Source: Unit 3 Calc AB Review, Texas A&M University
Tags: implicit differentiation, dy/dx, horizontal tangent, vertical tangent, inverse function theorem, derivative of inverse, AP Calculus AB
Implicit differentiation lets you find dy/dx when y is tangled up with x in an equation you cannot (or choose not to) solve for y explicitly. Derivatives of inverse functions use a clean formula that avoids finding the inverse altogether. Both techniques show up as multi-part exam problems, typically "find dy/dx, then find where tangents are horizontal or vertical."
Implicit differentiation
Differentiating both sides of an equation with respect to x, treating y as a function of x. Every time you differentiate a term containing y, you attach a factor of dy/dx (by the chain rule).
Explicit function
y is isolated on one side, e.g. y = x² + 3. You can differentiate directly.
Implicit equation / implicitly defined function
An equation relating x and y where y is not isolated, e.g. x² + xy + y² = 9. You differentiate "as is."
Horizontal tangent
A point on the curve where dy/dx = 0. The numerator of dy/dx equals zero (while the denominator does not).
Vertical tangent
A point on the curve where dy/dx is undefined because its denominator equals zero (while the numerator does not).
Inverse function theorem (for derivatives)
If f is differentiable and one-to-one, and f(a) = b, then (f⁻¹)'(b) = 1 / f'(a). In other words: the derivative of the inverse at a point is the reciprocal of the derivative of the original function at the corresponding point.
Start with the equation as given. Do not try to solve for y first.
Differentiate every term with respect to x:
Pure x terms: differentiate normally
Pure y terms: differentiate with respect to y, then multiply by dy/dx
Mixed terms (like xy): use the product rule, remembering that d/dx[y] = dy/dx
For ln(y): d/dx[ln(y)] = (1/y) · dy/dx
Collect all terms containing dy/dx on one side
Factor out dy/dx and solve
Once you have dy/dx as a fraction (numerator/denominator):
Horizontal tangent: set the numerator = 0, solve for x and y, then check that the point lies on the original curve
Vertical tangent: set the denominator = 0, solve for x and y, then check that the point lies on the original curve
You must substitute back into the original equation to confirm the points are valid. Solving the numerator or denominator alone may give you values that do not satisfy the curve.
If f and g are inverses (meaning g = f⁻¹), then:
g'(x) = 1 / f'(g(x))
Equivalently, (f⁻¹)'(b) = 1 / f'(a), where f(a) = b.
The steps for a typical problem:
You are given f(x) and told that f⁻¹(b) = a (or equivalently, f(a) = b)
Compute f'(x)
Evaluate f'(a)
The answer is 1 / f'(a)
When f and f' are given in a table and g = f⁻¹:
To find g'(b), first identify a such that f(a) = b (read the table)
Then g'(b) = 1 / f'(a)
Implicit differentiation (product rule on xy): d/dx [xy] = x · (dy/dx) + y · 1 = x(dy/dx) + y
Implicit differentiation (chain rule on y²): d/dx [y²] = 2y · (dy/dx)
Implicit differentiation (ln(y)): d/dx [ln(y)] = (1/y) · (dy/dx)
Inverse function derivative: (f⁻¹)'(b) = 1 / f'(f⁻¹(b)) = 1 / f'(a), where f(a) = b
⚠️ The most common implicit differentiation error is forgetting dy/dx on a y-term. Every single time you differentiate something involving y, the chain rule produces a dy/dx factor. Missing even one ruins the entire result.
⚠️ When using the product rule on xy, students often differentiate x and y separately and add them, forgetting that xy is a product. Write it out: d/dx[xy] = x·(dy/dx) + y·1.
⚠️ For horizontal/vertical tangent problems, always verify your candidate points lie on the original curve. The exam expects this.
⚠️ On inverse function derivative problems, the most frequent mistake is evaluating f' at the wrong point. You need f' at the x-value that maps to the given output, not at the output itself.
⚠️ The table-based inverse function problem is a near-certainty on exams. Read it carefully: g'(5) requires finding where f(x) = 5, not plugging 5 into f' directly.
Q: Given x² + xy + y² = 9, find dy/dx.
A: Differentiate each term: 2x + (x · dy/dx + y) + 2y · dy/dx = 0 2x + y + (x + 2y) · dy/dx = 0 dy/dx = −(2x + y) / (x + 2y)
Q: For x² + xy + y² = 9, find all points where the tangent is horizontal.
A: Horizontal tangent: numerator = 0, so 2x + y = 0, giving y = −2x. Substitute into x² + xy + y² = 9: x² + x(−2x) + (−2x)² = 9 x² − 2x² + 4x² = 9 3x² = 9, so x² = 3, x = ±√3.
Points: (√3, −2√3) and (−√3, 2√3).
Q: For x² + xy + y² = 9, find all points where the tangent is vertical.
A: Vertical tangent: denominator = 0, so x + 2y = 0, giving x = −2y. Substitute into (−2y)² + (−2y)y + y² = 9: 4y² − 2y² + y² = 9 3y² = 9, so y² = 3, y = ±√3.
Points: (−2√3, √3) and (2√3, −√3).
Q: Given x³ + y³ = 6xy, find dy/dx.
A: Differentiate: 3x² + 3y² · dy/dx = 6(x · dy/dx + y) 3x² + 3y² · dy/dx = 6x · dy/dx + 6y 3y² · dy/dx − 6x · dy/dx = 6y − 3x² dy/dx(3y² − 6x) = 6y − 3x² dy/dx = (6y − 3x²) / (3y² − 6x) = (2y − x²) / (y² − 2x)
Q: For x³ + y³ = 6xy, find where the tangent is horizontal.
A: Numerator = 0: 2y − x² = 0, so y = x²/2. Substitute into x³ + (x²/2)³ = 6x · (x²/2): x³ + x⁶/8 = 3x³ x⁶/8 = 2x³ x⁶ = 16x³ x³(x³ − 16) = 0 x = 0 or x = 2∛2.
At x = 0: y = 0 (but check: at the origin, the denominator y² − 2x also equals 0, so the derivative is indeterminate there). At x = 2∛2: y = (2∛2)²/2 = 2∛4.
Confirmed horizontal tangent at (2∛2, 2∛4).
Q: Given x²y + ln(y) = x + 3, find dy/dx.
A: Differentiate: 2xy + x² · dy/dx + (1/y) · dy/dx = 1 (x² + 1/y) · dy/dx = 1 − 2xy dy/dx = (1 − 2xy) / (x² + 1/y)
Multiply numerator and denominator by y: dy/dx = y(1 − 2xy) / (x²y + 1)
Q: For x²y + ln(y) = x + 3, find where the tangent is horizontal.
A: Numerator = 0: 1 − 2xy = 0, so y = 1/(2x). Substitute into x² · (1/(2x)) + ln(1/(2x)) = x + 3: x/2 + ln(1) − ln(2x) = x + 3 x/2 − ln(2x) = x + 3 −x/2 − ln(2x) = 3
This equation must be solved numerically (or left in this form on the exam). The key skill being tested is setting up the condition correctly.
Q: Let f(x) = x³ + 2x. Given that f⁻¹(5) = 1, find (f⁻¹)'(5).
A: f'(x) = 3x² + 2. Since f⁻¹(5) = 1, we need f'(1) = 3(1)² + 2 = 5. (f⁻¹)'(5) = 1/f'(1) = 1/5.
Q: Let f(x) = eˣ + x. Given that f⁻¹(2) = 0, compute (f⁻¹)'(2).
A: f'(x) = eˣ + 1. Since f⁻¹(2) = 0, we need f'(0) = e⁰ + 1 = 1 + 1 = 2. (f⁻¹)'(2) = 1/f'(0) = 1/2.
Q: The functions f and g are inverses. Using the table below, find g'(5).
x | 1 | 2 | 4 | 5 |
|---|---|---|---|---|
f(x) | 5 | 4 | 2 | 1 |
f'(x) | 3 | −2 | 6 | −4 |
A: g = f⁻¹, so g'(5) = 1/f'(g(5)). g(5) = f⁻¹(5). From the table, f(1) = 5, so f⁻¹(5) = 1. Therefore g'(5) = 1/f'(1) = 1/3.
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