Implicit
Functions
Most functions are normally
written 
where the 
can be isolated. These are called explicit functions. Equations such
as 
are called implicit functions.
We differentiate implicit functions
using the Chain Rule.
Recall that 
and 
Example 1: Differentiate 
with respect to x.
Answer: 
Solution: Since we can isolate the y in this example, let’s differentiate the function first as we would an explicit function.
Now differentiate the function as an implicit function. Differentiate both sides of the equation with respect to x.
Example 2: Differentiate 
with respect to x.
Answer: 
Solution: Since we can isolate the y in this example, let’s first differentiate the function as an explicit function.
Now differentiate the function as an implicit function. Differentiate both sides of the equation with respect to x.
Now let’s a look at some examples where either
it is very difficult to isolate the y or you cannot isolate the y.
Example 3: Differentiate
with respect to x.
Answer: 
Solution: This
is the equation of a circle centered at the origin with a radius of 3. If we
isolate the y, we get the two functions 
. Which one do we differentiate?
If we decide to differentiate 
,
If we decide to differentiate 
,
We have two different answers. Which one we decide to use will depend on the specific problem.
If we decide to differentiate as an implicit function,
we would differentiate both sides of with respect to x.
When we differentiate the function as an implicit function, we don’t have to decide which derivative to use.
The slope of any tangent line
to the circle is 
.
Let’s check this piece of
information. Let 
, then 
. This means there are
two points on the circle where . One point, 
, is located on the upper half of the circle and the other point, 
, is located on the
lower half of the circle.
The equation of the line
through is 
. The equation of the
line through is 
.
Example 4: Differentiate 
with respect to x.
Answer: 
Solution: Since we can isolate the y in this example, let’s differentiate the function as an explicit function.
Now differentiate the function
as an implicit
function. Differentiate both
sides of with respect to x. Note that the first term, 
, is a product
You might say “Wait a minute, looks
different than 
.” In fact,
the two look different; in reality, they are equivalent. Let’s illustrate why this is true.
Recall that the initial function
could be written as 
. Substitute this
value of y in the derivative derived from implicit differentiation.
Example 5: Find 
from the equation of 
.
Answer: 
Solution: Take the derivative with respect to x of both sides of the equation.
Each of the three terms is a
product. Therefore, use the product rule
to differentiate each term. Recall that 
and can also be written as
.
We are solving for y’, therefore isolate y’.
Example 6: Find from the equation of 
.
Answer: 
Solution: Take the derivative with respect to x of both sides of the equation.
Example 7: Find if 
.
Answer: 
Solution: The overall function is a natural log function; therefore, we take the derivative of the natural log function first, then the sine function, then the power function.
You can write the answer in
other ways. One way is to write 
as 
.
Example 8: Find if 
.
Answer: 
Solution: Take the derivative with respect to x of both sides of the equation.
Let us find the derivative of each term.
The first term is 
.
When you take the derivative of
, you have to ask yourself what function is operating
overall: a power function?, a sine function?, or a product of functions? Since is a short cut way of
writing 
, we can easily see that the exponent 2 operates on 
. Therefore, the first
term is a form of a power function.
- Use the Power Rule First:
. - Now let’s now find the derivative of . Use the sine
function rule.
- The derivative of the first term can now be written
.
- Let’s now find the derivative of
. We use the
product function.
- The derivative
of the first term is
The second term is 
.
- The term
is a shortcut way
of writing 
. Since the
exponent 3 operates on 
, we use the Power Rule first.
.
- The derivative of the tangent function is 1 divided by the cosine function squared. Therefore,
- The derivative of the second term can now be written as
- Let’s now take care of
.
- The derivative
of the second term is
The third term is 
.
- Overall we have a cosine function.
- The derivative
of the third term is
Now let’s write the derivative of the entire problem.
We are not done yet because we
are solving for or . To do that we must
isolate . Let’s do it by first
simplifying the following:
Let’s gather the terms that
contain the factor 
to the left side of
the equation and gather the terms without the factor to the other side.
Example 9: Find 
or 
if 
and write the answer
in terms of x and y.
Answer: 
Solution: To find the second derivative, we first find the first derivative and then find the derivative of the first derivative.
The first derivative of is 
. To find the second
derivative, we take the derivative of .
Since the directions were to write
the answer in terms of x and y, we need to substitute 
for y’.
Example 10. Find
the equations of the tangent lines to the circle whose radius is 8 inches and
is centered at the point 
origin at 
Answer: 
and 
Solution: The
equation of the circle is 
. When , we can find the values of y using substitution.
The two points of tangency are 
. The equations of the
two tangent lines are
The slopes of the two tangents lines are the value of y’ at the two points of tangency.
At the point 
, the slope 
and the equation of
the tangent line is 
which is simplified to
At the point 
, the slope 
and the equation of
the tangent line is 
which is simplified to
Example 11: Find
the point of intersection of the tangent lines to the ellipse 
where the points of
tangency are located at 
.
Answer: The point of intersection is 
.
Solution: When
, we can find the values of y using substituting 3 in the
equation
The two points of tangency are 
. The equations of the
two tangent lines are
The slopes of the two tangents lines are the value of y’ at the two points of tangency.
At the point 
, the slope
and the equation of the tangent line is
which is simplified to
At the point 
, the slope
and the equation of the tangent line is
which is simplified to
- The two tangent lines intersect when
To find the y-ordinate of the point, substitute this value for x in one of the tangent lines.
The point of intersection is .