C CONVERSATION: Voice level 0. No talking!

HHELP: Raise your hand and wait to be called on.

AACTIVITY: Whole class instruction; students in seats.

M MOVEMENT: Remain in seat during instruction.

PPARTICIPATION: Look at teacher or materials being discussed. Raise hand to contribute; respond to questions, write or perform other actions as directed.NO SLEEPING OR PUTTING HEAD DOWN, TEXTING, DOING OTHER WORK.   

S

Activity: Teacher-Directed Instruction

Calculus AB

2013 Implicit Differentiation

Objective

• C: The swbat differentiate implicitly equations in more than one variable.

• L: the sw explain to others how to find derivatives of multiple types of problems verbally and demonstratively

Implicit Differentiation

Equation for a line:

Explicit Form

<One variable given explicitly in terms of the other>

Implicit Form

<Function implied by the equation>

  Differentiate the Explicit

< Explicit: , y is function of x >

Differentiation taking place with respect to x. The derivative is explicit also.

y mx b

Ax By C

24 3 4y x x

8 3dy

xdx

Implicit Differentiation

Equation of circle:

 

To work explicitly; must work two equations

2 2 9y x

29y x

 Implicit Differentiation is a Short Cut - A method to handle equations that are not easily written explicitly.

( Usually non-functions)

29y x 𝑑𝑦𝑑𝑥

=12

(9−𝑥2 )−12 (−2 𝑥 )

𝑑𝑦𝑑𝑥

=−𝑥

√9−𝑥2

𝑑𝑦𝑑𝑥

=− 12

( 9− 𝑥2 )− 12 (−2𝑥 )

𝑑𝑦𝑑𝑥

=𝑥

√9−𝑥2

𝑥2𝑦+2 𝑦2𝑥+3 𝑦3=7

Don’t want to solve for y

Implicit Differentiation

Chain Rule Pretend y is some function like

so becomes

 (A)

(B)

(C)

Note: Use the Leibniz form. Leads to Parametric and Related Rates.

2 2 3y x x 2 4( 2 3)x x 4y

Find the derivative with respect to x

< Assuming - y is a differentiable function of x >

32y

4y

2 3x y

=

=

2𝑑𝑥𝑑𝑥

+3𝑑𝑦𝑑𝑥

=2+3𝑑𝑦𝑑𝑥

4 𝑦 3(2𝑥+2)

6 𝑦2(2 𝑥+2)

Implicit Differentiation

Find the derivative with respect to x

< Assuming - y is a differentiable function of x >

𝑥𝑦=¿ 𝑥𝑑𝑦𝑑𝑥

+𝑦𝑑𝑥𝑑𝑥 ¿ 𝑥

𝑑𝑦𝑑𝑥

+ 𝑦

𝑥2+ 𝑦2=¿2 𝑥𝑑𝑥𝑑𝑥

+2 𝑦𝑑𝑦𝑑𝑥

¿2 𝑥+2 𝑦𝑑𝑦𝑑𝑥

sin (𝑥𝑦) ¿cos (𝑥𝑦 )∗(𝑥 𝑑𝑦𝑑𝑥+𝑦 𝑑𝑥𝑑𝑥 )

¿ 𝑥cos (𝑥𝑦 ) 𝑑𝑦𝑑𝑥

+ 𝑦 cos (𝑥𝑦)

Implicit Differentiation

(D) Product Rule

 

 

2xy ¿ 𝑥 (2 𝑦 ) 𝑑𝑦𝑑𝑥

+ 𝑦2 𝑑𝑥𝑑𝑥

¿2 𝑥𝑦𝑑𝑦𝑑𝑥

+𝑦2

¿ 𝑥𝑑𝑦𝑑𝑥

+ 𝑦❑ 𝑑𝑥𝑑𝑥

¿ 𝑥𝑑𝑦𝑑𝑥

+ 𝑦❑

𝑥𝑦

Implicit Differentiation

 (E) Chain Rule 3( )xy

Product inside a chain

3 (𝑥𝑦 )2(𝑥 𝑑𝑦𝑑𝑥

+𝑦𝑑𝑥𝑑𝑥

)

3 𝑥3 𝑦2 𝑑𝑦𝑑𝑥

+3 𝑥2 𝑦3

¿3 𝑥2 𝑦2(𝑥 𝑑𝑦𝑑𝑥+𝑦 𝑑𝑥𝑑𝑥 )

sin (𝑥𝑦)

cos (𝑥𝑦 )∗(𝑥 𝑑𝑦𝑑𝑥 +𝑦 𝑑𝑥𝑑𝑥 )

𝑥cos (𝑥𝑦 ) 𝑑𝑦𝑑𝑥

+ 𝑦 cos(𝑥𝑦 )

Implicit Differentiation

 (E) Chain Rule

Product inside a chain

𝑢=𝑥𝑦 𝑑𝑢=𝑥𝑑𝑦𝑑𝑥

+𝑦𝑑𝑥𝑑𝑥

sin𝑢 (𝑑𝑢)

cos𝑢(𝑥 𝑑𝑦𝑑𝑥+𝑦 𝑑𝑥𝑑𝑥 )

cos 𝑥𝑦 (𝑥 𝑑𝑦𝑑𝑥 )+cos 𝑥𝑦 (𝑦 𝑑𝑥𝑑𝑥 )

Implicit Differentiation

To find implicitly.

 

EX: Diff Both Sides of equation with respect to x

  Solve for

 

dy

dx

2 2 9x y dy

dx2 𝑥𝑑𝑥𝑑𝑥

+2 𝑦𝑑𝑦𝑑𝑥

=0

2 𝑥+2 𝑦𝑑𝑦𝑑𝑥

=0

𝑑𝑦𝑑𝑥

=−2𝑥2 𝑦

=−𝑥𝑦

29y x

29y x

Need both x and y to find the slope.

C CONVERSATION: Voice level 0. No talking!

HHELP: Raise your hand and wait to be called on.

AACTIVITY: Whole class instruction; students in seats.

M MOVEMENT: Remain in seat during instruction.

PPARTICIPATION: Look at teacher or materials being discussed. Raise hand to contribute; respond to questions, write or perform other actions as directed.NO SLEEPING OR PUTTING HEAD DOWN, TEXTING, DOING OTHER WORK.   

S

Activity: Teacher-Directed Instruction

Objective

• C: The swbat differentiate implicitly equations in more than one variable.

• L: the sw explain to others how to find derivatives of multiple types of problems verbally and demonstratively

EX 1:3 2 25 4y y y x

(a) Find the derivative at the point ( 5, 3 ) , at ( -1,-3 )

(b) Find where the curve has a horizontal tangent.

 (c) Find where the curve has vertical tangents.

3 𝑦2 𝑑 𝑦𝑑𝑥

+2 𝑦𝑑𝑦𝑑𝑥−5

𝑑𝑦𝑑𝑥−2𝑥

𝑑𝑥𝑑𝑥

=0

(3 𝑦¿¿2+2 𝑦−5)𝑑𝑦𝑑𝑥

=2 𝑥¿

𝑑𝑦𝑑𝑥

=2𝑥

3 𝑦2+2 𝑦−5

𝑑𝑦𝑑𝑥 |¿ (5,3)=10

28𝑑𝑦𝑑𝑥 |¿ (−1 ,−3)=−2

16

EX 1:3 2 25 4y y y x

(b) Find where the curve has a horizontal tangent.

Horizontal tangent has a 0 slope

𝑎𝑏

=0∴𝑎=0

2 𝑥=0

𝑥=0

EX 1:3 2 25 4y y y x

 (c) Find where the curve has vertical tangents. Vertical tangent has an undefined slope𝑎

𝑏𝑢𝑛𝑑𝑒𝑓 𝑏=0

3 𝑦2+2 𝑦−5=0

(3 𝑦+5)(𝑦−1)3 𝑦+5=0

𝑦=−53

𝑦−1=0

𝑦=1

Ex 2:

3 3 2x y xy

< Folium of Descartes >

3 𝑥2 𝑑𝑥𝑑𝑥

+3 𝑦2 𝑑𝑦𝑑𝑥

=2(𝑥 𝑑𝑦𝑑𝑥+𝑦 𝑑𝑥𝑑𝑥 )

3 𝑥2+3 𝑦2 𝑑𝑦𝑑𝑥

=2𝑥𝑑𝑦𝑑𝑥

+2 𝑦

3 𝑦2 𝑑𝑦𝑑𝑥−2 𝑥

𝑑𝑦𝑑𝑥

=2 𝑦−3 𝑥2

𝑑𝑦𝑑𝑥 ( (3 𝑦

2−2 𝑥)(3 𝑦 2−2 𝑥))= 2 𝑦−3𝑥2

(3 𝑦2−2𝑥 )

𝑑𝑦𝑑𝑥

=2 𝑦−3𝑥2

3 𝑦2−2𝑥

3 𝑥2−2 𝑦=2 𝑥𝑑𝑦𝑑𝑥−3 𝑦2 𝑑𝑦

𝑑𝑥

3 𝑥2−2 𝑦=(2 𝑥−3 𝑦2)𝑑𝑦𝑑𝑥

3 𝑥2−2 𝑦(2 𝑥−3 𝑦2)

=𝑑𝑦𝑑𝑥

Why Implicit?

3 3 2x y xy

< Folium of Descartes > Explicit Form:

3 6 3 3 6 33 31

1 1 1 18 8

2 4 2 4y x x x x x x

3 6 3 3 6 33 32 1

1 1 1 1 13 8 8

2 2 4 2 4y y x x x x x x

3 6 3 3 6 33 33 1

1 1 1 1 13 8 8

2 2 4 2 4y y x x x x x x

2nd Derivatives

NOTICE:The second derivative is in terms of x , y , AND dy /dx.

The final step will be to substitute back the value of dy / dx into the second derivative.

EX: Our friendly circle. Find the 2nd Derivative.

2 2 9x y 2 𝑥

𝑑𝑥𝑑𝑥

+2 𝑦𝑑𝑦𝑑𝑥

=0

2 𝑥+2 𝑦𝑑𝑦𝑑𝑥

=0

2 𝑦𝑑𝑦𝑑𝑥

=−2 𝑥

𝑑𝑦𝑑𝑥

=−2𝑥2 𝑦

𝑑𝑦𝑑𝑥

=−𝑥𝑦

𝑑2 𝑦𝑑𝑥2 =¿

𝑦 (−1𝑑𝑥𝑑𝑥 )−(− 𝑥) 𝑑𝑦

𝑑𝑥

𝑦2

𝑑2 𝑦𝑑𝑥2 =

−𝑦+𝑥𝑑𝑦𝑑𝑥

𝑦 2

𝑑2 𝑦𝑑𝑥2 =

−𝑦+𝑥 (−𝑥𝑦 )𝑦2 ( 𝑦𝑦 )

−𝑦 2−𝑥2

𝑦3𝑑2 𝑦𝑑𝑥2 =¿

2nd Derivatives

EX: Find the 2nd Derivative.

23 5xy

0−(𝑥 2 𝑦 𝑑𝑦𝑑𝑥

+𝑦2 𝑑𝑥𝑑𝑥 )=0

−2 𝑥𝑦𝑑𝑦𝑑𝑥− 𝑦2=0

−2𝑥 𝑑𝑦𝑑𝑥− 𝑦 (−2

𝑑𝑥𝑑𝑥 )

(−2𝑥)2

𝑑2 𝑦𝑑𝑥2 =¿

𝑑2 𝑦𝑑𝑥2 =¿

𝑑2 𝑦𝑑𝑥2 =¿

−2𝑥𝑑𝑦𝑑𝑥

+2 𝑦

4 𝑥2

−2𝑥 ( 𝑦−2𝑥 )+2 𝑦

4 𝑥2𝑑𝑦𝑑𝑥

= 𝑦 2

−2𝑥𝑦= 𝑦−2𝑥

𝑑2 𝑦𝑑𝑥2 =¿

𝑦+2 𝑦4 𝑥2

3 𝑦4 𝑥2

𝑑2 𝑦𝑑𝑥2 =¿

Higher Derivatives

EX: Find the Third Derivative.

sin( )y x

cos (𝑦 ) 𝑑𝑦𝑑𝑥

=𝑑𝑥𝑑𝑥

𝑑𝑦𝑑𝑥

=1

cos(𝑦 )

𝑑𝑦𝑑𝑥

=sec (𝑦 )

𝑑2 𝑦𝑑𝑥2 =¿

𝑑2 𝑦𝑑𝑥2 =¿

𝑑2 𝑦𝑑𝑥2 =¿

sec (𝑦 ) tan (𝑦 )𝑑𝑦𝑑𝑥

sec (𝑦 ) tan ( 𝑦 ) sec (𝑦 )

𝑠𝑒𝑐2 (𝑦 ) tan (𝑦 )

Last update

• 10/19/10

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