chapter 2 (2.1) : advanced differentiation formula and implicit functions (part 2)

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


- 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)


u = 7x + 2 , dudx

= 7


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 )


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


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


u = x3 → dudx

= 3x2 v = sinh 2x → dvdx

= cosh 2 x ⋅ 2 = 2 cosh 2x


∴ 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


Differentiate 2x = 2

Differentiate sinh

ddx


Differentiate cosh

ddx


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


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


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


Differentiate ln:

ddx

( ln u )=1u⋅du

dx

Differentiate sinh:

ddx


cosh xsinh x

=coth x

2.1.3: Perform Differentiation of Inverse Hyperbolic Functions


- 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


x)

Solution:

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


u = 3 - 2x , dudx

= -2


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


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 )


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


y = (x2 - 1) tanh-1 x

u = x2 - 1 v = tanh-1 x


u = x2 - 1 →

dudx = 2x v = tanh-1 x →

dvdx =

1

1−x2 ⋅ 1


∴ 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

chapter 2 (2.1) : advanced differentiation formula and implicit functions (part 2)

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