Transcript Aim: What Is Implicit Differentiation and How Does It Work?

Aim: What Is Implicit Differentiation and
How Does It Work?
Implicit
Do Now:
Find the derivative of 4x 2 y  3 y  x 3  1
y  4x 2  3   x 3  1
x 1
y 2
4x  3
3
y' 
Explicit
4x  9x  8x
4
 4x
Aim: Implicit Differentiation
2
2
 3
2
Course: Calculus
Implicit vs. Explicit
Explicit Form
Implicit Form
1
xy  1
y
x
variable y is written as a function of x
1
2
 y'   x
derivative of y? y  x
Often you can solve for y in term of x
Not Always!
x2  2 y3  4 y  2
Implicit Differentiation is used
Aim: Implicit Differentiation
Course: Calculus
Differentiating with Respect to x
a.
d
 x 3   3 x 2
dx
Use Simple Power Rule
variables agree
un
nun-1 u’
d
3
2 dy
 y   3 y
b.
dx
dx
Use Chain Rule
variables disagree
d
dy
c.
x  3 y  1  3

dx
dx
Chain Rule
d
3 y  3 y '
dx
d
d
2
2
2 d
 xy   x  y   y
d.
x  Product Rule

dx
dx
dx
dy
 dy 
2
2

x
2
y

y
1

2
xy

y


 dx 
Chain Rule
dx


Aim: Implicit Differentiation
Course: Calculus
Simplify
Differentiating with Respect to x
a.
d
 x 3   3 x 2
dx
Use Simple Power Rule
variables agree
un
nun-1 u’
d
3
2 dy
 yCOMMON
  3 y
b.
dx
dx
Use Chain Rule
ERROR!
DON’T
FORGET
variables disagree
c.
d  6
5 dy
y

6
y
d
dy
 x  3 yd x 1  3 Chain
dxRule
dx
dx
d
3 y  3 y '
dx
d
d
2
2
2 d
 xy   x  y   y
d.
x  Product Rule

dx
dx
dx
dy
 dy 
2
2

x
2
y

y
1

2
xy

y


 dx 
Chain Rule
dx


Aim: Implicit Differentiation
Course: Calculus
Simplify
Guidelines for Implicit Differentiation
1. Differentiate both sides of the equation
with respect to x.
2. Collect all terms involving dy/dx on the
left side of the equation and move all
other terms to the right side of the
equation.
3. Factor dy/dx out of the left side of the
equation.
4. Solve for dy/dx by dividing both sides of
the equation by the left-hand factor that
does not contain dy/dx.
Aim: Implicit Differentiation
Course: Calculus
Model Problem
Find dy/dx given y3 + y2 – 5y – x2 = -4
1.
Differentiate both sides of the equation with respect to x.
d
d
3
2
2
 y  y  5 y  x  
4 

dx
dx
d
d
d
d
d
3
2
2
 y  
 y    5 y  
 x  
4

dx
dx
dx
dx
dx
dy
dy
2 dy
3y
 2y
 5  2x  0
dx
dx
dx
2. Collect all terms involving dy/dx on the left side of the equation
dy
dy
dy
3y
 2y
5
 2x
dx
dx
dx
2
Aim: Implicit Differentiation
Course: Calculus
Model Problem
Find dy/dx given y3 + y2 – 5y – x2 = -4
3. Factor dy/dx out of the left side of the equation.


dy
3 y2  2 y  5  2 x
dx
4. Solve for dy/dx by dividing by (3y2 + 2y – 5)
dy
2x
 2
dx 3 y  2 y  5
function?
NO
(1, 1)
(2, 0)
(1, -3)
slope at (1, 1) und
slope at (2, 0) -4/5
slope at (1, -3) 1/8
y3 + y2 – 5y – x2 = -4
Aim: Implicit Differentiation
Course: Calculus
Functions from Equations
If a segment of a graph can be represented
by a differentiable function, dy/dx will
have meaning as the slope.
x  y2  1
2
1.5
function?
NO
1
y  1  x 2 YES
0.5
-1
1
-0.5
y   1  x YES
2
-1
-1.5
Recall: a function is not differentiable at
points with vertical tangents nor at points
where the function is not continuous
Aim: Implicit Differentiation
Course: Calculus
Aim: What Is Implicit Differentiation and
How Does It Work?
Do Now:
Determine the slope of the tangent line to
the graph x2 + 4y2 = 4 at the point
1 

 2, 
.
2

Aim: Implicit Differentiation
Course: Calculus
Model Problem
Determine the slope of the tangent line to
the graph x2 + 4y2 = 4 at the point
dy
1 

Note:
 y'
 2, 
.
dx
2

2 x  8 yy '  0
implicit differentiation
2 x  x
y' 

8y
4y
solve for dy/dx
dy 22 2 2 1 1
y ' 

dx   1 1  2 2
8 8   
  2 2 
evaluate for the point
Slope of tangent at
Aim: Implicit Differentiation


2, 1 / 2 is 1/2
Course: Calculus
Model Problem
Determine the slope of the tangent line to
the graph 3(x2 + y2)2 = 100xy and the point
(3, 1). d
 3 x 2  y 2 2   d 100 xy 
 dx
dx 


Constant and General Power Rules

  2 x  2 yy '  100  xy ' y 1

  2 x  yy '  100 xy ' 100 y
3  2 x 2  y 2
6 x2  y2

12 x 2  y 2
  x  yy '  100 xy ' 100 y
FOIL and isolate dy/dx
Aim: Implicit Differentiation
Course: Calculus
Model Problem
Determine the slope of the tangent line to
the graph 3(x2 + y2)2 = 100xy and the point
(3, 1).




12 x 2  x   x 2 yy ' y 2  x   y 2 yy '  100 xy ' 100 y
12 x 2  x   y 2  x   x 2 yy ' y 2 yy '  100 xy ' 100 y
12
2
2
2
2
x
x

y

yy
'
x

y





   100 xy ' 100 y





 
12  x  x 2  y 2  12 yy ' x 2  y 2
   100 xy ' 100 y

12 yy ' x 2  y 2  100 xy '  100 y  12  x  x 2  y 2
Aim: Implicit Differentiation
Course: Calculus

Model Problem
Determine the slope of the tangent line to
the graph 3(x2 + y2)2 = 100xy and the point
(3, 1).



12 yy ' x 2  y 2  100 xy '  100 y  12  x  x 2  y 2
 




y ' 12 y x 2  y 2  100 x  100 y  12  x  x 2  y 2


y' 
12 y  x  y   100 x
100  1  12  3   3  1  13
y' 

12  1  3  1   100  3  9

100 y  12  x  x 2  y 2
2
2
2
2
2
Aim: Implicit Differentiation
2
substitute (3, 1)
Course: Calculus
Finding the 2nd Derivative Implicitly
2
Given
x2 +
y2
= 25, find
d y
dx 2
find first derivative implicitly:
dy
2x  2 y
0
dx
dy
2y
 2 x
dx
 dy 
y  1   x   

2
d y
dx 


2
2
dx
y
 x 
 y   x 
y 


y2
y2  x2

y3
sub 25 for x2+y2
Aim: Implicit Differentiation
dy 2 x  x


dx
2y
y
quotient rule
sub –x/y for dy/dx
dy 2
25
 3
2
dx
y
Course: Calculus
Model Problem
Find the tangent line to the graph given by
x2(x2 + y2) = y2 at the point  2 2 
,


 2 2 
implicit differentiation
x4  x2 y2  y2  0
 
4 x3  x2
2
2
y
y
'

2
xy
 2 yy '  0
 

4 x  2 xy  2 yx
3
2

2
 y ' 2 yy '  0



2 x 2 x 2  y 2  y ' 2 y  x 2  1  0
y' 

2 x 2 x 2  y 2
 2 y   x 2  1

Aim: Implicit Differentiation
Course: Calculus
Model Problem
Find the tangent line to the graph given by
x2(x2 + y2) = y2 at the point  2 2 
,


 2 2 
2
2
2
2
2 x 2 x  y
x 2x  y
y' 

2
2
2
y
x

1
y
1

x
 





2
2

2  2  2 
 2
 
 
2   2   2  


y' 
2

 2 
2
1 
 
2   2  



2
2
y
 3 x 

Aim:
Implicit
Differentiation
2
2 




substitute
=3=m
point-slope formula for equation
(y – y1) = m(x – x1)
y  3x  2
Course: Calculus
Model Problem
Find dy/dx implicitly for the equation sin y = x.
Then find the largest interval of the form
–a < y < a such that y is a differentiable function
d
d
of x.
sin y  
x


r y = sin y
dx
dx
dy
cos y
1
dx
 
 1, 2 
dy
1






dx cos y

1,



2

explicitly:
6
4
2
-2
2
-2
-4
-6
-8
for the interval -/2 < y < /2,
we use cos y  1  sin 2 y and
substitute the original equation
1
to arrive at dy
 Course: Calculus
Aim: Implicit Differentiation
dx
1  x2