In this section we differentiate equations that contain more than one variable on one side.

### Review of the chain rule

Implicit differentiation is really just an application of the chain rule. So recall:

Of particular use in this section is the following. If is a differentiable function of and if is a differentiable function, then

### Implicit differentiation

The functions we’ve been dealing with so far have been defined *explicitly* in terms of
the independent variable. For example: However, this is not always necessary or even
possible to do. Sometimes we choose to or we have to define a function *implicitly
*. In this case, the dependent variable is not stated explicitly in terms of
an independent variable. Some examples are: Your inclination might be
simply to solve each of these equations for and go merrily on your way.
However, this can be difficult or even impossible to do. Since we are often
faced with a problem of computing derivatives of such functions, we need
a method that will enable us to compute derivatives of implicitly defined
functions.

We’ll start with a basic example.

- (a)
- Find the slope of the line tangent to the circle at the point .
- (b)
- Find the slope of the line tangent to the circle at the point .
Notice that we had to differentiate twice, not to mention that we had to first solve for in terms of in order to compute these two slopes.

- (a)
- Compute .
- (b)
- Find the slope of the line tangent to the circle at .
- (c)
- Find the slope of the line tangent to the circle at .

The curve defined by the equation is not a graph of a function. If we solve for , we obtain two solutions: and . Therefore, we can say that any point on the curve lies on the graph of some function . Starting with we differentiate both sides of the equation with respect to to obtain Applying the sum rule we see Let’s examine each of these terms in turn. To start On the other hand, is somewhat different. Here we assume that for some function , defined on some open interval (this is true for all points ). Hence, by the chain rule

Putting this together, we are left with the equation At this point, we solve for . Write

Remark: Notice that the derivative is expressed in terms of both variables and . This should not come as a surprise. If you think about it, the function and its derivative are not determined solely by the value of . Recall what happens if .

The advantage of the expression for that we obtained above is that it can be used for computation of the slope of the tangent line at each of these two points, and at any other point on the curve, where defined.

So, for the second part of the problem, we simply plug and into the expression above, hence the slope of the tangent line at this point is . For the third part of the problem, we simply plug and into the expression above, hence the slope of the tangent line at this point is .

We can confirm our results by looking at the graph of the curve and our tangent line:

Let’s see another illustrative example:

- (a)
- Compute .
- (b)
- Find the slope of the line tangent to this curve at .

Considering the final term , we assume that , on some interval . Hence

Putting this all together we are left with the equation At this point, we solve for . Write

For the second part of the problem, we simply plug and into the formula above, hence the slope of the tangent line at is . We’ve included a plot for your viewing pleasure:

You might think that the step in which we solve for could sometimes be difficult. In
fact, *this never happens*. All occurrences arise from applying the chain rule, and
whenever the chain rule is used it deposits a single multiplied by some other
expression. Hence our expression is linear in , it will always be possible to group
the terms containing together and factor out the , just as in the previous
examples.

One more last example:

**not**. This means the variables have changed places! Not to worry, everything is exactly the same. We apply to both sides of the equation to get which gives us Distributing and multiplying by yields

Grouping terms, factoring, and dividing finally gives us

so, and now we see