Implicit Differentiation
Implicit and Explicit functions
An explicit
function is an function expressed as y
= f(x) such as
y =
2x^{3} + 5
y is defined
implicitly
if both x and y
occur on the same side of the equation such as
x^{2} +
y^{2 } = 4
we can think of y as function of x
and write:
x^{2} +
y(x)^{2} = 4
Implicit
differentiation
To find dy/dx, we proceed as
follows:

Take d/dx of both sides of the
equation remembering to multiply by y' each
time you see a y term.

Solve for y'
Example
Find dy/dx implicitly for the circle
x^{2} +
y^{2} = 4
Solution

d/dx(x^{2}
+ y^{2}) = d/dx (4)
or
2x +
2yy' = 0

Solving for y, we get
2yy'
= 2x
y' = 2x/2y
y' = x/y
Example:
Find y' at (2,2) if
xy + x/y
= 5
Solution:

(xy)' + (x/y)' = (5)'
Using the product rule and the quotient rule we have
y  xy'
xy' + y +
= 0
y^{2}

Now plugging in x = 2
and y = 2,
2y' + 2 +
(2  2y')/4 = 0
Multiply both sides by 4
8y' + 8 + 2  2y' = 0
6y' = 10
y' = 5/3
Exercises:

Let
3x^{2}
 y^{3 }^{ }= 4x + y^{2
}Find dy/dx

Find dy/dx at
(1,1) if
x + y
= x^{3} + y^{3}

Find dy/dx if
x^{2}
+ 3xy + y^{2} = 1

Find y'' if
x^{2}
 y^{2} = 4
Application
Example
Suppose that the demand function for a boat shop is given by
p
= 0.01x^{3} + x + 10,000 Find the rate of change of x
with respect to p when x = 20. A
boat craftsman can think of this question as who fast will the number of boats
she will need to build change as the price is increased. Solving for x
in terms of p is nearly impossible. Instead,
we can differentiate implicitly.
1 = 0.03x^{2} x' + x' Now plug in 20
for x to get
1 = 0.03(20)^{2} x' + x'
1 = 12x' + x' = 11x'
x' = 1/11 The rate of change is 1/11
boats per dollar increase.
Back
to the calculus home page
Back to the math
department home page
email
Questions and Suggestions 