Source: AB Calculus Unit 3 Notes, Texas A&M University
Tags: implicit differentiation, dy/dx, chain rule, implicit function, inverse trig derivatives, arcsin, arccos, arctan, inverse function derivative formula, AP Calculus AB
Implicit differentiation lets you find dy/dx when y is tangled up with x in an equation you cannot easily solve for y. You differentiate both sides with respect to x, attach dy/dx to every y-term via the chain rule, then isolate dy/dx. The same chain-rule logic underpins the derivative formulas for inverse trig functions and for the inverse of any differentiable function.
Implicit function
A relationship between x and y given by an equation (such as x² + y² = 16) where y is not isolated on one side. Contrast with an explicit function like y = 3x + 1.
Implicit differentiation
The technique of differentiating both sides of an implicit equation with respect to x, treating y as a function of x and applying the chain rule wherever y appears.
dy/dx (Leibniz notation)
The derivative of y with respect to x. In implicit differentiation, this is the quantity you solve for algebraically after differentiating.
Inverse trigonometric functions (arcsin, arccos, arctan, etc.)
Functions that reverse the standard trig functions. Also written sin⁻¹ x, cos⁻¹ x, tan⁻¹ x. Their derivatives are formulaic and come up frequently on the AP exam.
Derivative of an inverse function
If g is the inverse of f, then g'(x) = 1 / f'(g(x)). This connects the slope of an inverse to the slope of the original function, evaluated at the swapped point.
Differentiate every term on both sides with respect to x.
Whenever you differentiate a term containing y, multiply by dy/dx (chain rule, because y depends on x).
Collect all dy/dx terms on one side; move everything else to the other side.
Factor out dy/dx if there is more than one such term.
Solve for dy/dx. The answer will typically contain both x and y, and that is fine.
A quick example of the chain rule in action: d/dx (y⁴) = 4y³ · dy/dx. The 4y³ comes from the power rule; the dy/dx is tacked on because y is a function of x.
When x and y are multiplied together, you need the product rule as well. For instance, d/dx (x²y³) = 2x · y³ + x² · 3y² · dy/dx. The first piece differentiates x² while holding y³; the second differentiates y³ (chain rule included) while holding x².
Once you have dy/dx as a formula in x and y, plug in a specific point (x₀, y₀) that lies on the curve to get the slope there. Then use point-slope form: y − y₀ = m(x − x₀).
A tangent line is horizontal when dy/dx = 0, i.e. the numerator of the dy/dx expression equals zero (while the denominator is non-zero).
A tangent line is vertical when dy/dx is undefined, i.e. the denominator equals zero (while the numerator is non-zero).
Differentiate dy/dx again with respect to x. You will usually need the quotient rule, and you will encounter dy/dx inside the result. Substitute the first-derivative expression back in and simplify. This comes up in concavity questions and on the AP exam.
These six formulas are worth memorising outright. The chain-rule versions simply multiply by du/dx (or u') at the end.
Function | Derivative |
|---|---|
sin⁻¹ x | 1 / √(1 − x²) |
cos⁻¹ x | −1 / √(1 − x²) |
tan⁻¹ x | 1 / (1 + x²) |
cot⁻¹ x | −1 / (1 + x²) |
sec⁻¹ x | 1 / (|x|√(x² − 1)) |
csc⁻¹ x | −1 / (|x|√(x² − 1)) |
Notice the pattern: sin/cos share the √(1 − x²) denominator, tan/cot share (1 + x²), and sec/csc share |x|√(x² − 1). The "co-" version is always the negative of its partner.
When a composite function is inside (e.g. sin⁻¹(t²)), apply the chain rule: d/dt [sin⁻¹(t²)] = 1/√(1 − t⁴) · 2t.
sin⁻¹ x and arcsin x are the same thing. Domains are restricted so that each inverse trig function is properly a function.
If f and g are inverses, then f(g(x)) = x. Differentiating both sides with the chain rule gives f'(g(x)) · g'(x) = 1, so:
g'(x) = 1 / f'(g(x))
In words: to find the derivative of the inverse at a point, take 1 over the derivative of the original function, evaluated at the corresponding point.
Worked pattern (Example 5 from the notes): Let f(x) = x³ + 2x − 1. Find (f⁻¹)'(2).
You need the x-value where f(x) = 2. Given that f(1) = 2, so f⁻¹(2) = 1.
f'(x) = 3x² + 2, so f'(1) = 5.
Therefore (f⁻¹)'(2) = 1/5.
Implicit differentiation algorithm (boxed in the original notes)
Differentiate both sides with respect to x (y is a function of x, so use the chain rule).
Collect all dy/dx terms on the left; move everything else to the right.
Factor dy/dx out of the left side if there is more than one dy/dx term.
Solve for dy/dx. (It is fine to have both x's and y's in your answer.)
To find dy/dx at a specific point, you can alternatively plug in the point immediately after differentiating, then solve. This often saves algebra.
Inverse function derivative formula
g'(x) = 1 / f'(g(x))
⚠️ Forgetting the dy/dx factor when differentiating y-terms is the single most common mistake in implicit differentiation. Every y-term needs it.
⚠️ When finding tangent lines from implicit equations, confirm the given point actually satisfies the original equation before plugging in.
⚠️ The AP exam regularly asks "show that dy/dx = [given expression]." Work through the algebra step by step and match their form exactly.
⚠️ For inverse function derivatives, you must identify the correct corresponding point. If the question asks for (f⁻¹)'(a), find b such that f(b) = a, then compute 1/f'(b).
⚠️ The inverse trig derivative formulas appear on the AP formula sheet for BC but not AB. AB students should have them memorised.
Q: Find dy/dx for the equation x³ + y² − 3xy = 0.
A: Differentiate term by term: 3x² + 2y(dy/dx) − [3y + 3x(dy/dx)] = 0. Collect: (2y − 3x)(dy/dx) = 3y − 3x². So dy/dx = (3y − 3x²) / (2y − 3x).
Q: For the curve x² + y² = 4, at what points is the tangent line horizontal?
A: dy/dx = −x/y. Horizontal means dy/dx = 0, so x = 0. Substituting into x² + y² = 4 gives y = ±2. The points are (0, 2) and (0, −2).
Q: For the curve x² + y² = 4, at what points is the tangent line vertical?
A: Vertical means the denominator y = 0. Then x² = 4, so x = ±2. The points are (2, 0) and (−2, 0).
Q: Let f(x) = x⁵ + 2x − 1 and let g be the inverse of f. Given g(0) = −1, find g'(0).
A: f'(x) = 5x⁴ + 2. Since g(0) = −1, we need f'(−1) = 5(1) + 2 = 7. So g'(0) = 1/7.
Q: Find d/dx [tan⁻¹(√(x − 1))].
A: Using the chain rule with the tan⁻¹ derivative: [1 / (1 + (√(x−1))²)] · [1/(2√(x−1))] = 1 / [x · 2√(x−1)] = 1 / (2x√(x−1)).
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