NapkinCalc

Calculus 1B — The Derivative

Implicit differentiation

continues from lesson 4 — values defined earlier in the course stay live here

ELI5: sometimes y is tangled up with x and you simply can't solve for it — like the circle x² + y² = 25. No problem: differentiate both sides as they are, remembering that y is secretly a function of x (so every y-term picks up a dy/dx), then solve for dy/dx. For the circle you get the tidy dy/dx = −x / y.

xpt=3x_{pt} = 3
ypt=4y_{pt} = 4 the point (3, 4) sits on the circle x² + y² = 25
slopeimp=−xptyptslope_{imp} = \frac{-x_{pt}}{y_{pt}} = −0.7500-0.7500 implicit slope dy/dx = −x/y = −3/4
✓ pass abs(xpt2+ypt2−25)<10−9\mathrm{abs}\left(x_{pt}^{2} + y_{pt}^{2} - 25\right) < 10^{-9} (3, 4) really is on the circle
✓ pass abs(slopeimp⋅ypt+xpt)<10−9\mathrm{abs}\left(slope_{imp} \cdot y_{pt} + x_{pt}\right) < 10^{-9} it satisfies the differentiated circle: x + y·y′ = 0

Real-world hook: the slope along a constraint curve — an economics indifference curve, a pressure–volume contour, a level line on a topographic map — is exactly an implicit derivative.

Try it yourself: the point (−4, 3) is also on x² + y² = 25. What is dy/dx there? (Use dy/dx = −x/y.)

xq=−4x_{q} = -4
yq=3y_{q} = 3 another point on the same circle
slopeq=43slope_{q} = \frac{4}{3} = 1.33331.3333 ✏️ Your turn: replace 0 with dy/dx at (−4, 3). The check verifies x + y·y′ = 0 — so it never tells you the number.
✓ pass abs(slopeq⋅yq+xq)<10−9\mathrm{abs}\left(slope_{q} \cdot y_{q} + x_{q}\right) < 10^{-9} green when your slope satisfies the differentiated circle

Where next: Calculus 1C — Applications of the Derivative puts all this to work: related rates, optimization, and Newton's method.