Implicit differentiation, what's going on here? | Chapter 6, Essence of calculus
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Implicit differentiation finds tangent slopes for curves given by equations relating x and y, even when y is not explicitly a function of x.
Briefing
A calculus “weirdness” becomes manageable once tiny changes in two variables are given a geometric meaning: implicit differentiation is really about tracking how an expression like x² + y² changes under an infinitesimal move (dx, dy), then enforcing that the move stays on the curve—equivalently, that the change in the defining quantity is zero. That perspective turns the slope of a tangent line to an implicit curve from a mysterious algebra trick into a direct consequence of how derivatives approximate change.
The starting point is a circle of radius 5 centered at the origin, described by x² + y² = 5². To find the slope of the tangent at the point (3, 4), the usual “differentiate y as a function of x” approach fails because the circle is not given as y = f(x). Instead, x and y are linked by an equation, so the method differentiates both sides with respect to an infinitesimal change. Differentiating x² + y² = 25 produces 2x·dx + 2y·dy = 0 (since the right side is constant). Solving for dy/dx yields dy/dx = −x/y, and at (3, 4) the tangent slope becomes −3/4.
Why does the derivative produce terms like dx and dy “floating free”? The transcript resolves this by comparing the circle problem to a related-rates ladder problem, where the same algebraic structure appears but with a clear common parameter: time. For a 5-meter ladder, x(t)² + y(t)² = 25 always holds. Differentiating both sides with respect to time gives 2x·dx/dt + 2y·dy/dt = 0, and with dy/dt = −1 at the initial moment (y = 4, x = 3), the bottom’s speed comes out to dx/dt = 4/3. The circle case is analogous, except the “common variable” is not time; the common constraint is the curve equation itself.
To make that constraint concrete, the circle expression s = x² + y² is treated as a quantity assigned to every point (x, y). A tiny step (dx, dy) changes s by approximately ds = 2x·dx + 2y·dy. Staying on the circle means s must remain 25, so ds = 0. For sufficiently small steps, the condition ds = 0 aligns with moving along the tangent direction, which is why the same algebra yields the tangent slope.
The same machinery generalizes. For an implicit relation like sin(x)·y² = x, differentiating both sides uses the product rule and chain rule to relate the changes in each side during a tiny move. Often the goal is again to isolate dy/dx.
Finally, the method recovers a classic derivative: ln(x). By rewriting y = ln(x) as e^y = x, differentiating both sides under an infinitesimal change gives e^y·dy = dx. Rearranging yields dy/dx = 1/e^y, and since e^y = x on the curve, the slope becomes 1/x—confirming d/dx ln(x) = 1/x. The broader takeaway is that implicit differentiation is a disciplined way to translate “tiny nudges” into slope information while respecting the equation that ties x and y together.
Cornell Notes
Implicit differentiation works by treating the curve’s defining equation as a constraint on how x and y can change together. For the circle x² + y² = 25, a tiny move (dx, dy) changes the quantity x² + y² by about ds = 2x·dx + 2y·dy. Points on the circle keep x² + y² constant, so ds = 0, giving 2x·dx + 2y·dy = 0 and therefore dy/dx = −x/y. This viewpoint is clarified by comparing to related rates, where the same differentiation happens but with time as the shared variable. The same approach extends to other implicit equations, including deriving d/dx ln(x) = 1/x by rewriting y = ln(x) as e^y = x.
Why can’t the circle problem be solved by differentiating y as a function of x?
What does the expression 2x·dx + 2y·dy = 0 mean geometrically?
How does the ladder related-rates problem make the circle method feel less mysterious?
How do you use implicit differentiation on sin(x)·y² = x?
How does the method derive the derivative of ln(x)?
Review Questions
- For the circle x² + y² = 25, derive dy/dx at a general point (x, y) on the circle and then evaluate it at (3, 4).
- Explain why setting ds = 0 corresponds to staying on the circle (or more precisely, to the tangent direction for infinitesimal steps).
- In the relation sin(x)·y² = x, what rules (product rule, chain rule) are needed to differentiate the left-hand side, and what form does the resulting dy/dx equation take?
Key Points
- 1
Implicit differentiation finds tangent slopes for curves given by equations relating x and y, even when y is not explicitly a function of x.
- 2
For a constraint like x² + y² = 25, differentiating produces 2x·dx + 2y·dy = 0 because the right side is constant.
- 3
Interpreting s = x² + y² as a quantity assigned to points (x, y) makes dx and dy meaningful: ds ≈ 2x·dx + 2y·dy.
- 4
Staying on the curve requires ds = 0, which yields the tangent slope by solving for dy/dx.
- 5
Related-rates problems use time as the shared variable; the circle problem uses the curve constraint instead, but the differentiation logic is parallel.
- 6
The same approach generalizes to more complex implicit equations using the product rule and chain rule.
- 7
Rewriting ln(x) as e^y = x and differentiating both sides recovers d/dx ln(x) = 1/x.