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Implicit Differentiation
Lesson 3.6
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Introduction
Consider an equation involving both x
and y:
2
2
x  y  49
This equation implicitly defines a
function in x
It could be defined explicitly
y  x  49
2
(where x  7)
Differentiate
Differentiate both sides of the equation



each term
one at a time
use the chain rule for terms containing y
For
x  y  49 we get
2
2
dy
2x  2 y
0
dx
Now solve for dy/dx
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Differentiate
Then
dy
2x  2 y
 0 gives us
dx
dy 2 x x


dx 2 y y
We can replace the y in the results with
the explicit value of y as needed
This gives us
dy
x
the slope on the

2
dx
x  49
curve for any
legal value of x
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Slope of a Tangent Line
Given x3 + y3 = y + 21
find the slope of the tangent at (3,-2)
3x2 +3y2y’ = y’
2
3
x
Solve for y’
y' 
2
1 3y
Substitute x = 3, y = -2
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slope 
11
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Second Derivative
Given x2 –y2 = 49
x
y'
y
y’ =??
y’’ =
Substitute
d y y  x y'

2
2
dx
y
2
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Derivatives of Inverse Trig Functions
See Theorem 3.11, pg 182
Try this
y = cos-1(4x – 3)
dy
1

4
dx
1  (4 x  3)
Exponential & Log Functions
Given y = bx
where b > 0, a constant
dy
x
 ln b  b
dx
Given y = logbx
1
y'
ln b  x
Note: this is a constant
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Using Logarithmic Differentiation
Given
y  ( x  1) ( x  3)
18
10
3
7
8
Take the log of both sides, simplify
Now differentiate both sides with
respect to x, solve for dy/dx
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Implicit Differentiation on the TI Calculator
We can declare a function which will do
implicit differentiation:
Usage:
Assignment
Lesson 3.6
Page 186
5, 7, 9, 13, 15, 17, 19, 21, 29, 30, 33,
35, 39, 41, 51, 57
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