Practice: Implicit differentiation
Self-check
Section titled “Self-check”Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.
1. State the implicit-differentiation procedure in steps.
Show answer
(1) Treat y as an unknown function of x. (2) Differentiate both sides of the relation with respect to x. (3) Attach a dy/dx to every y term (that is the chain rule). (4) Solve the resulting equation for dy/dx algebraically. You never have to solve the original relation for y.
2. Why is d/dx(y²) = 2y · dy/dx rather than 2y?
Show answer
Because y is secretly a function of x, so y² is really (y(x))², a composition. The chain rule gives the outer derivative 2·y(x) times the inner derivative y'(x) = dy/dx. The dy/dx is the chain-rule factor that rides along on every y term; forgetting it is the number-one implicit-differentiation error.
3. Why is implicit differentiation “not a new technique”?
Show answer
Because it is just the chain rule applied to a relationship. Every y term is a composition (some function of y(x)), so differentiating it produces a dy/dx. The x terms differentiate normally. Collecting the dy/dx terms and solving is plain algebra. If you can do the chain rule, you can do this.
4. What extra rule do you need for a mixed term like xy?
Show answer
The product rule: d/dx(xy) = (1)·y + x·(dy/dx) = y + x·dy/dx. A term with both variables multiplied is a product, so it gets the product rule, and the y factor still contributes its dy/dx. Treating xy as a single variable’s derivative loses a term.
5. What is “related rates,” and how does it differ from finding dy/dx?
Show answer
Related rates is the time-based twin: when x and y both change over time t under a constraint, you differentiate the constraint with respect to t, linking their rates dx/dt and dy/dt. It is the same implicit-differentiation move, but you differentiate against time instead of x, so you get linked speeds rather than a slope.
6. Why don’t you solve the relation for y before differentiating?
Show answer
Usually you cannot (a tilted ellipse like x² + xy + y² = 7 has no clean y = f(x) form), and you do not need to. Implicit differentiation finds the slope directly from the relation as it stands, which is exactly its payoff: it does not care whether the relation can be untangled.
Try it yourself, part 1: find dy/dx implicitly
Section titled “Try it yourself, part 1: find dy/dx implicitly”Pen and paper, about 7 minutes.
(a) The circle x² + y² = 169. Find dy/dx, then evaluate it at the point (5, 12) and check it is perpendicular to the radius there.
(b) The relation x·y² = 8 (which you would not want to solve for y). Find dy/dx. (Hint: the left side is a product.)
Show answer
(a) Differentiate both sides with respect to x:
2x + 2y·dy/dx = 0 -> dy/dx = -x/yAt (5, 12) (on the circle, since 25 + 144 = 169): dy/dx = -5/12. Check: the radius from the origin to (5, 12) has slope 12/5, and the tangent must be perpendicular to it, with slope the negative reciprocal -5/12. The implicit derivative matches the geometry.
(b) The left side x·y² is a product, so use the product rule (and the chain rule on y²):
d/dx(x·y²) = (1)·y² + x·(2y·dy/dx) = y² + 2xy·dy/dx = 0Solve: 2xy·dy/dx = -y², so dy/dx = -y²/(2xy) = -y/(2x). No need to untangle y from the relation.
Try it yourself, part 2: a related-rate
Section titled “Try it yourself, part 2: a related-rate”About 4 minutes. A spherical balloon is being inflated. Its volume and radius are tied by V = (4/3)π·r³, and both change with time. The radius is growing at dr/dt = 2 cm/s. How fast is the volume growing when the radius is r = 5 cm?
Show answer
Differentiate the constraint with respect to time t (the r³ term needs the chain rule, depositing a dr/dt):
dV/dt = (4/3)π · 3r² · dr/dt = 4π·r²·dr/dtPlug in r = 5 and dr/dt = 2:
dV/dt = 4π · (5)² · 2 = 4π · 25 · 2 = 200π ≈ 628 cm³/sThe volume is growing at 200π cubic cm per second at that instant. Notice the volume’s rate depends on the current radius (through the r² factor): the bigger the balloon, the faster its volume grows for the same dr/dt, because surface area grows with r². The constraint links the two rates, and differentiating it with respect to time is what extracts the link.
Flashcards
Section titled “Flashcards”Nine cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.
Q. What is the implicit-differentiation procedure?
Treat y as a function of x, differentiate both sides with respect to x, attach a dy/dx to every y term (chain rule), then solve algebraically for dy/dx. You never solve the original relation for y.
Q. Why is d/dx(y²) = 2y·dy/dx?
Because y is a function of x, so y² is the composition (y(x))². The chain rule gives 2·y(x) times the inner derivative dy/dx. The dy/dx factor rides on every y term; dropping it is the number-one error.
Q. Find dy/dx for the circle x² + y² = 25.
Differentiate: 2x + 2y·dy/dx = 0, so dy/dx = -x/y. At (3,4) that is -3/4, perpendicular to the radius (slope 4/3), exactly the tangent slope.
Q. Why is implicit differentiation just the chain rule?
Every y term is a composition (a function of y(x)), so differentiating it deposits a dy/dx. The x terms differentiate normally. Collecting dy/dx terms and solving is plain algebra. No new technique.
Q. How do you differentiate a mixed term like xy implicitly?
With the product rule: d/dx(xy) = y + x·dy/dx. The y factor still carries its dy/dx. Skipping the product rule on mixed terms loses a term.
Q. How does implicit differentiation produce the derivative of ln(x)?
From e^y = x (which says y = ln x): differentiate to get e^y·dy/dx = 1, so dy/dx = 1/e^y = 1/x (since e^y = x). The derivative of ln x is 1/x, with no special log rule needed.
Q. What is related rates?
The time twin of implicit differentiation: when x and y both change with time under a constraint, differentiate the constraint with respect to t to link dx/dt and dy/dt. Same move, differentiated against time instead of x.
Q. Sliding ladder x² + y² = 100: how are the rates linked?
Differentiate w.r.t. time: 2x·dx/dt + 2y·dy/dt = 0, so dy/dt = -(x/y)·dx/dt. If the base slides out at dx/dt = 1, the top falls faster as x grows: at (6,8), -0.75 ft/s; at (8,6), -1.33 ft/s.
Q. Where does implicit differentiation show up in machine learning?
In methods using constraints or fixed points rather than explicit formulas: constrained optimization (Lagrange/KKT), deep equilibrium models (a layer defined as a fixed point f(z,x)=z), and differentiating through the ODEs in score-based diffusion.