2
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-1 x → u = x
ii. y = sin-1 5x → u = 5x
iii. y = tan-1 (2x-1) → u = 2x-1
iv. y = sin-1 (cos x) → u = cos x
v. y = sin-1 ( 13
x ) → u =13
x
vi. y = cosec-1 (1+x2) → u = 1+x2
Example (a)Differentiate the following functions with respect to x:
i. y = sin-1 (cos x)
11
ADVANCED DIFFERENTIATION
u
ddx
(cot−1 u )= −1
1+u2
dudx
ddx
(sec−1 u )= 1
u√u2−1
dudx
ddx
(cosec−1 u )= −1
u√u2−1
dudx
FORMULA:
ddx
( s in−1 u )= 1
√1−u2
dudx
ddx
(cos−1 u )= −1
√1−u2
dudx
ddx
( tan−1 u )= 1
1+u2
dudx
FORMULA:
cos2 x+sin2 x=1
sec2 x=1+ tan2 x
cosec2 x=1+cot2 x
sin 2 x=2 sin x cos x
cos 2 x=cos2 x−sin2 x
=1−2 sin2 x
=2 cos2 x−1
ii. y = sin-1 ( 13
x )iii. y = cosec-1 (1+x2)
Solution:
i. y = sin-1 (cos x)
Step 1: Identify the ‘u’ from the equation and find dudx
u = cos x , dudx
= - sin x
Step 2: Determine the formula and replace the ‘u’ and dudx
ddx
(sin−1u )= 1
√1−u2
dudx
→
ddx
(sin−1 cos x )= 1
√1−(cos x )2⋅(−sin x )
=
−sin x
√1−cos2 x
=
−sin x
√sin2 x =
−sin xsin x = -1
ii. y = sin-1 ( 13
x )u =
13
x , dudx
= 13
ddx
(sin−1u )= 1
√1−u2
dudx →
ddx (sin−1 1
3x)= 1
√1−( 13
x)2⋅( 1
3 )
=
1
3√1−( x2
9 ) =
1
3√ 9− x2
9
=
1
(3 )(√ 19 )√9−x2
=
1❑
iii. y = cosec-1 (1+x2)
u = 1+x2 , dudx
= 2x
12
cos2 x + sin2 x = 1
1 – cos2 x = sin2 x
ddx
(cosec−1 u )= −1
u√u2−1
dudx →
ddx
(cosec−11+x2)= −1
(1+ x2)√( 1+ x2 )2−1⋅(2 x )
=
−2 x
(1+x2)√ ( x4+2 x2)
=
−2x
(1+x2) (√ x2)√ x2+2 =
−2
(1+x2)√x2+2
USING PRODUCT RULE
- Calculations using product rule formula:
dydx
=u( dvdx )+v ( du
dx )
- Form of question → y = - Example:
i. y = x2 sin-1 x2
→ u = x2 , v = sin-1 x2
ii. y = 3 sin-1 (ln 2x) → u = 3 , v = sin-1 (ln 2x)iii. y = x cos-1 x → u = x , v = cos-1 x
Example (b)Differentiate the following functions with respect to x:
i. y = x2 sin-1 x2
ii. y = 3 sin-1 (ln 2x) iii. y = x cos-1 x
Solution:
i. y = x2 sin-1 x2
Step 1: Form of y = uv, identified the value of ‘u’ and ‘v’.
y = x2 sin-1 x2
u = x2 v = sin-1 x2
Step 2: Differentiate both ‘u’ and ‘v’. For ‘v’, refer formula from page 11.
u = x2 → dudx
= 2x v = sin-1 x2
→ dvdx
13
Differentiate x2
= 12
Try this!
Find dydx
for the following:
a) y = tan-1 (2x-1)b) y = sin-1 5x
uv
inverse trigonometric x
Differentiate
sin-1
ddx
( s in−1 u )= 1
√1−u2
dudx
= 1
√1−( x2 )2
⋅ 12
= 1
2√1− x2
4
Step 3: Use formula of product rule and simplify.
∴ dydx
=u( dvdx )+v ( du
dx )
dydx
=( x2 )( 1
2√1−x2
4 )+sin−1 x2
(2 x )
¿ x2
2√1−x2
4
+2 x sin−1 x2
¿ x2
2√ 4−x2
4
+2 x sin−1 x2
¿ x2
2√( 14 ) ( 4−x2 )
+2 x sin−1 x2 ¿
x2
√( 4−x2 )+2 x sin−1 x
2
ii. y = 3 sin-1 (ln 2x)
u = 3 →
dudx = 0 v = sin-1 (ln 2x) →
dvdx =
1
√1−( ln 2 x )2⋅
12 x
⋅2
=
1
x√1−( ln 2 x )2
∴ dydx
=u( dvdx )+v ( du
dx )
dydx
=3()+sin−1( ln2 x) (0 )
=
3
x √1−( ln 2 x )2
iii. y = x cos-1 x
14
Differentiate sin-1
ddx
( s in−1 u )= 1
√1−u2
dudx
Differentiate cos-1
ddx
(cos−1 u )= −1
√1−u2
dudx
u =
x → dudx = 1 v = cos-1 x →
dvdx =
−1
√1−x2⋅1
=
−1
√1−x2
∴ dydx
=u( dvdx )+v ( du
dx )
dydx
=x ( −1
√1−x2 )+cos−1 x (1 )
= ( −x
√1−x2 )+cos−1 x
USING QUOTIENT RULE
- Calculations using quotient rule formula:
dydx
=v ( du
dx )−u( dvdx )
v2
- Form of question → y =uv
- Example:
i. y = tan−1 x
x → u = tan-1 , v = x
ii. y = 1+tan−1 x
2−3 tan−1 x→ u = 1+ tan−1 x , v = 2−3 tan−1 x
Example (c)Differentiate the following functions with respect to x:
i. y = tan−1 x
x
15
Try this!
Find dydx
for the following:
a) y = (1 - x2) sin-1 xb) y = x tan-1 x
ii. y = 1+tan−1 x
2−3 tan−1 x
Solution:
i. y = tan−1 x
x
Step 1: Form of y = uv
, 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 →
dudx =
1
1+( x)2 ⋅ 1
v = x → dvdx
= 1
Step 3: Use formula of quotient rule and simplify.
dydx
=v ( du
dx )−u( dvdx )
v2
= x( 1
1+x2 )−tan−1
x (1 )
(x)2
= ( x
1+x2 )−tan−1
x
x2
16
Differentiate tan-1
ddx
( tan−1 u )= 1
1+u2
dudx
Differentiate x = 1
= x
x2 (1+x2 )− tan−1 x
x2 = 1
x (1+x2 )− tan−1 x
x2
ii. y = 1+tan−1 x
2−3 tan−1 x
u = 1+ tan−1 x
→
dudx = 0 +
1
1+x2⋅1
v = 2−3 tan−1 x →
dvdx = 0 -
3( 1
1+x2 ) = -3( 1
1+x2 )
∴ dydx
=v ( du
dx )−u( dvdx )
v2
¿
(2−3 tan−1
x )( 1
1+x2 )−( 1+ tan−1
x ) (−3 )( 1
1+ x2 )(2−3 tan−1 x)2
¿( 2−3 tan−1 x
1+x2 )−(−3−3 tan−1 x1+x2 )
( 2−3 tan−1 x )2
¿5
(1+x2) (2−3 tan−1 x )2
2.1.2: Perform Differentiation of Hyperbolic Functions
17
Differentiate tan-1
ddx
( tan−1 u )= 1
1+u2
dudx
Differentiate 3tan-1x (use product rule):
u = 3 v = tan-1 x
dudx = 0
dudx =
1
1+x2⋅1
Try this!
Find dydx
for the following:
y = cot−1 2 x1+4 x2
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 (ln x) → u = ln x
Example (d)Differentiate the following functions with respect to x:
i. y = cosh (7x + 2)ii. y = sinh 2x
iii. y = cosh (ln x)
Solution:
i. y = cosh (7x + 2)
Step 1: Identify the ‘u’ from the equation and find dudx
u = 7x + 2 , dudx
= 7
Step 2: Determine the formula and replace the ‘u’ and dudx
ddx
( cosh u )=sinh u dudx →
ddx
( cosh 7 x+2 )=sinh (7 x+2 )⋅7
= 7 sinh (7x + 2)
ii. y = sinh 2x
u = 2x , dudx
= 2
ddx
( sinh u )=cosh u dudx →
ddx
( sinh 2x )=cosh 2 x⋅2
18
u
ddx
( coth u )=−cosech2 ududx
ddx
(sech u )=−s ech u tanh ududx
ddx
(cosech u )=−cosech u coth ududx
FORMULA:
ddx
( sinh u )=cosh u dudx
ddx
( cosh u )=sinh u dudx
ddx
( tanh u )=sec h2 ududx
= 2 cosh 2x
iii. y = cosh (ln x)
u = ln x , dudx
= 1x
ddx
( cosh u )=sinh u dudx →
ddx
( cosh ln x )=sinh (ln x )⋅1x⋅1
= 1
xsinh ( ln x )
USING PRODUCT RULE
- Calculations using product rule formula:
dydx
=u( dvdx )+v ( du
dx )
- Form of question → y = - Example:
i. y = x sinh x → u = x , v = sinh xii. y = e3x cosh x2 → u = e3x , v = cosh x2
iii. y = x3 sinh 2x → u = x3 , v = sinh 2x
Example (e)Differentiate the following functions with respect to x:
i. y = x3 sinh 2xii. y = x sinh x
iii. y = e3x cosh x2
Solution:
i. y = x3 sinh 2x
Step 1: Form of y = uv, identified the value of ‘u’ and ‘v’.
y = x3 sinh 2x
u = x3 v = sinh 2x
19
Try this!
Find dydx
for the following:
a) y = ln (tanh x3)b) y = coth (ln x2)
c) y = sech( 12
x )
uv
hyperbolic functions x
Step 2: Differentiate both ‘u’ and ‘v’. For ‘v’, refer formula from page 16.
u = x3 → dudx
= 3x2 v = sinh 2x → dvdx
= cosh 2 x ⋅ 2 = 2 cosh 2x
Step 3: Use formula of product rule and simplify.
∴ dydx
=u( dvdx )+v ( du
dx )dydx
=x3 (2 cosh 2 x )+sinh 2 x (3 x2 )
¿2 x3 (cosh 2 x )+3 x2 sinh 2 x
ii. y = x sinh x
u = x → dudx
= 1 v = sinh x → dvdx
= cosh x ⋅ 1
∴ dydx
=u( dvdx )+v ( du
dx )dydx
=x (cosh x )+sinh x (1 )
¿ x cosh x+sinh x
iii. y = e3x cosh x2
u = e3x → dudx = 3e3x v = cosh x2 →
dvdx = sinh x2⋅2x
= 2x sinh x2
∴ dydx
=u( dvdx )+v ( du
dx )20
Differentiate sinh
ddx
( sinh u )=cosh u dudx
Differentiate 2x = 2
Differentiate sinh
ddx
( sinh u )=cosh u dudx
Differentiate cosh
ddx
( cosh u )=sinh u dudx
dydx
=e3 x ( 2 x sinh x2 )+cosh x2 (3 e3 x )
¿2 xe3 x (sinh x2 )+3 e3 x cosh x2
USING QUOTIENT RULE
- Calculations using quotient rule formula:
dydx
=v ( du
dx )−u( dvdx )
v2
- Form of question → y =uv
- Example:
i. y = cosh 2 x
x2 → u = cosh 2 x , v = x2
ii. y = ln (sinh x )
e3 x → u = ln (sinh x ) , v = e2x
iii. y = 3x2
cosh 4 x → u = 3 x2 , v = cosh 4 x
Example (f)Differentiate the following functions with respect to x:
i. y = cosh 2 x
x2
ii. y = ln (sinh x )
e3 x
Solution:
i. y = cosh 2 x
x2
Step 1: Form of y = uv
, identified the value of ‘u’ and ‘v’.
u = cosh 2 x , v = x2
Step 2: Differentiate both ‘u’ and ‘v’. Refer formula from page 16 for any hyperbolic functions.
21
u = cosh 2 x → dudx
= sinh 2x • 2 = 2 sinh 2x
v = x2 → dvdx
= 2x
Step 3: Use formula of quotient rule and simplify.
∴ dydx
=v ( du
dx )−u( dvdx )
v2
¿x2 (2sinh 2 x )−cosh2 x (2 x )
x4
¿2 x2 (sinh 2 x )−2x cosh2 x
x 4 ¿2 x (x si nh 2 x− cosh2 x)
x4 ¿2(x si nh 2 x− cosh 2 x)
x3
ii. y = ln (sinh x )
e2 x
u = ln (sinh x ¿
→
dudx =
1sinh x
⋅cosh x v = e
3x
→
dvdx = 3e3x
∴ dydx
=v ( du
dx )−u( dvdx )
v2
¿e
3 x ( cosh x
sinh x )−ln sinh x (3 e3 x )
( e3 x)2
¿e3 x coth x−3e3 x ln sinh x
e6 x
¿e3 x (coth x−3 ln sinh x¿ ¿e6 x ¿
coth x−3 ln sinh x
e3 x
22
Differentiate 2x = 2Differentiate tan-1
ddx
( cosh u )=sinh u dudx
Differentiate ln:
ddx
( ln u )=1u⋅du
dx
Differentiate sinh:
ddx
( sinh u )=cosh u dudx
cosh xsinh x
=coth x
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-1 x → u = x
ii. y = cosh-1 (3 - 2x) → u = 3 - 2x
iii. y = tanh-1 ( 34
x)→ u = 13
x
Example (g)Differentiate the following functions with respect to x:
i. y = cosh-1 (3 - 2x)ii. y = sinh-1 x
iii. y = tanh-1 ( 34
x)
Solution:
i. y = cosh-1 (3 - 2x)
Step 1: Identify the ‘u’ from the equation and find dudx
u = 3 - 2x , dudx
= -2
Step 2: Determine the formula and replace the ‘u’ and dudx
23
u
ddx
(coth−1 u )= 1
1−u2
dudx
, |u| ¿ 1¿
ddx
(sech−1u )= −1
u√1−u2
dudx
ddx
(cosech−1 u )= −1
|u|√1+u2
dudx
FORMULA:
ddx
(sinh−1 u )= 1
√1+u2
dudx
ddx
(cosh−1 u )= 1
√u2−1
dudx
ddx
( tanh−1 u )= 1
1−u2
dudx
, |u| ¿¿
Try this!
Find dydx
for the following: y = 3x2
cosh 4 x
ddx
(cosh−1 u )= 1
√u2−1
dudx
→
ddx
(cosh−1 3−2 x )= 1
√(3−2 x )2−1⋅(−2)
=
−2
√(3−2 x )2−1 =
−2
√4 x2−12 x+9−1
=
−2
√4( x2−3 x+2) =
−1
√( x2−3 x+2)
ii. y = sinh-1 x
u = x , dudx
= 1
ddx
(sinh−1 u )= 1
√1+u2
dudx
→
ddx
(sinh−1 x)= 1
√1+x2⋅1
=
1
√1+x2
iii. y = tanh-1 ( 34
x)u = ( 3
4x) ,
dudx
= 34
ddx
( tanh−1 u )= 1
1−u2
dudx
, |u| ¿¿→
ddx (tanh−1 3
4x )= 1
1−( 34
x )2⋅3
4
=
3
4 (1− 9 x2
16 ) =
12
16−9 x2
USING PRODUCT RULE
- Calculations using product rule formula:24
Try this!
Find dydx
for the following:
a) y = sech-1 (sin x)b) y = cosech-1 (tanh 2x)
dydx
=u( dvdx )+v ( du
dx )
- Form of question → y = - Example:
i. y = (x2 - 1) tanh-1 x → u = x2 - 1 , v = tanh-1 xii. y = e2x cosh-1 x → u = e2x , v = cosh-1 x
iii.
Example (h)Differentiate the following functions with respect to x:
i. y = (x2 - 1) tanh-1 xii. e2x cosh-1 x
Solution:
i. y = (x2 - 1) tanh-1 x
Step 1: Form of y = uv, identified the value of ‘u’ and ‘v’.
y = (x2 - 1) tanh-1 x
u = x2 - 1 v = tanh-1 x
Step 2: Differentiate both ‘u’ and ‘v’. For ‘v’, refer formula from page 21.
u = x2 - 1 →
dudx = 2x v = tanh-1 x →
dvdx =
1
1−x2 ⋅ 1
Step 3: Use formula of product rule and simplify.
∴ dydx
=u( dvdx )+v ( du
dx ) ¿ ( x2−1 )( 1
1−x2 )+( tanh−1
x ) (2x )
¿( x2−11− x2 )+2 x ( tanh−1 x )
25
uv
inverse hyperbolic x
Differentiate tanh-1
ddx
( tanh−1 u )= 1
1−u2
dudx
, |u| ¿¿
Differentiate x = 1
ii. e2x cosh-1 x
u = e2x →
dudx = 2e2x v = cosh-1 x →
dvdx =
1
√x2−1⋅1
=
1
√x2−1
∴ dydx
=u( dvdx )+v ( du
dx )¿e2 x ()+cosh−1 x (2 e2 x)
¿()+2 e2 x cosh−1 x
2.1.4: Perform Differentiation of Implicit Functions
Function of: f(xy) Differentiate functions of y with respect to x The chain rule must be used whenever the function y is being differentiated.
Example (i):
Use implicit differentiation to find dydx
for each equation:
i. y2 – 2x2 = 1ii. 2xy – y3 = x2
iii. xy + sin y = 1
26
Try this!
Find dydx
for the following:
y = 3 sinh-1 (√2 x2−1 )
Solution:
i. y2 – 2x2 = 1
Step 1: Differentiate functions of x with respect to x (straight forward).
y2−4 x=0
Step 2: But when differentiating a function of y with respect to x we must remember the rule: ddx
[ f ( y ) ]= ddy
[ f ( y ) ] • dydx
2 y ( dydx )−4 x=0
Step 3: Rearrange equation to collect all terms involving dydx
together and simplify.
dydx
=4 x2 y
= 2 xy
ii. 2 xy – y3 = x2
2[ x ( dydx )+( y )]− y3=x2
2 x ( dydx )+2 y−3 y2( dy
dx )=2 x
2 x ( dydx )−3 y2( dy
dx )=2x−2 y
dydx
(2 x−3 y2 )=2 x−2 y
dydx
= 2 x−2 y
2 x−3 y2
iii. xy + sin y = 1
27
Use product rule:
u = x v = y
dudx
=1 dvdx
=1( dydx )
∴ dydx
=u( dvdx )+v ( du
dx )
x ( dydx )+ y+cos y ( dy
dx )=0
dydx
( x+cos y )=− y
dydx
= − yx+cos y
28
Try this!
a) Determine the gradient of the curve:
x2 + y2 – 2x – 6y + 5 = 0 at (3,2)
b) Given x2 + 3xy – ln y = 3x. Find dydx
if x = 0 and y =
Use product rule:
u = x v = y
dudx
=1 dvdx
=1( dydx )
∴ dydx
=u( dvdx )+v ( du
dx )
Differentiate dydx
of sin y
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