MHL Implicit Differentiation 1
Implicit Differentiation
How do you find the gradient functions for the following:
These are called implicit functions because neither x or y are given explicitly.
Give some examples of explicit functions
We cannot always express an implicit function explicitly so we need a new
technique to find the gradient function.
Differentiate with respect to x
Step 1
Differentiate every term with respect to x
How do you differentiate?
Step 2
Use the chain rule for any y terms
2 2
( ) = ( )×
d d dy
y y
dx dy dx
2 3
+ = 5 ,y x
sin =1,x y
2 2
1 1
+ = xy
x y
2 3
+ = 5y x
2 3
( )+ ( ) = (5)
d d d
y x
dx dx dx
2 3
( )+ ( ) = (5)
d d d
y x
dx dx dx
= 2
dy
y
dx
MHL Implicit Differentiation 2
This gives
Now change the subject:
Practice questions
Differentiate with respect to x:
Find the equation of the tangent to the curve x2 + y2 – xy = 7 at the point (2, –1).
Find the equation of the normal to the curve 2x2 – (y + 3)2 = 1 at the point (5, 4).
3 2
2 +7 5 = 8x y y x 
dy
dx
2
+ 32 0=
dy
xy
dx

Implicit differentiation

  • 1.
    MHL Implicit Differentiation1 Implicit Differentiation How do you find the gradient functions for the following: These are called implicit functions because neither x or y are given explicitly. Give some examples of explicit functions We cannot always express an implicit function explicitly so we need a new technique to find the gradient function. Differentiate with respect to x Step 1 Differentiate every term with respect to x How do you differentiate? Step 2 Use the chain rule for any y terms 2 2 ( ) = ( )× d d dy y y dx dy dx 2 3 + = 5 ,y x sin =1,x y 2 2 1 1 + = xy x y 2 3 + = 5y x 2 3 ( )+ ( ) = (5) d d d y x dx dx dx 2 3 ( )+ ( ) = (5) d d d y x dx dx dx = 2 dy y dx
  • 2.
    MHL Implicit Differentiation2 This gives Now change the subject: Practice questions Differentiate with respect to x: Find the equation of the tangent to the curve x2 + y2 – xy = 7 at the point (2, –1). Find the equation of the normal to the curve 2x2 – (y + 3)2 = 1 at the point (5, 4). 3 2 2 +7 5 = 8x y y x  dy dx 2 + 32 0= dy xy dx