Implicit Differentiation
Homework: pg. 162 # 1, 3, 5, 7, 9
In Exercises 1-5, sketch the curve defined by the equation and
find two functions y1 and y2 whose graphs will combine to give
the curve.
1.
x  y2  0
2.
4 x 2  9 y 2  36
3.
x2  4 y 2  0
4.
x2  y 2  9
5.
x2  y 2  2x  3
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In Exercises 6-8, solve for y  in terms of y and x.
6.
x 2 y   2 xy  4 x  y
7.
y  sin x  x cos x  xy   y
8.
x  y 2  y  y  x 2  y 
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In Exercises 9 and 10, find an expression for the function using
rational powers rather than radicals.
9.
10.

x x 3 x

x  3 x2
x3
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In Exercises 1-5, sketch the curve defined by the equation and
find two functions y1 and y2 whose graphs will combine to give
the curve.
1.
x  y2  0
y1  x ,
2.
4 x 2  9 y 2  36
3.
x2  4 y 2  0
4.
x2  y 2  9
5.
x2  y 2  2 x  3
y2   x
2
2
9  x2 ,
y2  
9  x2
3
3
x
x
y1  ,
y2  
2
2
y1 
y1  9  x 2 ,
y2   9  x 2
y1  2 x  3  x 2 ,
y1   2 x  3  x 2
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In Exercises 6-8, solve for y  in terms of y and x.
6.
7.
y  sin x  x cos x  xy   y
y 
8.
x  y  y  y  x  y 
xy 2
y  2
x  yx
2
2
y 
4 x  y  2 xy
x2
x y   2 xy  4 x  y
2
y  x cos x
sin x  x
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In Exercises 9 and 10, find an expression for the function using
rational powers rather than radicals.
9.
10.

x x x
3
x  3 x2
x
3

3
2
x x
1
2
5
6
x x
5
6
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



Implicitly Defined Functions
Lenses, Tangents, and Normal Lines
Derivatives of Higher Order
Rational Powers of Differentiable Functions
… and why
Implicit differentiation allows us to find
derivatives of functions that are not defined
or written explicitly as a function of a single
variable.
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The graph of
(called a folium).
Although not the graph of a function, it is the
union of the graphs of three separate functions.
This particular curve dates to Descartes in 1638.
dy
Find
if 3 y 2  2 y  5 x
dx
dy
To find
differentiate both sides of the equation with respect
dx
to x, treating y as a differentiable function of x and applying the
Chain Rule.
3 y 2  2 y  5x
dy
dy
6y  2 5
dx
dx
dy
6 y  2  5
dx
dy
5

dx 6 y  2
d
d
2
2 dy
3
y

3
y



 dx 
dy
dx


 d  2 y   d  2 y  dy
 dx
dy
dx
1. Differentiate both sides of the equation with respect to x.
2. Collect the terms with
dy
on one side of the equation.
dx
dy
.
dx
dy
4. Solve for
.
dx
3. Factor out
Solve #2, 4, 12, 49 pg.162-163
In the law that describes how light changes direction
as it enters a lens, the important angles are the
angles the light makes with the line perpendicular to
the surface of the lens at the point of entry (angles A
and B in Figure 3.50). This line is called the normal
to the surface at the point of entry. In a profile view
of a lens, the normal is a line perpendicular to the
tangent to the profile curve at the point of entry.
Implicit differentiation is often used to find the
tangents and normals of lenses described as
quadratic curves.
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Find the equations of the tangent and normal lines to the graph
 2
2
given by x  x y  y  0 at the point 
,
.
2 
 2
Differentiate implicitly as
d 4
d 2 2
d 2
x

x
y

y 0





dx
dx
dx
dy
 dy 
4 x 3  x 2  2 y   2 xy 2  2 y  0
dx
 dx 
dy
2 y  x 2  1   2 x  2 x 2  y 2 
dx
2
2

2
x
2
x

y


dy

dx
2 y  x 2  1
4
2
2
2
dy

dx
x  2x2  y 2 
y  x  1
2
2
treat x y
 slope of the tangent line
2
as a product
x  2x2  y 2 
 2 2
dy

Find
the
slope
of
the
tangent
line
at
,


dx
2
2 
y  x 2  1

2  2  2 
2   2   2 
 2
3
 
 
2    

2   2   2  

   2   4   4   2  3
m= 
2
1

2  2  
2  2 


1
 

  1
2
2   4  

2  2 


The equation of the tangent line at this point is
2
y
2

2
2
 3 x 
  y  3x  2.
2
2


The equation of the normal line at this point is
y
2
1
2
1
2
2
  x 

  y x 
2
3
2 
3
6
2
1
2 2
y x
3
3
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d 2x
Find
if x 2  y 2  25
2
dy
Differentiating each term with respect to x gives
2x  2 y
dy
0
dx
dy
  2x
dx
dy
x

dx
y
Differentiating again with respect to x gives
2y
2
d x

dy 2
 y 1   x  
dy 

 dx 
y2
 x
y   x  
 y

y2
y2  x2

y3
25
 3
y
Quotient Rule
Substitute 
x
dy
for
y
dx
Simplify
Substitute 25 for x
2
 y
2
Slide 3- 16







Pg. 146 #8, 9, 10 (u/v)’,
Pg.162-163 # 17, 24 (implicit dif, tangent,
normal), 52, 53
Pg. 92-94 #1, 6, 38
Pg.77 # 4 (sandwich theorem)
Pg.67-68 # 51, 53, 57, 58 (right-left hand
limits)
Pg.81 removing discontinuity, # 25 pg. 85
Pg. 107 # 26


Standardized Test Questions pg.164
Homework:pg. 170 quick review # 1-9 (odd)
Derivatives of Inverse Trigonometric
Functions
In Exercises 1-5, give the domain and range of the function,
and evaluate the function at x 1.
1. y  sin 1 x
2. y  cos 1 x
3. y  tan 1 x
4. y  sec 1 x
5. y  tan  tan 1 x 
Slide 3- 22
In Exercises 6-10, find the inverse of the given function.
6. y  3 x  8
7. y  3 x  5
8
x
3x  2
9. y 
x
8. y 
x
10. y  arctan  
3
Slide 3- 23
In Exercises 1-5, give the domain and range of the function,
and evaluate the function at x 1.
1. y  sin 1 x
2. y  cos 1 x
3. y  tan 1 x
4. y  sec 1 x
5. y  tan  tan 1 x 

  
Domain: 1,1 Range: - ,  At 1:
2
 2 2
Domain: 1,1 Range: 0,   At 1:0

  
Domain:All Reals Range:  - ,  At 1:
4
 2 2
Domain:  ,  1 ∪ 1,  
    
Range: 0,  ∪  ,   At 1:0
 2 2 
Domain:All Reals Range:All Reals At 1:1
Slide 3- 24
In Exercises 6-10, find the inverse of the given function.
x 8
3
6. y  3 x  8
f 1  x  
7. y  3 x  5
f 1  x   x 3  5
8
x
3x  2
9. y 
x
f 1  x  
8
x
f 1  x  
2
3 x
 x
10. y  arctan  
3
f 1  x   3tan x, 
8. y 

2
x

2
Slide 3- 25





Derivatives
Derivatives
Derivatives
Derivatives
Derivatives
of
of
of
of
of
Inverse Functions
the Arcsine
the Arctangent
the Arcsecant
the Other Three
… and why
The relationship between the graph of a function and
its inverse
allows us to see the relationship between their
derivatives.
Slide 3- 26
dy
If f is differentiable at every point of an interval I and
dx
is never zero on I , then f has an inverse and f 1 is differentiable
at every point on the interval f  I  .
Slide 3- 27
If u is a differentiable function of x with u  1, we apply the
Chain Rule to get
d
1 du
sin 1 u 
, u  1.
2
dx
1  u dx
Slide 3- 29