2.5 Implicit Differentiation
Don' t
You can do it!!!
How would you find the derivative in the equation
x2 – 2y3 + 4y = 2
where it is very difficult to express y as a function
of x?
To do this, we use a procedure called implicit
differentiation.
This means that when we differentiate terms
involving x alone, we can differentiate as usual.
But when we differentiate terms involving y, we
must apply the Chain Rule.
Watch the examples very carefully!!!
Differentiate the following with respect to x.
3x2
6x
2y3
6y2 y’
x + 3y
1 + 3y’
xy2
x(2y)y’ + y2(1) = 2xyy’ + y2
Product rule
Find dy/dx given that
y3 + y2 – 5y – x2 = -4
dy
dy
dy
3y
 2 y  5  2x  0
dx
dx
dx
2
dy
dy
dy
3y
 2y 5
 2x
dx
dx
dx
dy
3y2  2 y  5  2x
dx
dy
2x

2
dx 3 y  2 y  5
2




Isolate dy/dx’s
Factor out dy/dx
What are the slopes at the following points?
(1,-3)
4
m
5
1
m
8
x=0
m0
(1,1)
undefined
(2,0)
Determine the slope of the tangent line to the
graph of x2 + 4y2 = 4 at the point 2, 1 2 .

dy
2x  8 y
0
dx
dy
8y
 2 x
dx
dy  2 x

dx 8 y

dy
 2
 2  2
m



dx
 1 
1
4
4

 2
2 1
m 
4 2
dy  x

dx 4 y
-2
-1
1

2
2, 1 2

Differentiate sin y = x
dy
cos y
1
dx
dy
1

dx cos y
Differentiate x sin y = y cos x
Product Rule
x cos y (y’) + sin y (1) = y (-sin x) + cos x (1)(y’)
x cos y (y’) - cos x (y’) = -sin y - y sin x
y’(x cos y - cos x) = -sin y - y sin x
 sin y  y sin x
y' 
x cos y  cos x
Given x2 + y2 = 25, find y”
2 x  2 yy'  0
 2x
x
y' 

2y
y
y (1)  xy'
y"  
y2
x

y  x
y 

y"  
2
y
x
Now replace y’ with 
y
Multiply top and bottom by y
 y
 
y
 
25
x y



3
3
y
y
2
What can we substitute in for x2 + y2?
2