1 implicit differentiation. 2 introduction consider an equation involving both x and y: this...
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Implicit Differentiation
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2
Introduction
• Consider an equation involving both x and y:
• This equation implicitly defines a function in x
• It could be defined explicitly
2 2 49x y
2 49 ( 7)y x where x
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3
Differentiate
• Differentiate both sides of the equation– each term– one at a time– use the chain rule for terms containing y
• For we get
• Now solve for dy/dx
2 2 49x y
2 2 0dy
x ydx
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Differentiate• Then gives us
• We can replace the y in the results with the explicit value of y as needed
• This gives usthe slope on the curve for any legal value of x
2 2 0dy
x ydx
2
2
dy x x
dx y y
2 49
dy x
dx x
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Guidelines for Implicit Differentiation
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Slope of a Tangent Line
• Given x3 + y3 = y + 21find the slope of the tangent at (3,-2)
• 3x2 +3y2y’ = y’
• Solve for y’2
2
3'
1 3
xy
y
Substitute x = 3, y = -2 27
11slope
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Second Derivative
• Given x2 –y2 = 49
• y’ =??
• y’’ =
'x
yy
2
2 2
2 2
2 2
2 2 3 3
'
49
d y y x y
dx y
x xy y
y xy yy y y y
Substitute
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Find the derivative with respect to x.
22 2 2 2dy dy dx
y x y ydx dx
dy dx dy dxx y x x y
dx dx xdx dx d
2 2 23. x y y xy xy 22 2x y xyy xy
2 22 2 2dy dy dy dy
x xy y x y xy ydx dx dx dx
2 22 2 2dy dy dy dy
x y x xy y xy ydx dx dx dx
2 22 2 2dy
x y x xy y xy ydx
2
2
2
2 2
dy y xy y
dx x y x xy
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1. Find the equation of the tangent line to at the point (2,1).
2 4 8x y
2 4 2 0dy
x y xdx
2
2
4
dy xy
dx x
42,1
8
dy
dx
1
2
11 2
2y x
12
2y x
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2. Find the second derivative with respect to x.
2 2 16x y
2 2 0dy
x ydx
dy x
dx y
2
2
d y
dx
2
1d
x
y
ydx
y
2
xy
y x
y
2
2
xy
yy
y
y
2 2
3
y x
y
3
16
y
2 2 16x y
Look for a substitution of the original.
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3. Find the points at which the graph has horizontal and vertical tangents.
2 225 16 200 160 400 0x y x y
0slope undefinedslope
50 32 200 160 0dy dy
x ydx dx
50 200
32 160
dy x
dx y
Horizontal: Vertical:
50 200x 32 160y
0 0
4x 5y
2 225 4 16 200 4 160 400 0y y 2225 16 5 200 160 5 400 0x x 216 160 0y y 16 10 0y y
0,10y
and 4,0 4,10
225 200 0x x 25 8 0x x
0, 8x
and0,5 8,5
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We need the slope. Since we can’t solve for y, we use
implicit differentiation to solve for .dy
dx
Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
2 2 7x xy y
2 2 0dydy
x yx ydxdx
Note product rule.
2 2 0dy dy
x x y ydx dx
22dy
y xy xdx
2
2
dy y x
dx y x
2 2 1
2 2 1m
2 2
4 1
4
5
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Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
4
5m tangent:
42 1
5y x
4 42
5 5y x
4 14
5 5y x
normal:
52 1
4y x
5 52
4 4y x
5 3
4 4y x
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Higher Order Derivatives
Find if .2
2
d y
dx3 22 3 7x y
3 22 3 7x y
26 6 0x y y
26 6y y x
26
6
xy
y
2xy
y
2
2
2y x x yy
y
2
2
2x xy y
y y
2 2
2
2x xy
y
x
yy
4
3
2x xy
y y
Substitute back into the equation.
y