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AP Calculus AB & BC Intermediate

Finding the Second Derivative Implicitly

How to differentiate an implicitly defined equation twice, keep track of the first derivative, and evaluate the second derivative at a point without solving explicitly for y.

Why Find a Second Derivative Implicitly?

Some curves are much easier to describe with an equation involving both and than by solving explicitly for . A circle such as is a simple example.

Implicit differentiation lets you find the first derivative without solving for . If a problem asks for the second derivative, however, you must differentiate your first-derivative equation one more time.

The important idea is that is itself a function of . When you differentiate a term containing again, the chain rule appears automatically.

The Key Formula

Suppose an equation defines implicitly. After the first differentiation, you obtain an expression for . Differentiate that entire equation with respect to again:

A particularly important result occurs when the first derivative appears in the equation. Because is a function of , its derivative is:

That is why you will often see appear when finding the second derivative implicitly. You are differentiating using the chain rule.

A Step-by-Step Method

  1. Differentiate the original equation once with respect to , treating as a function of .
  2. Solve the first-derivative equation for , whenever practical. This makes the second differentiation much easier.
  3. Differentiate the first-derivative equation again. Remember that must be differentiated with the chain rule.
  4. Solve for . The second derivative may appear on both sides of the equation.
  5. Substitute the coordinates of the requested point, along with the corresponding value of , if needed.

Worked Examples

Example 1: A Circle

Find for the curve . Then find its value at the point .

Step 1: Differentiate once. Differentiating with respect to gives .
Step 2: Solve for the first derivative.
Step 3: Differentiate again. Starting from , use the quotient form and the product/chain rules:
Step 4: Simplify.
Step 5: Evaluate at the point. At , , so .
Result:
Example 2: A More Algebraic Curve

Find for .

Step 1: Differentiate once.
Step 2: Solve for .
Step 3: Differentiate again.
Step 4: Solve for the second derivative.
Result:
Example 3: Evaluating at a Specific Point

The curve passes through . Find the value of at that point.

Step 1: Differentiate once.
Step 2: Find the first derivative at the point. At , , so .
Step 3: Differentiate the first-derivative equation.
Step 4: Substitute the point and the known first derivative.
Result:
Example 4: The General Pattern

Suppose an implicit equation produces the first derivative

To find the second derivative, differentiate both sides. The term involving requires the chain rule.

Step 1:
Step 2: Apply the product rule to the numerator if necessary, and remember that .
Step 3: Collect all terms containing on one side and solve.

Why the Chain Rule Matters the Second Time

The first implicit differentiation is usually straightforward: every time you differentiate a term involving , a factor of appears.

The second differentiation is where students often lose track of what is happening. For example, if your first derivative contains , then differentiating that expression gives:

The extra factor of is not optional. It comes from the chain rule because is not an independent constant; it depends on .

Common Mistakes

Forgetting that y' must be differentiated

If your first derivative contains , do not treat it like a constant. Its derivative is . This is the single most important idea in second-order implicit differentiation.

Using the first derivative but forgetting to substitute it

After differentiating twice, your equation may contain both and . If the problem asks for a numerical value, you usually need the first derivative at the point before you can finish evaluating the second derivative.

Solving explicitly for y when you do not need to

You can sometimes solve an equation for , but doing so may create unnecessary square roots and multiple branches. Implicit differentiation is specifically designed to avoid that work.

Stopping after the first differentiation

Finding is only half the job when the question asks for . Read the requested derivative carefully before stopping.

Practice Problems

Question 1

Find for the curve .

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Question 2

Find for at the point .

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Question 3

The curve passes through . Find at that point.

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Question 4 (Combined Rules)

For , find in terms of , , and .

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