IMPLICIT DIFFERENTIATION
By now, it should be easy to take the derivative of an equation such as y  3x5  7 x . If you’re
given an equation such as y 2  3x5  7 x , you can still figure out the derivative by taking the
square root of both sides, which gives you y in terms of x. This is known as finding the
derivative explicitly. It’s messy, but it’s possible.
If you have to find the derivative of something like y 2  y  3x5  7 x , you don’t have an easy
way to get y in terms of x, so you can’t differentiate this equation using any of the techniques
you’ve learned so far. That’s because each of those previous techniques needs to be used on an
equation in which y is in terms of x. When you can’t isolate y in terms of x, (or if isolating y
makes taking the derivative a nightmare), it’s time to take the derivative implicitly.
Implicit differentiation is one of the simpler techniques you need to learn to do in calculus, but
for some reason it gives many students trouble. Suppose you have the equation y  x 2  x . This
means that the value of y is a function of the value of x. When we take the derivative,
dy
, we’re
dx
looking at the rate at which y changes as x changes. Thus, for y  x 2  x , when we write
dy
 2 x  1,
dx
what we are saying is that “the rate” at which y changes, with respect to how x changes, is
2x  1 .
Suppose now that we wanted to find the rate that y changes with respect to t (for time). Then we
would have:
y  x2  x
dy
dx
dx
 2x
1
.
dt
dt
dt
Here, we have to use the Chain Rule because we are taking the derivative with respect to t, which
differs from the variable used in the formula.
We could take the derivative with respect to r to find the rate in which y changes with respect to
r, say, the radius of some circle:
y  x2  x
dy
dx
dx
 2x
1
.
dr
dr
dr
We can find the derivative with respect to any variable. But if the variable is different than the
formula’s variable, we must use the chain rule.
Example 1
a)
d
3 x 3   9 x 2
dx  
(The variables agree. Use simple integration rules.)
b)
d
dy
3 y 3   9 y 2
dx
dx
(The variables disagree. Use the Chain Rule.)
c)
d
dy
 x  3y  1 3
dx
dx
d)
d
d
d
 xy 2   x
 y 2   y 2
 x
dx
dx
dx
dy 

2
 x2y
  y 1
dx


 2 xy
dy
 y2
dx
(Product Rule)
(Chain Rule)
(Simplify)
GUIDELINES FOR IMPLICIT DIFFERENTIATION
1. Differentiate both sides of the equation with respect to x.
dy
2. Collect all terms involving
on the left side of the equation and move all other terms
dx
to the right side of the equation.
dy
3. Factor
out of the left side of the equation.
dx
dy
4. Solve for
.
dx
Example 2
Find
dy
if y 2  y  3x5  7 x .
dx
(Problem presented at the beginning of the lesson.)
Example 3
Find
dy
if y3  4 y 2  x5  3x 4 .
dx
Example 4
Find
dy
if sin y 2  cos x 2  cos y 2  sin x 2 .
dx
Example 5
Find
dy
if 3x 2  5 xy 2  4 y3  8 .
dx
Example 6
Find the derivative of 3x 2  4 y 2  y  9 at the point  2,1 .
Example 7
Find the derivative of
2x  5 y2
  x at the point 1,1 .
4 y3  x2
Sometimes you’ll be asked to find the second derivative implicitly.
Example 8
Find
d2y
if y 2  2 y  4 x 2  2 x .
2
dx
Your Turn:
1.
Find
dy
if x 2  y 2  6 xy .
dx
2.
Find
dy
if x  cos y  xy .
dx
3.
Find the derivative with respect to t of x 2  y 2  z 2 .
4.
1
Find the derivative with respect to t of V   r 2 h .
3
5.
Find
d2y
if y 2  x 2  2 x .
dx 2
Homework. Use implicit differentiation to find the following derivatives.
1.
Find
dy
if x3  y 3  y .
dx
2.
Find
dy
if x 2  16 xy  y 2  1 .
dx
3.
Find
dy
x y
at the point  2,1 if
 3.
dx
x y
4.
Find
dy
if cos y  sin x  sin y  cos x .
dx
5.
Find
dy
if 16 x 2  16 xy  y 2  1 at 1,1 .
dx
6.
Find
1
1
dy
if x 2  y 2  2 y 2 at 1,1 .
dx
7.
Find
dy

  
if x sin y  y sin x 
at  ,  .
dx
2 2
4 4
8.
Find
d2y
if x 2  4 y 2  1 .
dx 2
9.
Find
d2y
if sin x  1  cos y .
dx 2
10.
Find
d2y
if x 2  4 x  2 y  2 .
2
dx
Answers:
1.
3x 2
1 3y2
2.
8y  x
y  8x
3.
1
2
4.

5.
8
7
6.
1
7
7.
1
8.

9.
sin x sin 2 y  cos y cos 2 x
sin 3 y
10.
1
sin x  cos x
sin y  cos y
1
16 y 3