In this explainer, we will learn how to use implicit differentiation to differentiate functions defined implicitly.

By this point in a calculus course, we are able to differentiate many types of functions where one variable is explicitly expressed in terms of another. For example,

Generally, they are functions of the type . However, some functions are not defined explicitly like this but are instead defined implicitly by a relation between and . One familiar example would be the equation of a circle which is implicitly defining a function. In this case, it is possible to solve for to get two explicit functions of ; namely,

However, this might not be possible in general. For example, consider the relation

It is not possible to find a simple closed-form expression explicitly defining in terms of . However, by using implicit differentiation, we are still able to find an expression for the derivative of with respect to even when we cannot rearrange a relation to get an explicit function. How will we be able to do this? We will need to apply the chain rule.

### Rule: Chain Rule

Given a function of variable , its derivative of with respect to the variable is given by

Here is the derivative of the function obtained using the rules of differentiation.

When the variable of the function we are differentiating is , then the derivative can be immediately computed using the rules of differentiation. However, if the variable of the function we are differentiating is , then the chain rule stated above tells us that the derivative involves an extra terms, .

We will demonstrate the method by example.

### Example 1: Implicit Differentiation

Consider the equation .

- Using implicit differentiation, find an expression for in terms of and .
- For the semicircle where , express explicitly in terms of ; then, differentiate this expression to get an expression for in terms of .

### Answer

**Part 1**

To implicitly differentiate, we need to differentiate both sides of the equation with respect to as follows:

Using the linearity of the derivative, we can rewrite this as

Using the power rule, we can differentiate the first term:

As for the second, since is a function of and we can consider to be a function of , we can use the chain rule to get

Substituting this back into equation (1), we have

We can now rearrange to make the subject. By subtracting from both sides, and dividing by two, we have

For all , we can rewrite this as

**Part 2**

We begin by rearranging the equation we have been given to make the subject. By subtracting from both sides of the equation, we have

Since we are only interested in the semicircle where , we can take the positive square root of both sides of the equation to get

Now we have an explicit formula for in terms of , we can differentiate using the familiar rules of derivatives. Notice that we have a composition of functions, so we will need to apply the chain rule. Setting , we have . We now differentiate to get

Applying the chain rule, we have

Notice that since , we could express this as , which is the same answer we got in the first part of the question.

The last example demonstrates a number of important points:

- Even when it is possible to rewrite a relation as an explicit function, it is often simpler to use implicit differentiation.
- When we differentiate implicitly, in general, we will get an expression for in terms of both and .

It is fairly common to use implicit differentiation to find the equation of a tangent to a curve defined implicitly. In the next couple of examples, we will demonstrate how we can use implicit differentiation to address problems of this type.

### Example 2: Finding Tangents to Implicitly Defined Curves

The equation describes a curve in the plane.

- Find the coordinates of two points on this curve, where .
- Determine the equation of the tangent at the points where and the -coordinate is positive.
- Find the coordinates of another point, if it exists, at which the tangent meets the curve.

### Answer

**Part 1**

To find the points on the curve where , we substitute this value into the equation and solve for . Hence,

Hence, , which implies and are the two -values where , which we can write as coordinates as and .

**Part 2**

We need to find the equation of the tangent to the curve at . To do this, we start by finding the derivative at this point because this will be equal to the slope of the tangent. Hence, differentiating implicitly, we have

Hence,

Since, for the point we are interested in, , we can rewrite this as

Substituting in and , we have

Therefore, the slope of the tangent at is . Hence, using the slopeβpoint form of the equation of a line, we can find the equation of the tangent as follows:

We can simplify this to

**Part 3**

To find the coordinates of another point where this tangent crosses the curve, we can substitute into the equation for the curve and try to solve for . We can then find the corresponding value of . Therefore,

Expanding the parentheses, we have

Hence,

We already know that the tangent passes through the point . Hence, is one of the roots of this equation. We can, therefore, factor this out. We can accomplish this via long division or by matching the coefficients of the terms. We start by writing

We can see that the first term in the parentheses must be to ensure we get the correct coefficient for the -term. Hence,

Similarly, we can see that the constant term in the parentheses must be . Therefore,

Finally, we need to find the middle term in the parentheses. We can do this by comparing the coefficients of the -term on both sides of the equation. On the right-hand side, we have a coefficient of , whereas on the left-hand side, we have a coefficient of . Therefore, and

It is often useful to check that this value gives us the correct coefficient for the final term on the right-hand side, which is . Therefore, considering the coefficient of on the left-hand side, we have as required.

We now have a quadratic equation we can solve. Using the quadratic formula or otherwise, we find this has solutions and . Therefore, the tangent intersects the curve again at the point where . Using the equation of the tangent, we can substitute in this value of to find the corresponding value of as follows:

Hence, the tangent at meets the curve again at the point .

Sometimes, we may be asked to find the equation of a tangent from its slope as opposed to the point of tangency. In the next example, we will tackle a problem of this type.

### Example 3: Finding Tangents with Given Slopes

Find, for , the tangent to that has slope , giving your equation in terms of .

### Answer

We have been given the slope of a tangent and would like to find the values of and at the point that has this slope. To do this, we begin by differentiating the equation to find an expression for the derivative . Using implicit differentiation,

Since the derivative is equal to the slope of the tangent, we can substitute in for as follows:

Simplifying, we get

Hence, .

In previous examples, we computed the first derivative and applied the derivative to find the slope of the tangent line at a point. Sometimes, we are interested in finding the second, or higher-order, derivative. To find the second derivative, , we first find the first derivative . Then, we can find the second derivative by differentiating the first derivative. Letβs consider an example where we find the second derivative from an implicit equation.

### Example 4: Second Derivatives Using Implicit Differentiation

Given that , determine by implicit differentiation.

### Answer

We begin by differentiating implicitly to find the first derivative as follows:

At this point, we can either differentiate again or first rearrange to make the subject. Generally, when we differentiate again, we will get an expression for in terms of . For this reason, it is often helpful to rearrange the equation first. By subtracting from both sides of the equation, we get

If , we can divide by to get

At this point, we can use the quotient rule, to differentiate again to get

We can now substitute in the expression for that we calculated earlier to get

Expressing the numerator as a single fraction, we have

This formula is valid for all and such that .

Letβs look at an example where we compute the third derivative, , from an implicit equation.

### Example 5: Higher-Order Derivatives Using Implicit Differentiation

Find , given that .

### Answer

We begin by differentiating implicitly to get

At this point, we can either differentiate again or first rearrange to make the subject. Generally, when we differentiate again, we will get an expression for in terms of . For this reason, it is often helpful to rearrange the equation first. Therefore, by subtracting from both sides of the equation, we have

Hence, when ,

We can now use the quotient rule, to differentiate again to get

Substituting in the expression for , we have

Since , we can rewrite this as

We can now differentiate again to get an expression for as follows:

Substituting into the equation yields

Letβs recap a few important concepts from this explainer.

### Key Points

- When given a function defined implicitly, we can differentiate each term of the equation using the following version of the chain rule for the terms including our dependent variable :
- Even when it is possible to rewrite a relation as an explicit function, it is often simpler to use implicit differentiation.
- When we differentiate implicitly, in general, we will get an expression for in terms of both and .
- We can find higher-order derivatives with implicit differentiation. In such cases, we will often need to substitute expressions for lower-order derivatives, and even the equation which defines the function implicitly, to simplify the expressions.