5.12 Exploring Behaviors of Implicit Relations
Welcome to the last key topic of unit five! We know that a function's derivative can be used to understand the behavior of a function. This holds true for implicit functions too!
📣 This topic is a bit complicated, but it isn’t asked often on the AP Calculus AB exam.
🧠 Understanding Implicit Functions
An implicit function is a type of function that is not defined in the usual manner! Before this unit, you would see explicit functions that pass the vertical line test and take the form of:
In contrast, an implicit function is represented by an equation that has multiple variables on the same side of an equation. Although they look different and sometimes don’t pass the vertical line test, you can still determine the behavior of an implicit function using its derivatives.
Here’s an example of an implicit function:
Remember that implicit differentiation allows us to work with functions defined implicitly by taking d/dx of both sides of the equation to find , which is usually defined in terms of both x and y.
🥇 First Derivative Applications of Implicit Functions
🎯 Critical Points on Implicit Functions
If you need a refresher on critical points, click here! In summary, all you have to do is to take the first derivative and determine when it is 0 or undetermined to find potential critical points. These points represent potential minima/maxima.
🪜 Steps To Find Critical Points on Implicit Relations
- Differentiate both sides of the implicit equation with respect to x or take .
- Solve for . This will give you the derivative of the function in implicit form.
- Set to identify any potential critical points. In addition, determine where is undefined
- Make a conclusion about each critical point by determining how changes signs around that point.
🧠 Behavior of Implicit Functions Using First Derivative
Let’s quickly recap how to analyze a function, since the rules apply for an implicit function as well!
- If is positive, then the graph of is increasing.
- If is negative, then the graph of is decreasing.
- If is increasing or has a positive slope, then the graph of is concave up.
- If is decreasing or has a negative slope, then the graph of is concave down.
- If changes from positive to negative at , then has a relative maximum at .
- If changes from negative to positive at , then has a relative minimum at .
Remember that for implicit functions, relative minimums and relative maximums may not be as intuitive to find when compared with viewing explicit functions!
📣 To review any of these concepts, check out the following study guides:
- 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
- 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
🥈 Second Derivative Applications of Implicit Functions
After finding the first derivative implicitly, differentiate again with respect to x to find the second derivative. Be aware that the second derivative may involve , , and .
Once you have the second derivative, you can use the second derivative test to determine the concavity of the function and identify the points of inflection.
Here’s a quick recap regarding how to use the second derivative to determine the behavior of a function:
- If is positive, then the function is concave up.
- If is negative, then the function is concave down.
- A point of inflection is a point where changes concavity and . Make sure to check that is changing signs at this point by finding the value of just before and after the potential inflection point!
To review how to determine concavity, check out 5.6 Determining Concavity.
🧩 Behavior of Implicit Functions: Practice Problems
Now that we know how to interpret the behavior of implicit functions, we can do some practice!
Practice Problem 1
Consider the implicit equation . Find the critical points and determine their nature (local minimum, maximum, or neither).
When you do the implicit differentiation, you should get:
Solve for !
We now have the first derivative! Find the critical points by setting and solving for x.
The original equation was so if , and , so 2 potential critical points are and .
Also, don’t forget to set undefined and solve! This function is undefined when the denominator is 0, or when . Plug back in to find out the x values for these critical points.
First Derivative Test Chart
Now you can take your 4 critical points and make a conclusion about the behavior of the function around those points.
| The Point | Dy/Dx To the Left | Dy/Dx To The Right | Conclusion |
|---|---|---|---|
| (0,5) | + | - | Maxima |
| (0,-5) | - | + | Minima |
| (5,0) | + OR - | + OR - | Undefined |
| (-5,0) | + OR - | + OR - | Undefined |
Why are two of the critical points undefined as maxima or minima? Let’s look at a graph of the function to see!
As you can see here, below the critical point of , the function has a negative slope or negative derivative in the 3rd quadrant of the graph. Above the same critical point, in the 2nd quadrant, the function has a positive slope and derivative. The critical point has similar changes in the derivative sign from positive in the fourth quadrant to negative in the first quadrant.
However, despite this change in sign, these points are not conclusively maxima or minima under the definitions of Calculus AB. This is because to find maxima and minima, we need our critical points to either be endpoints of explicit functions or be in the “middle” or an implicit function (have valid points with x values greater and less than the critical point).
Here, there are no points to the left of (-5,0) and no points to the right of (5,0). Due to the function’s implicit nature, this makes it difficult to determine whether either point is a minimum or a maximum.
💡 Think of it this way! If you traveled the circle clockwise, (5,0) would be a minimum, and (-5,0) would be a maximum since traveling from the 1st quadrant to the 4th quadrant changes the slope of the circle from negative to positive, and traveling from the 3rd quadrant to the 2nd makes the slope go from positive to negative. However, if you travel counter-clockwise, (5,0) is now a maximum, and (-5,0) is now a minimum!
Phew! You made it through this question. 😎
Practice Problem 2
A ladder is leaning against a wall. The bottom of the ladder is sliding away from the wall at a constant rate of 2 ft/s. If the ladder is 10 feet long and its bottom is originally 6 feet away from the wall, how fast is the top of the ladder sliding down the wall when the bottom is 8 feet away from the wall?
First, let’s list out the given information!
- Horizontal rate of change of the bottom of the ladder = 2 ft/sec.
- Length of the ladder = 10 ft.
- Distance from the wall = 6 ft.
- What to Find: Vertical rate of change of top of the ladder when the distance from the wall is 8ft
Next, we need to make an implicit equation that represents the ladder. We will use x for the ladder’s horizontal distance from the wall and y for its vertical distance from the ground. Combining this with the Pythagorean Theorem, we get
Use implicit differentiation to get
Solve for to get
Don’t stop here! Instead of finding the change of y with respect to x, or , we instead want to find the change in y with respect to time, or .
To do this, we will be using the Chain Rule that states the following:
You can also look at this algebraically to see that the dx factor in the numerator and denominator cancel to give us .
In the problem, we are given the information that , throughout the entire ladder’s movement (constant rate).
Substitute this value into the chain rule, to find the final equation:
Now, we need to substitute x and y, from our target scenario, when the bottom of the ladder is 8 feet away from the wall. To find this, use the Pythagorean Theorem.
- x = 8 (distance from the wall)
- y = ? (height of the ladder)
- Solve for y:
Plugging the values of and into our earlier equation gives us
Therefore, the top of the ladder is sliding down the wall at a rate of when the bottom is 8 feet away from the wall.
Remember that the directions up and right are considered positive while directions down and left are considered negative. Since is negative, the ladder is sliding down and it’s vertical distance from the floor is increasing.
Pat yourself on the back for getting through another practice problem! 🥳
⭐️ Conclusion
To solve any implicit differentiation problem, follow these simple steps!
- List out any information given and write down your final goal.
- Draw a picture of your scenario or graph the equation given.
- Make an equation to represent the situation given. Some problems give this formula to you while others require you to use basic mathematical models such as the Pythagorean Theorem.
- Use implicit differentiation in your equation.
- Either find critical points by setting or use the chain rule with earlier information to find or .
- Check your critical points using minima/maxima tests or plugin earlier information into the equation given by the Chain Rule.
- Double-check your work, making sure any positive/negative signs make logical sense.
Now you’ll be able to work through problems where you have to use a function’s or implicit function’s derivative to understand the behaviors the function. Great job and see you in the next guide! 👋🏽
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| critical point | A point in the domain of a function where the derivative is zero or undefined, which are candidates for local and absolute extrema. |
| derivative | The instantaneous rate of change of a function at a specific point, representing the slope of the tangent line to the function at that point. |
| first derivative | The derivative of a function, denoted f', which describes the rate of change and indicates where a function is increasing or decreasing. |
| implicit differentiation | A technique for finding the derivative of a function defined implicitly by differentiating both sides of an equation with respect to the independent variable. |
| implicit relation | A relation defined by an equation in which the dependent variable is not explicitly solved in terms of the independent variable. |
| implicitly defined function | A function defined by an equation relating x and y, where y is not explicitly solved in terms of x. |
| second derivative | The derivative of the first derivative, denoted f'', which describes the concavity of a function and indicates where it is concave up or concave down. |
Frequently Asked Questions
How do I find critical points when I have an implicit function?
Take the implicit equation F(x,y)=0 and differentiate implicitly to get dy/dx in terms of x and y. Critical points are where dy/dx = 0 or dy/dx does not exist (FUN-4.D.1). Steps: 1. Implicitly differentiate: d/dx[F(x,y)]=F_x + F_y (dy/dx)=0 so dy/dx = −F_x / F_y (or another algebraic form). 2. Solve for points (x,y) on the curve where dy/dx = 0 (numerator = 0 while denominator ≠ 0). Those are horizontal tangents. 3. Solve where dy/dx is undefined (denominator = 0 while numerator ≠ 0): vertical tangents. 4. If both numerator and denominator = 0 at a point, it’s singular (possible cusp or crossing). Use higher derivatives, examine limits of dy/dx, or use the implicit function theorem to classify. 5. To test concavity or inflection, compute d^2y/dx^2 via implicit differentiation (it will involve x, y, and dy/dx) and evaluate at candidate points (FUN-4.E.2). This is exactly what the AP CED expects: find dy/dx in terms of x and y, locate points where it’s zero or fails, and justify behavior using first/second derivatives. For worked examples and practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and unit practice problems (https://library.fiveable.me/practice/ap-calculus).
What's the difference between finding critical points for regular functions vs implicit relations?
For an explicit function y = f(x) you find critical points by solving f′(x) = 0 or where f′(x) does not exist (but x must be in the domain). Those x-values are candidates for local extrema or inflection tests use f″(x). For an implicit relation given by F(x,y)=0 you do the same idea but with dy/dx found by implicit differentiation. A point (x,y) on the curve is “critical” when dy/dx = 0 (horizontal tangent) or dy/dx does not exist (often a vertical tangent when the denominator = 0 and numerator ≠ 0). Important extra cases: if both numerator and denominator of dy/dx are 0 at (x,y), the point is singular (cusp or other non-regular behavior) and you can’t conclude from dy/dx alone—you may need to analyze the curve locally (use higher derivatives or parametrize). Second derivatives for implicit curves will involve x, y, and dy/dx, so use them to justify concavity or inflection (CED FUN-4.D and FUN-4.E). The implicit function theorem tells you when you can locally solve for y(x). For AP prep, be ready to compute dy/dx in terms of x and y, set it to 0 or check DNE, and use second-derivative evidence to justify behavior. See the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and practice problems (https://library.fiveable.me/practice/ap-calculus).
When does the first derivative equal zero or not exist for implicit functions?
For an implicit relation F(x,y)=0, you get dy/dx by implicit differentiation; typically dy/dx = -F_x / F_y. So: - dy/dx = 0 when the numerator (−F_x) = 0 while the denominator F_y ≠ 0. That gives a horizontal tangent and is a critical point (FUN-4.D.1). - dy/dx does not exist when the denominator F_y = 0 while the numerator ≠ 0. That gives a vertical tangent (also a critical point). - If both numerator and denominator are 0 at the same point, you have a singular point: the tangent could be vertical, horizontal, slanted, or you might have a cusp or other non-smooth behavior. You must do higher-order analysis (solve for slopes, use limits or the second derivative via implicit differentiation) to classify it. Note second derivatives for implicit relations can involve x, y, and dy/dx (FUN-4.E.2). On the AP exam, treat any point where dy/dx = 0 or DNE as a candidate critical point, justify with algebra/derivatives, and use the second derivative or geometry to classify. For a quick review see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and try practice problems (https://library.fiveable.me/practice/ap-calculus).
Can someone explain step by step how to find where dy/dx = 0 in implicit differentiation?
Start with the implicit equation F(x,y)=0 (e.g. x^2 + y^2 = 25). Steps to find where dy/dx = 0 (horizontal tangents / critical points): 1. Differentiate both sides with respect to x, treating y as a function y(x) and using chain rule for y-terms. That gives an expression for dy/dx in terms of x and y. Example: differentiate x^2 + y^2 = 25 → 2x + 2y·(dy/dx) = 0. 2. Solve that equation algebraically for dy/dx. From the example, dy/dx = −x/y. 3. Set the expression for dy/dx equal to 0 and solve for (x,y) pairs that also satisfy the original implicit equation. For dy/dx = −x/y = 0, numerator must be 0 so x = 0. Plug x = 0 into the original relation to find y (here y = ±5). Those points are where the slope is zero. 4. Check domain issues: if dy/dx is undefined (denominator 0) those are vertical tangents or singular points—treat separately. 5. Use second derivative (implicit) or sign test of dy/dx nearby to classify as local max/min or inflection if the AP question asks (FUN-4.D and FUN-4.E). For more examples and AP-style practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and try practice problems (https://library.fiveable.me/practice/ap-calculus).
I'm confused about implicit relations - how do I know if a point is actually a critical point?
First check the point actually lies on the implicit curve. Then find dy/dx by implicit differentiation (you’ll get dy/dx as an expression in x and y). According to the CED (FUN-4.D.1), a point is a critical point of the implicitly defined function if either - dy/dx = 0 at that point (horizontal tangent), or - dy/dx does not exist at that point (DNE)—e.g., denominator = 0 giving a vertical tangent, or a true singular/cusp where the implicit function theorem fails. So: plug the point into your dy/dx expression. If the numerator = 0 while denominator ≠ 0, you have dy/dx = 0 (horizontal tangent). If the denominator = 0 (and numerator ≠ 0) you get DNE—a vertical tangent (also a critical point). If both numerator and denominator are 0, investigate further: the point may be singular or a cusp (use higher-order analysis or check partial derivatives—see FUN-4.E for second-derivative/concavity work). For classification use the second derivative (implicit form) or geometric evidence. Want more practice? See the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL), the Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5), and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).
How do I find the second derivative when I'm doing implicit differentiation?
Do the implicit steps twice: 1) Differentiate the equation with respect to x, treating y as y(x) and using the chain rule (every dy/dx appears when you differentiate y-terms). Solve that result algebraically for dy/dx. 2) Differentiate that dy/dx expression with respect to x to get d^2y/dx^2. Use product/quotient and chain rules again, and whenever you get a dy/dx inside, substitute the expression you found in step 1. Finally, solve algebraically for d^2y/dx^2 so it’s isolated. Quick example: x^2 + y^2 = 25. First: dy/dx = -x/y. Differentiate: d^2y/dx^2 = d/dx(-x/y) = [(-1·y) - (-x)·(dy/dx)]/y^2 = (-y + x·y')/y^2. Substitute y' = -x/y to get d^2y/dx^2 = -(x^2 + y^2)/y^3 = -25/y^3. Remember second derivatives for implicit relations can depend on x, y, and dy/dx and are used to test concavity and inflection points (CED FUN-4.E). For extra practice and examples see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and more problems (https://library.fiveable.me/practice/ap-calculus).
What's the formula for finding d²y/dx² when y is defined implicitly?
If an implicit relation is F(x,y) = 0, first differentiate w.r.t. x to get y' = dy/dx = -F_x / F_y (provided F_y ≠ 0). Differentiate y' again (treat y as y(x) and use product/chain rule). The general formula is y'' = d²y/dx² = - [ F_y (F_xx + F_xy·y') - F_x (F_yx + F_yy·y') ] / (F_y)², where subscripts denote partial derivatives (e.g., F_xx = ∂²F/∂x², F_xy = ∂²F/∂x∂y). Since F_xy = F_yx, you can simplify if needed. Note y' appears on the right, so y'' is usually expressed in terms of x, y, and y'. Why this matters for AP: second derivatives for implicit relations can involve x, y, and dy/dx and are used to test concavity and inflection points (FUN-4.E.2). For walkthroughs and practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and more practice problems at (https://library.fiveable.me/practice/ap-calculus).
When do I use the second derivative test on implicit functions?
Use the second-derivative test for an implicit relation exactly like you would for an explicit function—but only after you’ve found a critical point. Steps in short: - Find critical points: solve for dy/dx (by implicit differentiation) and locate points where dy/dx = 0 (horizontal tangent) or dy/dx does not exist (possible vertical tangent or singular point)—FUN-4.D.1. - If dy/dx = 0 at (x0,y0), compute d2y/dx2 (by differentiating implicitly; it may depend on x, y, and dy/dx). - If d2y/dx2(x0,y0) > 0, the curve is locally concave up → local minimum on that branch. - If d2y/dx2(x0,y0) < 0, concave down → local maximum. - If d2y/dx2 = 0, test fails—examine higher derivatives or use sign changes of dy/dx/graphical evidence. - If dy/dx is undefined at the point (vertical tangent or singular point), the usual second-derivative test doesn’t apply; instead analyze the curve’s behavior (look for sign changes in dy/dx on either side, use higher-order derivatives, or detect cusps/vertical tangents). This matches FUN-4.E (use derivatives to justify behavior). For examples and practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and work practice problems (https://library.fiveable.me/practice/ap-calculus).
Why does my second derivative have x, y, and dy/dx all mixed together?
That’s exactly what should happen. When you implicitly differentiate once you get dy/dx expressed in x and y. Differentiating again means you must differentiate that expression—and any y in it is a function of x, so you apply the chain/product/quotient rules and introduce more dy/dx terms. So the second derivative often ends up containing x, y, and dy/dx mixed together. What to do about it: - Solve the first-derivative formula for dy/dx (in terms of x and y). - When you compute d²y/dx², substitute that dy/dx expression wherever dy/dx appears. That eliminates the extra dy/dx and gives d²y/dx² in x and y only (when possible). - If substitution is messy or impossible at a singular point (vertical tangent, cusp, or where dy/dx DNE), treat the point separately—the CED notes second derivatives for implicit relations may include dy/dx and you may need the implicit function theorem or geometry/limits to conclude concavity or inflection (FUN-4.E). For practice and step-by-step examples, check the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and unit review (https://library.fiveable.me/ap-calculus/unit-5). For lots of problems, use Fiveable practice (https://library.fiveable.me/practice/ap-calculus).
How do I determine if an implicitly defined function is increasing or decreasing?
Take the implicit equation F(x,y)=0, differentiate both sides implicitly to get dy/dx expressed in x and y. Then: - Critical points: solve dy/dx = 0 (horizontal tangent) or where dy/dx does not exist (vertical tangent, cusp, or singular point)—those are your critical points per FUN-4.D.1. - Increasing vs decreasing: pick test points (x,y) on the curve (or parameter values) and evaluate the sign of dy/dx (in terms of x and y). If dy/dx > 0 the implicitly defined branch is increasing there; if dy/dx < 0 it’s decreasing (FUN-4.E.1). - Use intervals along the curve (or level-curve pieces) separated by critical points to determine where the sign of dy/dx changes—that gives you increasing/decreasing intervals (first-derivative test logic). - For concavity/inflection: compute d^2y/dx^2 via implicit differentiation (may involve x, y, and dy/dx) and check its sign (FUN-4.E.2). For AP exam work, show algebra for dy/dx, list critical points, test sign of dy/dx on each interval, and justify conclusions. See the Topic 5.12 study guide for examples (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL). For broader review, check Unit 5 (https://library.fiveable.me/ap-calculus/unit-5) and practice problems (https://library.fiveable.me/practice/ap-calculus).
I don't understand how to apply the derivative tests to implicit relations - can someone help?
Think of implicit relations like regular functions except dy/dx is solved in terms of x and y, then you apply the same derivative tests. Steps (quick): 1. Differentiate implicitly: if F(x,y)=0, find dy/dx by Fx + Fy·(dy/dx)=0 so dy/dx = -Fx/Fy (or whatever your equation gives). This is your first derivative in terms of x and y (CED: dy/dx in terms of x and y). 2. Critical points (FUN-4.D): solve where dy/dx = 0 (horizontal tangent) or where dy/dx does not exist (vertical tangent or singular point). Evaluate on the curve only (points (x,y) satisfying F(x,y)=0). 3. First-derivative test for local behavior (FUN-4.E.1): check sign of dy/dx on either side along the curve (you may parametrize locally or test nearby points on the level curve) to decide increasing/decreasing. 4. Concavity and inflection (FUN-4.E.2): compute d2y/dx2 by differentiating dy/dx implicitly; it may include x, y, and dy/dx. Use its sign to determine concavity or set =0 for possible inflection points (verify sign change). 5. Special cases: if both Fx and Fy = 0 at a point, the implicit function theorem fails—possible cusp or singular point; treat separately. For worked examples and AP-style practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and more Unit 5 review (https://library.fiveable.me/ap-calculus/unit-5). For lots of practice problems, go to (https://library.fiveable.me/practice/ap-calculus).
What's the step by step process for analyzing the behavior of an implicit function using derivatives?
1) Start with the implicit equation F(x,y)=0. Differentiate both sides with respect to x using implicit differentiation to get dy/dx expressed in x and y (and maybe dy/dx terms)—this is your slope formula. (CED: FUN-4.E) 2) Find critical points: solve dy/dx = 0 and identify where dy/dx does not exist. Points on the curve where either holds are critical (FUN-4.D.1). Check they lie on F(x,y)=0. 3) Classify tangents: if dy/dx = 0 at a point you have a horizontal tangent; if dy/dx is undefined because denominator = 0 but numerator ≠ 0, suspect a vertical tangent. If both numerator and denominator → 0, investigate further (possible cusp or singular point). 4) Use the second derivative: differentiate dy/dx implicitly to get d2y/dx2 in terms of x, y, and dy/dx. Evaluate at critical points to determine concavity (concave up/down) and locate inflection points (where concavity changes). Note second-derivative expressions may include y′ (FUN-4.E.2). 5) Confirm geometry: use the implicit function theorem or check nearby points numerically/graphically to verify behavior (level curve + gradient vector give tangent direction). For practice and AP-focused examples see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5). For extra problems try Fiveable practice sets (https://library.fiveable.me/practice/ap-calculus).
How do I find local max and min points when my function is given implicitly?
Quick recipe—steps you can follow every time: 1. Differentiate the implicit equation with respect to x to get dy/dx in terms of x and y (implicit differentiation). 2. Find candidate critical points: solve dy/dx = 0 and also find where dy/dx does not exist. Each solution (x,y) must satisfy the original implicit equation (so plug back in). (This matches FUN-4.D: critical points occur where the first derivative equals zero or fails to exist.) 3. Classify each candidate: - First-derivative test: examine sign of dy/dx on either side of the point along the curve (if dy/dx changes +→− it’s a local max, −→+ a local min). - Or compute the second derivative implicitly (d^2y/dx^2 will involve x, y, and dy/dx) and use its sign for concavity (FUN-4.E). 4. Watch special cases: vertical tangent (dy/dx undefined but x′ behavior), cusps or singular points may not be extrema. This is exactly what AP asks for in Topic 5.12—see the study guide for examples (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and more unit review (https://library.fiveable.me/ap-calculus/unit-5). Practice similar problems at (https://library.fiveable.me/practice/ap-calculus).
Can I use the same derivative applications on implicit functions as I do on regular functions?
Yes—almost everything you do with derivatives for explicit y = f(x) carries over to implicit relations, but you work with dy/dx expressed in x and y and watch for places it doesn’t exist. How to apply the same tests (CED terms): - Find dy/dx by implicit differentiation. Critical points occur where dy/dx = 0 (horizontal tangent) or dy/dx does not exist (vertical tangent, cusp, or singular point)—FUN-4.D.1. - Use dy/dx(x,y) to test increasing/decreasing: sign(dy/dx) tells you whether the implicit curve rises or falls. - Compute d2y/dx2 (it may involve x, y, and dy/dx) to test concavity and locate inflection points—FUN-4.E.2. - If denominator of dy/dx = 0 while numerator ≠ 0, you typically have a vertical tangent; if both = 0, investigate further (possible cusp or singular point). The CED explicitly says derivative applications extend to implicit relations (FUN-4.E.1). For examples and practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL), the Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5), and hundreds of practice problems (https://library.fiveable.me/practice/ap-calculus).
Why do I keep getting stuck when trying to solve dy/dx = 0 for implicit equations?
You’re getting stuck because dy/dx for an implicit relation is usually an expression in both x and y—you can’t just set a symbol equal to zero without also enforcing the original relation. Quick checklist that fixes most student errors: - Write dy/dx from implicit differentiation and simplify to a single fraction (or expression) in x and y. - For horizontal tangents set the numerator = 0 AND require those (x,y) satisfy the original implicit equation. Points that make numerator 0 but don’t lie on the curve are not solutions. - Also check where the denominator = 0—those are places dy/dx DNE (vertical tangents or singular points), so they’re critical too. - After you find candidates, use the second-derivative relation (it will involve x, y, and dy/dx) or the implicit function theorem to classify them (horizontal tangent, cusp, vertical tangent, inflection, etc.). - If algebra gets nasty, solve the system numerically or parametrize if possible. For step-by-step examples and AP-aligned practice, see the Topic 5.12 study guide (https://library.fiveable.me/ap-calculus/unit-5/exploring-behaviors-implicit-relations/study-guide/yRk3bNSDXIb7YidT78RL) and more practice problems (https://library.fiveable.me/practice/ap-calculus).