Lesson: Derivative Techniques - 4 Objective – Implicit Differentiation.
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Transcript of Lesson: Derivative Techniques - 4 Objective – Implicit Differentiation.
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Lesson: Derivative Techniques - 4
Objective – Implicit Differentiation
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These functions are relatively easy to differentiate because they are defined “EXPLICITLY”. (meaning they are solved for y, in other words y is by itself on one side of the equation.)
Differentiate:
23 5 11y x x
(4 )y Cos x
75(3 4)y x
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1. Implicit Differentiation – The process of finding the derivative of a function that is not solved for y.
Implicit functions do not have y isolated on one side of the equation.
To do this, you need to use the chain rule, with a little creativity mixed in.
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EX 1: Differentiate
1xy y x
This function is not defined explicitly, but can be with the aid of a little algebra.
Group all y terms on1 side of the = sign.
1xy y x
Factor out a y( 1) 1y x x Divide by (x + 1) to isolate y
1
1
xy
x
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1
1
xy
x
Now use the quotient ruleTo differentiate
'
2
' 'f g f f g
g g
2
( 1)(1) ( 1)(1)'
( 1)
x xy
x
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2
( 1) ( 1)'
( 1)
x xy
x
2
1 1'
( 1)
x xy
x
2
2'
( 1)y
x
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Question:
O.K. but what about those that cannot be expressed explicitly or that are a nightmare to do so? I know you will be giving us some of those, won’t you, Mr. Winter!
Well, let’s work our way into those. Let meshow you first how to do the problem fromexample 1 “IMPLICITLY”.
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Process of Implicit Differentiation
• Differentiate both sides of the equation with respect to x.• Apply Chain Rule whenever you have an expression involving y.• Move all terms involving dy/dx (or y’) to the left side of the equation, and everything else to the right side.• Factor out dy/dx (or y’) on the left.• Solve for dy/dx (or y’)
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1xy y x Now let’s try it.Differentiate:
' 1 ' 1x y y y
' ' 1x y y y
' ' 1x y y y
'( 1) 1y x y
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1'
1
yy
x
But that answer doesn’t look the same as theOne we got by defining it explicitly!
Remember from the first method that:
1
1
xy
x
Substitute that in for y andSee what happens.
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EX. 2: Differentiate
5 2 33 4y x Sinx y 4 25 ' 6 12 'y y x Cosx y y 4 25 ' 12 ' 6y y y y Cosx x
4 2'(5 12 ) 6y y y Cosx x
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4 2
6'
5 12
Cosx xy
y y
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EX. 3: Use implicit differentiation to find dy/dx if
2 25 ( )y Sin y x
2 2[5 ( )] [ ]d d
y Sin y xdx dx
10 ' ( ) ' 2y y Cos y y x
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10 ' ( ) ' 2y y Cos y y x
'(10 ( )) 2y y Cos y x
2'
10 ( )
xy
y Cos y
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EX. 4: Use implicit differentiation to find the second derivative d2y/dx2 of
2 24 2 9x y
Solution: 3
9''y
y
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EX. 5: Find the slopes of the tangent lines to the curve y2 – x + 1 = 0 at the points (2, -1) and (2, 1).
Solution:2
1
1
2xy
dy
dx
21
1
2xy
dy
dx
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EX. 6: (a)Use implicit differentiation to find dy/dx for the “Folium of Descartes: x3 + y3 =3xy
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3 3 3x y xy 2 23 3 ' 3( ' 1)x y y x y y
2 23 3 ' 3 ' 3x y y x y y Divide all terms by 3 to simplify
2 2 ' 'x y y x y y
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2 2 ' 'x y y x y y
2 2' 'y y xy y x
2 2'( )y y x y x 2
2'
y xy
y x
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EX. 7:
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HW 4.1
Pg. 241(1 – 3, 5, 11 – 15, 21 – 24, 30*)