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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
USING PRODUCT RULE- Calculations using product rule formula: - Form of question y =- Example:
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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Step 2: Differentiate both u and v. For v, refer formula from page 17.
u =x3 = 3x2 v = sinh 2x = 22cosh x = 2 cosh 2x
Step 3: Use formula of product rule and simplify.
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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USING PRODUCT RULE- Calculations using product rule formula: - Form of question y =- Example:
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
Step 2: Differentiate both u and v. For v, refer formula from page 21.
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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Step 3: Use formula of product rule and simplify.
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