Download - Activity: Teacher-Directed Instruction
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.
P PARTICIPATION: 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
β’ C: The swbat differentiate implicitly equations in more than one variable.
β’ 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 ππ¦ππ₯ =1
2(9βπ₯2 )
β12 (β2 π₯ )
ππ¦ππ₯ =
βπ₯β9βπ₯2
ππ¦ππ₯=β 1
2( 9β π₯2 )
β 12 (β2π₯ )
ππ¦ππ₯ =
π₯β9βπ₯2
π₯2π¦+2 π¦2π₯+3 π¦3=7Donβt 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
=
=
2 ππ₯ππ₯ +3 ππ¦ππ₯=2+3 ππ¦ππ₯
4 π¦ 3(2π₯+2)
6 π¦2(2 π₯+2)
Implicit Differentiation
Find the derivative with respect to x
< Assuming - y is a differentiable function of x >
π₯π¦=ΒΏ π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯ ΒΏ π₯ ππ¦ππ₯ + π¦
π₯2+ π¦2=ΒΏ2 π₯ ππ₯ππ₯ +2 π¦ ππ¦ππ₯ ΒΏ2 π₯+2 π¦ ππ¦ππ₯
sin (π₯π¦) ΒΏcos (π₯π¦ )β(π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯ )ΒΏ π₯cos (π₯π¦ ) ππ¦ππ₯ + π¦ cos (π₯π¦)
Implicit Differentiation
(D) Product Rule
2xy ΒΏ π₯ (2 π¦ ) ππ¦ππ₯ + π¦2 ππ₯ππ₯
ΒΏ2 π₯π¦ ππ¦ππ₯ +π¦2
ΒΏ π₯ ππ¦ππ₯+ π¦β ππ₯ππ₯
ΒΏ π₯ ππ¦ππ₯ +π¦β
π₯π¦
Implicit Differentiation
(E) Chain Rule 3( )xy
Product inside a chain
3 (π₯π¦ )2(π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯ )
3 π₯3 π¦2 ππ¦ππ₯+3 π₯2 π¦3
ΒΏ3 π₯2 π¦2(π₯ ππ¦ππ₯ + π¦ ππ₯ππ₯ )
sin (π₯π¦)
cos (π₯π¦ )β(π₯ ππ¦ππ₯ + π¦ ππ₯ππ₯ )
π₯cos (π₯π¦ ) ππ¦ππ₯ + π¦ cos(π₯π¦ )
Implicit Differentiation
(E) Chain Rule
Product inside a chain
π’=π₯π¦ ππ’=π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯
sinπ’ (ππ’)
cosπ’(π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯ )
cos π₯π¦ (π₯ ππ¦ππ₯ )+cos π₯π¦ (π¦ ππ₯ππ₯ )
Implicit Differentiation
To find implicitly.
EX: Diff Both Sides of equation with respect to x
Solve for
dydx
2 2 9x y dydx2 π₯ ππ₯ππ₯+2 π¦ ππ¦ππ₯=0
2 π₯+2 π¦ ππ¦ππ₯=0
ππ¦ππ₯=
β2π₯2 π¦ =
βπ₯π¦
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.
P PARTICIPATION: 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
β’ C: The swbat differentiate implicitly equations in more than one variable.
β’ 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 π¦2 π π¦ππ₯ +2 π¦ ππ¦ππ₯ β5 ππ¦ππ₯ β2π₯ ππ₯ππ₯=0
(3 π¦ΒΏΒΏ2+2 π¦β5)ππ¦ππ₯=2 π₯ΒΏ
ππ¦ππ₯ =
2π₯3 π¦2+2 π¦β5
ππ¦ππ₯ |ΒΏ (5,3)=10
28ππ¦ππ₯ |ΒΏ (β1 ,β3)=β2
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
ππ=0β΄π=0
2 π₯=0π₯=0
EX 1:3 2 25 4y y y x
(c) Find where the curve has vertical tangents. Vertical tangent has an undefined slopeπ
ππ’ππππ π=0
3 π¦2+2 π¦β5=0(3 π¦+5)(π¦β1)
3 π¦+5=0π¦=
β53
π¦β1=0π¦=1
Ex 2:
3 3 2x y xy
< Folium of Descartes >
3 π₯2 ππ₯ππ₯ +3 π¦2 ππ¦
ππ₯=2(π₯ ππ¦ππ₯ +π¦ ππ₯ππ₯ )3 π₯2+3 π¦2 ππ¦
ππ₯=2π₯ ππ¦ππ₯ +2 π¦
3 π¦2 ππ¦ππ₯ β2 π₯ ππ¦ππ₯=2 π¦β3 π₯2
ππ¦ππ₯ ( (3 π¦
2β2 π₯)(3 π¦ 2β2 π₯))= 2 π¦β3π₯2
(3 π¦2β2π₯ )
ππ¦ππ₯ =
2 π¦β3π₯2
3 π¦2β2π₯
3 π₯2β2 π¦=2 π₯ ππ¦ππ₯ β3 π¦2 ππ¦ππ₯
3 π₯2β2 π¦=(2 π₯β3 π¦2)ππ¦ππ₯
3 π₯2β2 π¦(2 π₯β3 π¦2)
=ππ¦ππ₯
Why Implicit?
3 3 2x y xy
< Folium of Descartes > Explicit Form:
3 6 3 3 6 33 31
1 1 1 18 82 4 2 4
y x x x x x x
3 6 3 3 6 33 32 1
1 1 1 1 13 8 82 2 4 2 4
y y x x x x x x
3 6 3 3 6 33 33 1
1 1 1 1 13 8 82 2 4 2 4
y 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 π₯ ππ₯ππ₯ +2 π¦ ππ¦ππ₯=0
2 π₯+2 π¦ ππ¦ππ₯=0
2 π¦ ππ¦ππ₯ =β2 π₯
ππ¦ππ₯=
β2π₯2 π¦
ππ¦ππ₯=
βπ₯π¦
π2 π¦ππ₯2 =ΒΏ
π¦ (β1 ππ₯ππ₯ )β(β π₯) ππ¦ππ₯π¦2
π2 π¦ππ₯2 =
βπ¦+π₯ ππ¦ππ₯π¦ 2
π2 π¦ππ₯2 =
βπ¦+π₯ (βπ₯π¦ )π¦2 ( π¦π¦ )
βπ¦ 2βπ₯2
π¦3π2 π¦ππ₯2 =ΒΏ
2nd DerivativesEX: Find the 2nd Derivative.
23 5xy
0β(π₯ 2 π¦ ππ¦ππ₯ + π¦2 ππ₯ππ₯ )=0
β2 π₯π¦ ππ¦ππ₯ β π¦2=0
β2π₯ ππ¦ππ₯ β π¦ (β2 ππ₯ππ₯ )(β2π₯)2
π2 π¦ππ₯2 =ΒΏ
π2 π¦ππ₯2 =ΒΏ
π2 π¦ππ₯2 =ΒΏ
β2π₯ ππ¦ππ₯ +2 π¦
4 π₯2
β2π₯ ( π¦β2π₯ )+2 π¦
4 π₯2ππ¦ππ₯ =
π¦ 2
β2π₯π¦=π¦β2π₯
π2 π¦ππ₯2 =ΒΏ
π¦+2 π¦4 π₯2
3 π¦4 π₯2
π2 π¦ππ₯2 =ΒΏ
Higher DerivativesEX: Find the Third Derivative.
sin( )y x
cos (π¦ ) ππ¦ππ₯=ππ₯ππ₯
ππ¦ππ₯ =
1cos(π¦ )
ππ¦ππ₯ =sec (π¦ )
π2 π¦ππ₯2 =ΒΏ
π2 π¦ππ₯2 =ΒΏ
π2 π¦ππ₯2 =ΒΏ
sec (π¦ ) tan (π¦ )ππ¦ππ₯
sec (π¦ ) tan ( π¦ ) sec (π¦ )
π ππ2 (π¦ ) tan (π¦ )
Last update
β’ 10/19/10
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