Section 2.5 – Implicit
Differentiation
Explicit Equations
The functions that we have differentiated
and handled so far can be described by
expressing one variable explicitly in terms
of another variable. For example:
y  x 1
3
Or, in general, y = f(x).
y  x sin x
Implicit Equations
Some functions, however, are defined
implicitly ( not in the form y = f(x) ) by a
relation between x and y such as:
x  y  25
2
2
It is possible to solve some Implicit Equations for y:
x  y  25
2
2
y  25  x
2
y 
  25  x
2
2
x  y  6xy
3
3
Yet, it is difficult to rewrite most
Implicit Equations explicitly.
Thus, we must be introduced
to a new technique to
differentiate these implicit
functions.
*Reminder*
Technically the Chain Rule can be applied to every
derivative:
yx
dy
dx

3
d
dx
x 
3
3
u
f u 
ux
f ' u  3u




dy
dx
2
u'  1
 3u 1
2

2
  3x
Derivatives Involving the Dependent
Variable (y)
Find the derivative of each expression
y
a.
d
dx
b.
y
y'
y
3
d
dx
The derivative of y with
respect to x is…
the derivative of y.
dy
dx
This is another way
to write y prime.


y 
3
f u  u
2
f ' u  3u
u y
3




d
dx
The Chain
Rule is
Required.
u'  dy
dx
y   3u 
3
2

2
  3y 

dy
dx
dy
dx
Instructions for Implicit Differentiation
If y is an equation defined implicitly as a differentiable
function of x, to find the derivative:
1. Differentiate both sides of the equation with respect to
x. (Remember that y is really a function of x for part of
the curve and use the Chain Rule when differentiating
terms containing y)
2. Collect all terms involving dy/dx on the left side of the
equation, and move the other terms to the right side.
3. Factor dy/dx out of the left side
4. Solve for dy/dx

Example 1
If y  f x is a differentiable function of x such that
.
x 2 y  2y 3  3x  2y find dy
dx
2
3
d
Differentiate
x
y

2y
  dxd 3x  2y
dx 
both sides.
Product AND
Constant
Multiple Rules
2
3
d
x
y

2y
  dx   
3x   dxd 2y 
x 2 dxd y 
 y dxd x 2  2 dxd y 3   3 dxd x   2 dxd y 

d
dx
Chain
Rule
3
u
f u 
f1' u  3u 2
d
dx
uy
u'  dy
dx
dy
dy
dy
x 2 dx  y 2x  2 3u2  dx  3 1 2 dx
dy
2 dy
2 dy

x

2xy

6y

3
2
  dx
dx
dx

dy
dy
dy
2
2

x

6y
dx 2 dx  3  2xy
dx

dy
2
2


dx

x
 6y  2  3  2xy
dy
32xy

dx
x 2 6y 2 2
Example 2
Find y' if sin x  y   y 2 cos x .
Differentiate both sides
 Chain


Rule
Twice
Product Rule
f1 u1   sin u1
f ' u  cosu


1
1

1
d
dx
d
dx
sin x  y  
d
dx
2
y
 cos x
sin x  y   y 2 dxd cos x   cos x dxd y 2 
u1  x  y
dy
u1'1 dx
f 2 u2   u22 u2  y
dy
f 2 ' u2   2u2 u2 '  dx
dy
 y 2  sin x   cos x 2u2  dx
 
cos x 2y
sinx 
cosx  y  1  y 
dy
cosu1  1 dx
2

dx
dy



cosx  y   cosx  y  dx  y 2
sin x
 2y 
cos x
dy
dy
dx
dy
dy

cosx  y  dx  2y cos x dx  y 2 sin x  cosx  y 
dy
2
 dx cosx  y   2y cos x   y sin x  cosx  y 

dy
dx

y 2 sin x cosx y 
cosx y 2y cosx
dy
dx
Example 3
Find
d 2y
dx
2
if x 2  y 2  10.
d 2y
dx
Find the first derivative by Differentiating both sides.
x  y  
x  y  
d
dx


Chain
Rule




d
dx
2
2
2
d
dx
2
f u  u
f1' u  2u
Quotient Rule dx 2
d 2y
u' 

x
y
Now Find the Second Derivative
dx 2

dy
dx
2x  2u  0
 2x 
2y   0


2y    2x
 2x
2y

d 2y
d
dx
uy
2
dy
dx
dy
dx
dy
dx
dy
dx
dy
dx
10
d
dx 10
2
d 2y

Remember:
dy
dx
 
x
y
dx 2
d 2y
dx 2
2



Remember:
x  y  10
2
2
d y
dx 2
dy
dx
dy
dx
dy
dx










d
dx

x
y
d
d
y dx
y
xx dx
y2
dy
y1x dx
y2
dy
y x dx
y2
y x
 
x
y
y2
y   yx
2
y2
y 2 x 2
y3
 y2 x2

y3
10
y3


y
y
More with Derivatives
Now evaluate the limits and find tangent
lines.
Example 1
Find the slope of a line tangent to the circle x 2  y 2  5x  4 y
at the point P5,4.
2
2
d
Find the
x

y
  dxd 5x  4 y 
dx 
derivative by
differentiating
both sides.
2
2
d
x

y
  dx   
5x   dxd4 y 
 dxd x 2  dxd y 2   5 dxd x   4 dxd y 

Chain
Rule






d
dx
f u  u 2
f '  u  2u
d
dx
uy
u'  dy
dx
dy
dy
2x  2u dx  5  4 dx
dy
dy
2x

2y

5

4
  dx
dx
dy
dy
2y
 4 dx  5  2x
dx 
dy
dx 2y  4  5  2x
dy
dx

52x
2y 4
Evaluate the derivative
at x=5 and y=4.
dy
dx  5,4 

525
24 4

5
4

Example 2
If x  f x   xf x   6 and f 3 1 , find f ' 3.
3
Find the derivative by
differentiating both
sides.
x
d
dx
f x  f x

Chain
Rule
3
3
f u  u
2
f '  u   3u
3
xf x  xf x
xf x  xf x 
3
d
dx

d
dx
d
dx
3
d
dx
x  x dxd f x  f x  dxd x 
6
d
dx 6
d
dx
d
dx
6
u  f x 
u'  f ' x 
x 3u  f ' x   f x   1 x f ' x   f x  1  0
2
3
 3x f 
x  f ' x    f x  xf ' x   f x  0

2
3

3x f x  f ' x  xf ' x  f x   f x 


2
3
f ' x  3x  f x   x   f x   f x 
2
3


f ' x  
 f x   f x 
3
3x  f x  x
2
Example 2 (continued)
If x  f x   xf x   6 and f 3 1 , find f ' 3.
3
f ' x  
 f x   f x 
3
3x  f x  x
2
Evaluate the derivative with the given information.


f ' 3 



 3
 f 3  f 3
33 f 3 3
2
3
1 1
2
91 3
11
93
2
12
1
6



Example 3
Find an equation of the tangent to the circle x 2  y 2  25
at the point 3,4 .
Find the derivative by differentiating both sides.
2
2
x

y


25
2
2
d
d
d
x

y





dx
dx
dx 25

d
dx
2
u
f
u


Chain
Rule
 f '  u   2u



d
dx
dy
dx  3,4 
uy
u' 
dy
dx
dy
dx

x
y

3
4
Use the Point-Slope
Formula to find the
equation of the tangent
line
dy
2x  2u dx  0
dy
 2x 
2y 
dx  0
dy


2y  
dx  2x
dy
2x

dx
2y
Now evaluate the
derivative at x=3 and y=4.

y 4 
3
4
x  3