chapter 2: advanced differentiation

7/27/2019 CHAPTER 2: ADVANCED DIFFERENTIATION

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2.1: APPLY ADVANCED DIFFERENTIATION FORMULAE AND IMPLICIT FUNCTIONS

2.1.1: Perform Differentiation of Inverse Trigonometric Functions

USING FORMULA PROVIDED- Calculations directly using formula- Form of question y = inverse trigonometric- Example:

i. y = tan-1x u =xii. y = sin-1 5x u = 5x

iii. y = tan-1 (2x-1) u = 2x-1iv. y = sin-1 (cosx) u = cosxv. y = sin-1 u =

vi. y = cosec-1 (1 +x2) u = 1 +x2

Example (a)Differentiate the following functions with respect tox:

i. y = sin-1 (cosx)ii. y = sin-1

iii. y = cosec-1 (1+x2)

ADVANCED DIFFERENTIATION

u

FORMULA:

FORMULA:


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Solution:

i. y = sin-1 (cosx)Step 1: Identify the u from the equation and find

u = cosx ,

= - sinxStep 2: Determine the formula and replace the u and

dx

du

u

udx

d

2

1

1

1sin

)sin(

)(cos1

1cossin

2

1x

x

xdx

d

=

x

x

cos1

sin

2

=x

x

sin

sin

2

=

x

x

sin

sin= -1

ii. y = sin-1 u = , =

dx

du

u

udx

d

2

1

1

1sin

3

1

3

11

1

3

1sin

2

1

x

xdx

d

=

913

1

2x

=

9

9

3

1

2x

= 29

9

13

1

x

=

29 x

iii. y = cosec-1 (1 +x2)u = 1 +x2 ,

= 2x

dxdu

uuudx

d

1

1

cosec 2

1

xxxxdxd

2111

1

1cosec 222

21

=

242 212

xxx

x

= 212

222

xxx

x=

212

22

xx

cos2x + sin2x = 1

1cos2x = sin2x

Try this!

Find

for the following:

a) y = tan-1 (2x-1)

b) y = sin-1 5x


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USING PRODUCT RULE- Calculations using product rule formula: - Form of question y =- Example:

i. y =x2 sin-1 u =x2 , v = sin-1ii. y = 3 sin-1(ln 2x) u = 3 , v = sin-1(ln 2x)

iii. y =x cos-1x u =x , v = cos-1x Example (b)

Differentiate the following functions with respect tox:

i. y =x2 sin-1ii. y = 3 sin-1(ln 2x)

iii. y =x cos-1x

Solution:

i. y =x2 sin-1Step 1: Form of y = uv, identified the value of u and v.

y = x2 sin-1

u =x2 v = sin-1

Step 2: Differentiate both u and v. For v, refer formula from page 11.

u =x2 = 2x v = sin-1

4

2

12

1

x

uv

inverse trigonometricx

Differentiate sin-

Differentiate = 212

21

1

x


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Step 3: Use formula of product rule and simplify.

ii. y = 3 sin-1

(ln 2x)

u = 3 = 0 v = sin-1(ln 2x) =

2

2

1

21

1

2ln

x

x

=

21

2ln

1

xx

21 2ln1 xx

=

22ln1

3

xx

iii. y =x cos-1x

u =x

= 1 v = cos-1x

= 1

21

1

x

=2

1

1

x

=

Differentiate sin-

Differentiate cos-

Try this!

Find

for the following:

a) y = (1 -x2) sin-1x

b) y =x tan-1x


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USING QUOTIENT RULE- Calculations using quotient rule formula: - Form of question y =- Example:

i. y = u = tan-1 x , v =xii. y = u = , v =

Example (c)Differentiate the following functions with respect tox:

i. y = ii. y =

Solution:

i. y = Step 1: Form ofy =

, identified the value of u and v.u = tan-1 x , v =x

Step 2: Differentiate both u and v. Refer formula from page 11 for any inverse trigonometric.

u = tan-1 x = 12

)(1

1

x

v =x = 1

Differentiate tan-

Differentiatex = 1


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Step 3: Use formula of quotient rule and simplify.

=

= =

= ii. y =

u =

= 0 + 1

21

1

x v =

= 0 -

= -

( )( )

Differentiate tan- Differentiate 3 tan- x (use product rule):

u = 3 v = tan-1x

= 0

=

Try this!

Find

for the following:

y =o


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2.1.2: Perform Differentiation of Hyperbolic Functions

USING FORMULA PROVIDED- Calculations directly using formula- Form of question y = hyperbolic functions- Example:

i. y = cosh (7x + 2) u = 7x + 2ii. y = sinh 2x u = 2x

iii. y = cosh (lnx) u = lnx

Example (d)Differentiate the following functions with respect tox:

i. y = cosh (7x + 2)ii. y = sinh 2x

iii. y = cosh (lnx)

Solution:

i. y = cosh (7x + 2)Step 1: Identify the u from the equation and find

u = 7x + 2 ,

= 7Step 2: Determine the formula and replace the u and

dx

duuu

dx

dsinhcosh 727sinh27cosh xx

dx

d

= 7 sinh (7x + 2)

ii. y = sinh 2xu = 2x ,

= 2

dx

duuu

dx

dcoshsinh 22cosh2sinh xx

dx

d

= 2 cosh 2x

u

FORMULA:


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iii. y = cosh (lnx)u = lnx ,

=

dx

duuu

dx

dsinhcosh 11lnsinhlncosh

xxx

dx

d

x

x

lnsinh1


i. y =x sinhx u =x , v = sinhxii.

y = e

3x

cosh x

2

u = e

3x

, v = cosh x

2

iii. y =x3 sinh 2x u =x3 , v = sinh 2x

Example (e)Differentiate the following functions with respect tox:

i. y =x3 sinh 2xii. y =x sinhx

iii. y = e3x cosh x2

Solution:

i. y =x3 sinh 2xStep 1: Form of y = uv, identified the value of u and v.

y = x3 sinh 2x

u =x3 v = sinh 2x

Try this!

Find

for the following:

a) y = ln (tanhx3)

b) y = coth (lnx2)

c) y = ech

uv

hyperbolic functionsx


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u =x3 = 3x2 v = sinh 2x = 22cosh x = 2 cosh 2x


ch h ch h

ii. y =x sinhxu =x

= 1 v = sinhx = 1cosh x

() () ch h ch h

iii. y = e3x cosh x2

u = e3x = 3e3x v = cosh x2 = xx 22sinh

= 2x sinhx2

(

) (

) h ch h ch

Differentiate sinhDifferentiate 2x = 2

Differentiate sinh

Differentiate cosh


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USING QUOTIENT RULE- Calculations using quotient rule formula: - Form of question y =

- Example:i. y = o u = ch , v =x2

ii. y = h u = h , v = iii. y = o u = , v = ch

Example (f)Differentiate the following functions with respect to x:

i. y = o

ii. y = h Solution:

i. y = o Step 1: Form of y =

, identified the value of u and v.u = ch , v =

Step 2: Differentiate both u and v. Refer formula from page 17 for any hyperbolic functions.

u = ch =sinh 2x 2 = 2 sinh 2x

v = = 2xStep 3: Use formula of quotient rule and simplify.

h ch h ch ch ch

Differentiate 2x = 2Differentiate tan-


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ii. y = h u = h = x

x

coshsinh

1 v = = 3e3x

h

ch

ch ch

2.1.3: Perform Differentiation of Inverse Hyperbolic Functions

USING FORMULA PROVIDED- Calculations directly using formula- Form of question y = inverse hyperbolic- Example:

i. y = sinh-1x u =xii. y = cosh-1 (3 - 2x) u = 3 - 2x

iii. y = tanh-1 u =

Differentiate ln:Differentiate sinh:

ch h ch

u

FORMULA:

Try this!

Finddyd for : y = o


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Example (g)Differentiate the following functions with respect tox:

i. y = cosh-1 (3 - 2x)ii. y = sinh-1x

iii. y = tanh-1 Solution:

i. y = cosh-1 (3 - 2x)Step 1: Identify the u from the equation and find

u = 3 - 2x ,

= -2Step 2: Determine the formula and replace the u and

dx

du

u

u

dx

d

21

11sinh

)2(1

2)23(

123

1cosh

x

x

dx

d

=

12)23(

2

x

=

19122

4

2

xx

=)23

2(4

2

xx

=

)232

(

1

xx

ii. y = sinh-1xu =x ,

= 1

dx

du

u

u

dx

d

21

11sinh

12

1

11sinh

x

x

dx

d

= 21

1

x

iii. y = tanh-1 u =

,

=

dx

du

u

u

dx

d

12

11cosh

4

3

2

4

31

1

4

31tanh

x

x

dx

d

=

16

291

3

4x

=2

916

12

x


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i.

y = (x

2

- 1) tanh

-1

x u =x2

- 1 , v = tanh

-1

xii. y = e2x cosh-1x u = e2x , v = cosh-1x

Example (h)Differentiate the following functions with respect to x:

i. y = (x2 - 1) tanh-1xii. y = e2x cosh-1x

Solution:

i. y = (x2 - 1) tanh-1xStep 1: Form of y = uv, identified the value of u and v.

y = (x2 - 1) tanh-1x

u =x2 - 1 v = tanh-1x


u =x2 - 1 = 2x v = tanh-1x = 12

1

1

x

Try this!

Find

for the following:

a) y = sech-1 (sinx)

b) y = cosech-1 (tanh 2x)

uv

inverse hyperbolicx

Differentiate tanh-1

Differentiatex = 1


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h h

ii. y = e2x cosh-1x

u = e2x = 2e2x v = cosh-1x = 1

12

1

x

=

12

1

x

e

12

1

x

ch

12

2

x

x

e ch

2.1.4: Perform Differentiation of Implicit Functions

Function of: f(xy) Differentiate functions ofy with respect tox The chain rule must be used whenever the functiony is being differentiated. Example (i):

Use implicit differentiation to find for each equation:

i. y22x2 = 1ii. 2xyy3 =x2

iii. xy + siny = 1

Try this!

Find

for the following:

y = 3 sinh-1


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Solution:

i. y22x2 = 1Step 1: Differentiate functions ofx with respect tox (straight forward).

Step 2: But when differentiating a function ofy with respect tox we must remember the rule:

[] []

Step 3: Rearrange equation to collect all terms involving

together and simplify. =

ii. 2xyy3 =x2

* + () ()

(

) (

)

Use product rule:

u =x v =y

dyd


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iii. xy + siny = 1

c

c

o

2.2: UNDERSTAND PARTIAL DIFFERENTIATION

2.2.1: Define Partial Differentiation

Function of multiple variables (more than one variable). Applications that require functions with more than one variable:

i. Ideal of gas law (Changes of pressure with respect to volume).ii. Changes of oil level in cylinder.iii.

Changes of weather forecast.

Symbol is delta Since partial differentiation is essentially the same as ordinary differentiation, the product, quotient and chain rules may be applied.

(

) (

)

Use product rule:

u =x v =y

dyd

Differentiate

of siny

Try this!

a) Determine the gradient of the curve:

x2 +y22x6y + 5 = 0 at (3,2)

b) Given x

2

+ 3xylny = 3x. Find

if x = 0 and y = 1.


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2.2.2: Perform Partial Differentiation to Find First Order Partial Differentiation

Use this symbol and . Example (j):

Find and for following functions:

i. z= (x2 + 3y)5ii. z= sin 2x + cos 3yiii. z=y4 + 2x2y2 + 6x7

Solution:

i. z= (x2 + 3y)5Step 1: When finding

, y is treated as constant and differentiate the xexactly the same as for ordinary differentiation.z= (x2 + 3y)5

= 5(x2 + 3y)5-12x= 10x(x2 + 3y)4

Step 2: When finding , x is treated as constant and differentiate the yexactly the same as for ordinary differentiation.

z= (x2 + 3y)5

= 5(x2 + 3y)5-1 3= 15(x2 + 3y)4

ii. z= sin 2x + cos 3y= (cos 2x) 2 + 0

= 2 cos 2x

= 0 + (-sin 3y) 3= -3 sin 3y

iii. z=y4 + 2x2y2 + 6x7= 0 + 4(x)2-1(y2) + 60

= 4xy2 + 6

= (4y)4-1 + 2(x2)(2y)2-1 + 00= 4y3 + 4x2y

Try this!

a) Given z= . Find and

b) Givenz=x3 ln (5y), Find

and


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2.2.3: Second Order Partial Differentiation

Example (k):i. Givenz=x2 siny +y2 cos x. Find and ii. Given z= 4x3 + 6xy23y2. Find , and

Solution:

i. Givenz=x2 siny +y2 cos x. Find and Step 1: Find

and first= (2x)2-1 siny +y2 (-sinx)

= 2x sinyy2 sinx

=x2 (cosy) + (2y)2-1 cosx= x2 cosy + 2y cosx

Step 2: Find other derivatives

=

(x2 cosy + 2y cosx)

= 2x cosy + 2y (-sinx) = 2x cosy - 2y sinx

=

(2x sinyy2 sinx)= 2x cosy2y sinx

The Second-Order Partial Derivatives ofz =f(x, y)


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ii. Given z= 4x3 + 6xy23y2. Find , and = 12x2 + 6y20 = 0 + 12xy6y

=(12x2 + 6y2) = 24x

=

(12xy6y) = 12x6

=

(12x2 + 6y2) = 12y

2.3: CONSTRUCT TOTAL DIFFERENTIATION

2.3.1: Define Total Differentiation, dz

Try this!

a) Given z= cos (2x2 +y). Find and b) Given z= 2xy cosx, Find

and

c) Given z= (3x + 2y) (4x5y). Find

and

(

) (

)DEFINITION

Letz= f(x,y) then the total differential forzis:


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2.3.2: Use Partial Differentiation to Produce Total Differentiation

Example (l):Given Define total differentiation for belowz:

i. z= 2x3y +x2y2 +xy3ii. z=x siny -y sinx

Solution:

i. z= 2x3y +x2y2 +xy3

Step 1: Find and first

Step 2: Calculate the value using total differentiation equation

ii. z=x siny -y sinx

c c

c c

2.3.3: Solve Problems Regarding Rate of Changes

Example (m):The base radius of a cone, r, is decreasing at the rate of 0.2 mm/sec while the perpendicular height, h, is increasing at the rate of 0.4

mm/sec. Solve the rate at which the volume V, is changing when r= 8 mm and h = 10 mm.

Solution:

Step 1: List out data from the question.

h = 10 r= 8 Step 2: Differentiate the volume/area formula.


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Step 3: Complete the rate of change equation.

= -6.702 mm/s

Example (n):The power (P) used by one resistor (R) is given by formulaP=I2R (unit watt). IfI= 15A andR = 100k, find the value which the

powerPis changing whenIincrease at 4A andR increase at 20 k.

Solution:

I= 15 R = 100

dI= 4 dR = 20

P=I2R

= 16500

Try this!

a) Temperature (T), a flat metal is changing due to its position. Given T(x.y) =60. Find at the point (2,1).

b) The height of a cone is 3 mm and increases at the rate of 0.2 mm/s. The radius of the base is 2 mm and decreases at

the rate of 0.1 mm/s. Solve the rate of change for the volume of the cone where v =r2hc) The radius of an opened-cylinder increases at 0.2 cm/s and its height decreases at 0.5 cm/s. Discover the rates of

changes on its side area if radius is 8 cm and height is 12 cm.d)

i. If a = 8 cm, b = 10 cm and measurement error for cm and cm, calculate themaximum error of ABC area.

ii. If a increase with the rate of 1 cm/s and b decrease with the rate of 0.5 cm/s, calculate the change of

ABC area when a = 8 cm and b = 10 cm.

A

B C

a cm

b cm

chapter 2: advanced differentiation

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