Implicit Differentiation
Department of Mathematics
University of Leicester
www.le.ac.uk
What is it?
• The normal function we work with is a function
of the form:
y  f (x)
• This is called an Explicit function.
Example:
y  2x  3
What is it?
• An Implicit function is a function of the form:
f ( x, y )  0
• Example:
y x y x 0
3
4
3
What is it?
• This also includes any function that can be
rearranged to this form: Example:
y  x  y  x  2y
3
4
3
• Can be rearranged to:
y  x  y  x  2y  0
3
4
3
• Which is an implicit function.
What is it?
• These functions are used for functions that
would be very complicated to rearrange to the
form y=f(x).
• And some that would be possible to rearrange
are far to complicated to differentiate.
Implicit differentiation
• To differentiate a function of y with respect to
x, we use the chain rule.
• If we have:
z  z ( y ),
and
y  y (x)
• Then using the chain rule we can see that:
dz dz dy


dx dy dx
Implicit differentiation
• From this we can take the fact that for a
function of y:
d
d
dy
( f ( x)) 
( f ( y )) 
dx
dy
dx
Implicit differentiation: example 1
• We can now use this to solve implicit
differentials
• Example:
y  x  y  x  2y
3
4
3
Implicit differentiation: example 1
• Then differentiate all the terms with respect to
x:
d
d
d
d 3
d
3
4
(y ) 
( x)  ( y ) 
(x ) 
(2 y )
dx
dx
dx
dx
dx
Implicit differentiation: example 1
• Terms that are functions of x are easy to
differentiate, but functions of y, you need to
use the chain rule.
Implicit differentiation: example 1
• Therefore:
d 3
d
dy
3
2 dy
(y ) 
(y )
 3y
dx
dy
dx
dx
d
d
dy
4
4
3 dy
(y ) 
(y )
 4y
dx
dy
dx
dx
• And:
d
d
dy
dy
(2 y ) 
(2 y ) 
2
dx
dy
dx
dx
Implicit differentiation: example 1
• Then we can apply this to the original function:
d
d
d
d 3
d
3
4
(y ) 
( x)  ( y ) 
(x ) 
(2 y )
dx
dx
dx
dx
dx
• Equals:
dy
dy
3 dy
2
3y
1 4y
 3x  2
dx
dx
dx
2
Implicit differentiation: example 1
dy
• Next, we can rearrange to collect the
terms:
dx
dy
dy
3 dy
2
3y
1 4y
 3x  2
dx
dx
dx
2
• Equals:
dy
3x  1  (4 y  3 y  2)
dx
2
3
2
Implicit differentiation: example 1
• Finally:
dy
3x  1  (4 y  3 y  2)
dx
2
3
2
• Equals:
dy
3x  1

3
2
dx 4 y  3 y  2
2
Implicit differentiation: example 2
• Example:
cos y  2 x  sin y  0
2
• Differentiate with respect to x
d
d
cos y  4 x  sin y  0
dx
dx
Implicit differentiation: example 2
• Then we can see, using the chain rule:
d
d
dy
dy
cos y 
(cos y ) 
  sin y
dx
dy
dx
dx
• And:
d
d
dy
dy
sin y 
(sin y ) 
 cos y
dx
dy
dx
dx
Implicit differentiation: example 2
• We can then apply these to the function:
d
d
cos y  4 x  sin y  0
dx
dx
• Equals:
dy
dy
 sin y
 4 x  cos y
0
dx
dx
Implicit differentiation: example 2
• Now collect the
dy
terms:
dx
dy
4 x  (sin y  cos y )
dx
• Which then equals:
dy
4x

dx sin y  cos y
Implicit differentiation: example 3
• Example:
x x y y 2
2
2
3
2
• Differentiate with respect to x
d 2 d 2 3 d 2
d
x 
x y 
y 
2
dx
dx
dx
dx
Implicit differentiation: example 3
• To solve:
d 2 3
x y
dx
• We also need to use the product rule.
d 2 3
2 d
3
3 d
x y x
(y )  y
(x2 )
dx
dx
dx
dy
3
 x .3 y
 y .2 x
dx
2
2
Implicit differentiation: example 3
• We can then apply these to the function:
d 2 d 2 3 d 2
d
x 
x y 
y 
2
dx
dx
dx
dx
• Equals:
dy
dy
3
2 x  x .3 y
 y .2 x  2 y
0
dx
dx
2
2
Implicit differentiation: example 3
• Now collect the
dy
terms:
dx
dy
 (2 y x  2 x)  (2 y  3x y )
dx
3
2
• Which then equals:
dy
2 y x  2x

2 2
dx
2 y  3x y
3
2
Implicit differentiation: example 4
• Example:
x  y  arcsin( y )  2 x
2
2
• Differentiate with respect to x
d 2 d
d
2
x  arcsin( y ) 
2x
dx
dx
dx
Implicit differentiation: example 4
• To solve:
d
2
arcsin( y )
dx
• We also need to use the chain rule, and the
fact that:
d
1
arcsin( x) 
2
dx
1 x
Implicit differentiation: example 4
• This is because:
y  arcsin( x) , so x  sin( y )
• Therefore:
dx
2
2
 cos y  1  sin y  1  x
dy
• As x=sin(y), Therefore:
dy d
1

arcsin( x) 
dx dx
1 x2
Implicit differentiation: example 4
• Using this, we can apply it to our equation :
d
2
arcsin( y )
dx
1
dy
1
dy
 2 y.

 2 y.
2 2
4 dx
dx
1 (y )
1 y
Implicit differentiation: example 4
• We can then apply these to the function:
d 2 d
d
2
x  arcsin( y ) 
2x
dx
dx
dx
• Equals:
2y
dy
2x 
2
4 dx
1 y
Implicit differentiation: example 4
• Now collect the
dy
terms:
dx
2y
dy
2  2x 
4 dx
1 y
• Which then equals:
dy (2  2 x)( 1  y )

dx
2y
4
Conclusion
• Explicit differentials are of the form:
y  f (x)
• Implicit differentials are of the form:
f ( x, y )  0
Conclusion
• To differentiate a function of y with respect to
x we can use the chain rule:
d
d
dy
( f ( x)) 
( f ( y )) 
dx
dy
dx