The Chain Rule
Applications of The Chain Rule
UBC Calculus Online Course Notes

Implicit Differentiation

Up to this point, we have been discussing graphs of functions $  y 
= f(x)  $ . In this case, it is explicitly clear how the dependent variable $  y  $ depends on the independent variable $  x 
 $ . However, sometimes we are faced with an implicit relationship between and .

For example, consider the equation for the circle of radius $  \sqrt{2}  $ : $  x^2 + 
y^2 = 2  $ . Here we can no longer write as a function of : for values of between $  -\sqrt{2}  $ and $  \sqrt{2}  $ , there are two values of on the circle. Nevertheless, when we look at the circle, we might expect that the notion of a tangent line and hence a derivative still make sense.

However, we can no longer expect the derivative $ 
\frac{dy}{dx}  $ to depend only on $  x  $ , but rather on the point $  (x,y)  $ on the circle in which we are interested. Implicit differentiation is a process which will clarify this for us.

Implicit Functions

In spite of the fact that the circle cannot be described as the graph of a function, we can describe various parts of the circle as the graphs of functions. For instance, the upper semi-circle is the graph of $  y = f(x) = \sqrt{2-x^2}  $ (this is an implicit function defined by the equation). For the time being, let's forget that we can explicitly solve for $  f(x)  $ . In terms of the original equation of the circle, we have

\[  x^2 + [f(x)]^2  = 2. 

Now we can differentiate this expression as a function of by applying the chain rule :

\frac{d}{dx} (x^2 + [f(x)]^2) & =  & \frac{d}{dx}(2) = 0 \\ \\ 
2x + 2f^\prime(x) f(x) & = & 0 \\ 
x + f^\prime(x) f(x) & = & 0 

Now in terms of , we have

x + y^\prime y & = & 0 \\ 
y^\prime & = & -\frac xy 

This final expression may appear dissatisfying to you since the derivative of is expressed in terms of both and . However, this is a very useful expression: if we know a point on the circle $  (x,y)  $ , then we know that the slope of the tangent line there is $  y^\prime = 
-\frac xy  $ . For instance, at the point $ 
(1,1)  $ , we know that $  y^\prime = -1  $ . This feels right since, from the picture above, we expect the slope of the tangent line to be -1 at that point.

Notice something else important: we did not need to know explicitly what $  f(x)  $ was. Instead, we only needed to know that there was such a function. Seen in this light, the computation we have done still holds for the bottom half of the circle which may be described as the graph of $  f(x) = -\sqrt{2-x^2} 
 $ . For instance, at the point $  (1,-1)  $ we compute that $  y^\prime = 1  $ which again agrees with our intuition.

Notice that the derivative is not defined when $  y = 0 
 $ . This again makes sense because, from the picture above, the tangent line becomes vertical and then has a slope which is not defined.

This process is known as implicit differentiation.

The power rule

Now let's have a look at another example: $  y = 
\sqrt{x} = x^{1/2}  $ . This is actually given to us in an explicit way---that is, we know exactly how depends on . Nevertheless, we have not yet seen how to differentiate this type of function.

However, we can rewrite this implicitly as $  y^2 = x  $ and implicitly differentiate as follows:

\frac{d}{dx} y^2 & = & \frac{d}{dx} x \\ \\ 
\frac{d}{dy} (y^2) \frac{dy}{dx} & = & 1 \\ \\ 
2y\frac{dy}{dx} & = & 1 \\ \\ 
\frac{dy}{dx} & = & \frac 12 \frac 1y = \frac 12 x^{-1/2} 

Notice that this result

\[  \frac{d}{dx} x^{1/2} = \frac 12 x^{-1/2} 

still looks like the power rule which we have seen for integer exponents. In fact, we can use implicit differentiation to verify the power rule for any fractional exponenent. This means that we can now write

\[  \frac{d}{dx} x^{r} = r x^{r-1} 

whenever $  r  $ is a rational number (i.e. a quotient of two integers).

Another example

Here we will consider the implicit relationship

\[  x^{2/3} + y^{2/3} = 2^{2/3} 

This is a famous curve called the astroid and it arises in an interesting way. Suppose that a ball of radius $  \frac 12 
 $ is rolling inside a ball of radius 2. The astroid is the curve traced out by a point on the inner circle. You can see this below by either dragging the red ball around or by starting the animation using the "Start" button.

We can use implicit differentiation to understand its tangent lines:

\frac{d}{dx}(x^{2/3} + y^{2/3}) & = & \frac{d}{dx} 2^{2/3}\\ \\ 
\frac 23 x^{-1/3} + \frac{d}{dy}(y^{2/3}) \frac{dy}{dx} & = & 0 \\ \\ 
\frac 23 x^{-1/3} + \frac 23 y^{-1/3} \frac{dy}{dx} & = & 0 \\ \\ 
x^{-1/3} + y^{-1/3} \frac{dy}{dx} & = & 0 

First of all, notice that if either $  x = 0  $ or $  y 
= 0 $ , then the derivative cannot be defined. You can see this by looking at the animation above. When the point on the inner circle comes to one of these points, it comes to a stop and changes its direction. At these points, there is no tangent line and hence the derivative cannot be defined.

At the point $  (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})  $ , however, we find that $  \frac{dy}{dx} = -1  $ .

For your consideration:

  1. Near which points on the circle of radius $  \sqrt{2}  $ can the circle be described as the graph of a function?

  2. Near which points on the curve $  y^2 = x^3 - 4x  $ can the curve be described as the graph of a function?

  3. What happens to the tangent line and the derivative in both of these examples when $  y = 0  $ ?

  4. Answer these questions for the curve $  y^2 = x^3  $ . (Hint: to draw a sketch of this curve, notice that $  y = \pm 
x^{3/2}  $ .)

The Implicit Function Theorem

In the examples we've seen above, we can use graphs of functions to describe various pieces of the curves. This means that we would like to know when a portion of the curve can be described as the graph of a function. The basic fact is this: if the derivative $ 
\frac{dy}{dx}  $ which we find through implicit differentiation is defined at a point $  (x_0,y_0)  $ , then the curve near this point may be described as the graph of a function. This fact is called the Implicit Function Theorem.

How do we use the Implicit Function Theorem? Basically, by forgetting about the issue all together. If we compute the derivative at a point through implicit differentiation and it is defined at this point, then the Implicit Function Theorem justifies the calculation we have just done and we need not worry about anything. If the derivative at this point is not defined, then we cannot guarantee that there is a tangent line at that point. Simply said:

If a point on a curve has a tangent line with finite slope, it can be calculated by implicit differentiation.