long, 4ft across the top and 4ft deep. If the water flows in at the rate of 3 cu. ft/min., find how fast the surface is rising when the water is 6 inches deep?

87

v=

lwh 2

v=

lh2 2

10 (2h)dh dv = 2 dt dt 3=

10(0.5)dh dt

dh 3 = dt 5 dh = 0.6ft/min dt

5. A triangular through is 10ft long, 6ft across the top and 3ft deep. If the water flows in at the rate of 12 cu. ft/min., find how fast the surface is rising when the water id 6in deep. 6 x = 3 h x = 2h w = 2h lwh v= 2 l(2h)(h) v= 2 v = lh² dv 2hdh = 10 ( ) dt dt 20(0.5)dh 12 = dt 10dh 12 = dt dh ft = 1.2 dt min

88

Square root √8.73 y = √x dy = ½x-1/2 dx dx dy = 2√x x=9 ∆x = 8.73 − 9 dx = −0.27 0.27 dy = − 2√9 0.27 dy = − 6 dy = −0.045 √8.73 = y + dy √8.73 = 3 − 0.045 √8.73 = 2.995 √15 dx dy = 2√x x = 16 ∆x = 15 − 16 dx = −1 89

dy = −

1

2√16 1 dy = − 8 dy = −0.125 √15 = y + dy √15 = 4 − 0.125 √15 = 3.875 √17 x = 16 ∆x = 17 − 16 dx = 1 1 dy = 2√16 1 dy = 8 dy = 0.125 √17 = y + dy √17 = 4 + 0.125 √17 = 4.125 cube root 3

√26 3 y = √26 y = x1/3 dy = 1/3x-2/3dx dx dy = 3 3√x 2 x = 27 dx = 26 − 27 dx = −1 1 dy = − 3 3 √272 1 dy = − 3(9) 1 dy = − 27 dy = −0.037 3 √26 = 3 − 0.037 3 √26 = 2.963 Fourth root

90

4√17

y =4√x y = x1/4 dy = 1/4x-3/4dx dx dy = 4 4√x 3 x = 16 dx = 17 − 16 dx = 1 1 dy = 4 4√163 1 dy = 32 dy = 0.03125 4 √17 = 2 + 0.03125 4 √17 = 2.03125

Newton’s Law of Approximation y =2x 3 + 3x 2 − 12x + 5 y’ = 6𝑥 2 + 6𝑥 − 12 F(2)= 2(2)3 + 3(2)2 − 12(2) + 5 = 9 F’(2)= 6(2)2 + 6(2) − 12 = 24 f (x )

9

x2 = x1 - f(x 1) = 2-24 =1.625 1

F(1.625)=2.0039 F’(1.625)= 13.5937 x2 = 1.625 −

2.0039 = 1.4775 13.5937

F(1.4775)=0.0066 F’(1.4775)=9.9630 x2 = 1.4775 −

0.0066 = 1.4768 9.9630

91

F(1.4768)=0.2628 F’(1.4768)=6.9928 x2 = 1.4768 −

0.2638 = 1.4392 6.9928

F(1.4392)=0.0944 F’(1.4392)=9.0630 x2 = 1.4392 −

0.0944 = 1.446 9.0630

F(1.4496)=0.0010 F’(1.4496)=9.3056 x2 = 1.446 −

0.001 = 1.4494 9.3056

Derivatives of Trigonometric Functions d dθ = sin θ = cos θ dx dx d dθ = cos θ = − sin θ dx dx d dθ = tan θ = sec ²θ dx dx d dθ = cot θ = − csc ²θ dx dx d dθ = sec θ = sec θ tan θ dx dx d dθ = sec θ = − csc θ cot θ dx dx Examples 1. )y = sin 3x

92

dy = 3 cos 3x dx 2. )x = cot 4t dx = −4 sin 4t dt 3. )w = tan 2θ dw = 2 sec² 2θ dθ 4. )z = sec

y 2

dz y y = 1/2 sec tan dy 2 2 5. )y = cot 5x dy = −5 csc ² 5x dx 6. )y = csc 7x dy = −7 csc 7x cot 7x dx 7. )v = 3 cos 2u dv = −6 sin 2u du 8. )x = sec 3 2t dx = 6 sec² 2t sec 2t tan 2t dt dx = 6 sec³ 2t tan 2t dt 9. )y = x 2 sin

x 2

93

dy sin y x 2 x = 2x + cos dt 2 2 2 10. )y = cos4 t − sin4 t dy = 4cos3 t(− sin t) − 4sin3 t cos t dt dy = −4cos3 t sin t − 4cost sin3 t dt dy = −4 sin t cos t(cos2 t + sin2 t) dt dy = −4 sin t cos t dt sin(2θ) = 2sinθcosθ sin(2θ) = −2(2 sin t cos t) sin(2θ) = −2 sin 2t

11. )y = sec 2 θ − tan2 θ dy = 2 secθsecθtanθ − 2 tanθsecθ dt 𝐝𝐲 =𝟎 𝐝𝐭 12. )r = cosθcotθ dr = − sin θcotθ + cosθ (−csc 2 θ) dt dr = −sinθcotθ − cosθcsc 2 θ dt dr cosθ = sinθ − cosθcsc 2 θ dt sinθ

94

dr = −cosθ − cosθcsc 2 θ dt dr = −cosθ (1 + csc 2 θ) dt Derivatives of Logarithmic Functions Exponential dv d lnv = dx dx v dv m d log10 v = dx dx v Derivatives of Exponential Functions d v dv a = av ln d dx dx d dv = ey = ev dx dx

Examples: 1. )y = ln(7 − 3x) dy −3 = dx 7 − 3x 2. )y = ln(2x 3 + x − 1) dy 6x 2 + 1 = 3 dx 2x + x − 1 3. )w = ln √a2 − x 2 1

w = ln(a2 − x 2 )2 −1 1 ln 2 (a2 − x 2 ) 2 (−2x) w= 1

(a2 − x 2 )2

w=

a2

−x − x2

95

4. )y = log10

sin x a

m x dy a cos a = x dx sin a x dy m cos a = x dx a sin a dy m x = cos dx a a 5. )y = e2x

3

3

y = e2x (6x 2 ) 3 y = 6x 2 e2x y = e−3x y = e−3x (−3)

dy = −3e−3x dx 𝟔. ). x = 102t dy = 102t ln 10 (2) dx dy = 2(102t ln 10) dx

96

Name: Course/Year & Section: Assignment no.9 Newton’s Law of Approximation 1. 𝑥 3 + 5𝑥 + 2 = 0

2. 𝑥 = −4 tan 3𝑥

3. 𝑤 = 2 csc(1 − 3𝑥) 97

𝑥

4. 𝑦 = 4 cot 4

5. 𝑦 = 6 sec 3𝑥

𝑥

6. 𝑦 = 12 sin 2

𝜋 𝑥 7. 𝑦 = tan( − ) 4 3

𝜋 𝑥 8. 𝑦 = sin ( − ) 4 3

9. 𝑧 = cos ³ 2𝑥

98

10. 𝑣 = sin ² 3𝑡

11. 𝑦 = tan ² 4𝑡

99