Difficulty: Intermediate | Prerequisites: Partial derivatives, gradient, chain rule (MATH 215 §14.1–14.4)
Big picture
This material ties together two threads from earlier in the course: how to differentiate implicitly when variables are tangled inside one equation, and how to use the gradient to write tangent planes to surfaces. If you are comfortable with partial derivatives and the chain rule for multivariable functions, you have what you need. These tools show up constantly in optimisation (the next unit) and in any applied setting where you cannot solve for one variable explicitly.
TL;DR
When F(x, y, z) = 0 defines variables implicitly, you can extract each partial derivative using the ratio of F's own partials, and the cyclic product of those partials always equals −1. Tangent planes to a surface F = k are built from the gradient, and their intercepts encode geometric information about the surface.
Implicit differentiation (multivariable)
Finding a partial derivative of one variable with respect to another when all variables are bound together by an equation F(x, y, z) = 0, rather than by an explicit formula y = g(x).
In simple terms, this means you never have to solve the equation for one variable before differentiating.
Gradient vector (∇F)
The vector of all first-order partial derivatives of F: ∇F = (F_x, F_y, F_z). It points in the direction of steepest increase of F.
Think of it as the arrow that is always perpendicular to the level surface F = constant at any point.
Tangent plane
The plane that just touches a surface at a given point and shares the surface's slope there in every direction.
In simple terms, zoom in far enough on a smooth surface and it looks flat; that flat sheet is the tangent plane.
Normal vector to a surface
A vector perpendicular to the tangent plane at a point. For a surface defined by F(x, y, z) = k, the gradient ∇F at that point serves as a normal vector.
Cyclic chain rule identity
The result that, for three implicitly related variables, (∂z/∂x) · (∂x/∂y) · (∂y/∂z) = −1. The minus sign is the surprise; students often expect +1.
Think of it as the round-trip rule: go from z to x to y and back to z, and the product of the three partial derivatives picks up exactly one sign flip.
Setup: one equation, three unknowns
Suppose F(x, y, z) = 0 with F differentiable and all three partial derivatives F_x, F_y, F_z nonzero. Then each variable can, in principle, be written as a function of the other two:
z = f(x, y)
y = g(x, z)
x = h(y, z)
Extracting each partial derivative
Differentiate F(x, y, z) = 0 implicitly with respect to x, holding y fixed. Because z = f(x, y):
F_x + F_z (∂z/∂x) = 0, so ∂z/∂x = –F_x / F_z
Similarly, differentiate with respect to y holding z fixed:
∂x/∂y = –F_y / F_x
And with respect to z holding x fixed:
∂y/∂z = –F_z / F_y
The cyclic identity
Multiply the three results:
(∂z/∂x)(∂x/∂y)(∂y/∂z) = (–F_x/F_z)(–F_y/F_x)(–F_z/F_y) = (−1)^3 · 1 = −1
Each F-ratio cancels telescopically, and the three minus signs produce −1. This is the result to show in Q1(a).
General tangent-plane formula
For a surface given by F(x, y, z) = 0, the tangent plane at a point (x₀, y₀, z₀) is:
F_x(x₀, y₀, z₀)(x – x₀) + F_y(x₀, y₀, z₀)(y – y₀) + F_z(x₀, y₀, z₀)(z – z₀) = 0
This is the dot product of ∇F at the point with the displacement vector (x – x₀, y – y₀, z – z₀), set equal to zero.
Q1(b): Every tangent plane to x² + y² = z² passes through the origin
Define F(x, y, z) = x² + y² – z². Then:
F_x = 2x, F_y = 2y, F_z = –2z
The tangent plane at (x₀, y₀, z₀) on the cone is:
2x₀(x – x₀) + 2y₀(y – y₀) – 2z₀(z – z₀) = 0
Expand:
2x₀ x + 2y₀ y – 2z₀ z – 2(x₀² + y₀² – z₀²) = 0
Because (x₀, y₀, z₀) lies on the cone, x₀² + y₀² – z₀² = 0, so the constant term vanishes:
2x₀ x + 2y₀ y – 2z₀ z = 0, i.e. x₀ x + y₀ y – z₀ z = 0
Plugging in (x, y, z) = (0, 0, 0) gives 0 = 0, confirming the origin lies on the plane. This works because the cone is a level surface of a homogeneous-degree-2 function, so the constant term in the tangent-plane equation always cancels.
Q1(c): Sum of intercepts for √x + √y + √z = √c
Define F(x, y, z) = √x + √y + √z – √c. Compute the partials:
F_x = 1/(2√x), F_y = 1/(2√y), F_z = 1/(2√z)
The tangent plane at a point (a, b, d) on the surface is:
(x – a)/(2√a) + (y – b)/(2√b) + (z – d)/(2√d) = 0
Finding the intercepts
Set y = z = 0 to find the x-intercept. After simplification (using the constraint √a + √b + √d = √c):
x-intercept = a + 2√a(√b + √d) = a + 2√a(√c – √a) = (√a + (√c – √a))² is not quite the clean route. A cleaner approach:
Multiply the tangent-plane equation through by 2 and rearrange to the intercept form x/A + y/B + z/C = 1, where A, B, C are the intercepts.
The x-intercept (set y = z = 0):
x_int = a + 2√a · (√b + √d) · (using the surface equation, √b + √d = √c – √a)
So x_int = (√a + (√c – √a))²? No. Let us be more careful.
Rewrite the tangent plane as:
x/(2√a) + y/(2√b) + z/(2√d) = a/(2√a) + b/(2√b) + d/(2√d) = √a/2 + √b/2 + √d/2 = √c/2
So the equation is x/(2√a) + y/(2√b) + z/(2√d) = √c/2.
Divide both sides by √c/2:
x/(√(ac)) + y/(√(bc)) + z/(√(dc)) = 1
The three intercepts are √(ac), √(bc), √(dc). Their sum:
√(ac) + √(bc) + √(dc) = √c(√a + √b + √d) = √c · √c = c
The sum of intercepts is c, a constant, regardless of the point of tangency.
\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}\frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} = -1\text{Tangent plane at } (x_0, y_0, z_0): \quad F_x(x-x_0) + F_y(y-y_0) + F_z(z-z_0) = 0\text{For } \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{c}: \quad \text{sum of intercepts} = cImplicit differentiation of multivariable equations is the backbone of thermodynamics, where state variables (pressure, volume, temperature) are linked by equations of state you can rarely solve explicitly. The cyclic relation (∂P/∂V)(∂V/∂T)(∂T/∂P) = −1 appears in every physical chemistry textbook.
Tangent planes show up whenever you need to linearise a surface locally: in computer graphics for shading (the normal vector determines how light bounces off a surface), and in engineering whenever a constraint surface is approximated by a flat patch near an operating point.
Students often assume (∂z/∂x)(∂x/∂y)(∂y/∂z) = +1 by analogy with cancelling fractions. It does not. Partial derivatives are not fractions, and the cyclic product is −1.
When computing the tangent plane to the cone, students sometimes forget that the point (x₀, y₀, z₀) lies on the cone and fail to substitute x₀² + y₀² = z₀² to cancel the constant term. Without that substitution, the proof does not close.
Students confuse ∂z/∂x (holding y fixed, z depends on x) with ∂x/∂z (holding y fixed, x depends on z). These are not reciprocals in general when a third variable is held fixed.
In Q1(c), a common error is trying to find intercepts before simplifying the tangent-plane equation into intercept form. The clean route is to rewrite the right-hand side using the surface equation first.
⚠️ The cyclic chain rule identity is a standard exam question. Be ready to derive it from scratch, not just state it.
⚠️ "Show that every tangent plane passes through the origin" is a classic proof-style question. The key move is recognising that the constant term vanishes because the point satisfies the surface equation.
⚠️ For intercept-sum problems, know how to write the tangent plane in intercept form x/A + y/B + z/C = 1 and then read off A, B, C. Substituting the surface constraint to simplify is the step most students miss under time pressure.
True or False: (∂z/∂x)(∂x/∂z) = 1 when both partials come from F(x, y, z) = 0 with y held fixed. (Answer: True in this case, because you are holding the same variable fixed. The cyclic identity involves three different "held fixed" variables.)
Fill in the blank: The gradient ∇F at a point on the surface F = 0 is ______ to the tangent plane there. (Answer: perpendicular / normal)
True or False: The tangent plane to x² + y² = z² at the point (3, 4, 5) has the equation 3x + 4y – 5z = 0. (Answer: True)
Fill in the blank: For the surface √x + √y + √z = √c, the sum of the x, y, z intercepts of any tangent plane equals ______. (Answer: c)
Q: Suppose F(x, y, z) = x²y + yz³ – 2 = 0. Compute ∂z/∂x at a point where F_z ≠ 0.
A: ∂z/∂x = –F_x/F_z = –(2xy)/(3yz²). Simplify to –2x/(3z²).
Q: Write the equation of the tangent plane to the surface x² + y² – z² = 0 at the point (3, 4, 5).
A: ∇F = (2x, 2y, –2z) = (6, 8, –10) at (3, 4, 5). The tangent plane is 6(x−3) + 8(y−4) – 10(z−5) = 0, which simplifies to 6x + 8y – 10z = 0, or 3x + 4y – 5z = 0.
Q: Verify the cyclic identity for F(x, y, z) = x + y + z – 1 = 0.
A: F_x = F_y = F_z = 1. So ∂z/∂x = −1, ∂x/∂y = −1, ∂y/∂z = −1. Product: (−1)(−1)(−1) = −1. Confirmed.
Q: For √x + √y + √z = 3 (so c = 9), find the tangent plane at (1, 4, 4) and verify the intercept sum.
A: F_x = 1/2, F_y = 1/4, F_z = 1/4 at that point. Tangent plane: (x−1)/2 + (y−4)/4 + (z−4)/4 = 0. Rearranging to intercept form gives intercepts 9, 0 is wrong. Let us compute properly: the tangent plane equation is x/2 + y/4 + z/4 = 1/2 + 1 + 1 = 5/2. Intercepts: x-int = 5, y-int = 10, z-int = 10. Sum is not 9. Recheck: √1 + √4 + √4 = 1+2+2 = 5 ≠ 3, so (1,4,4) is not on the surface √x+√y+√z = 3. Use (1,1,1) instead: √1+√1+√1 = 3, so c = 9. F_x = F_y = F_z = 1/2. Tangent plane: (x−1)/2 + (y−1)/2 + (z−1)/2 = 0, i.e. x + y + z = 3. Intercepts: 3, 3, 3. Sum = 9 = c. Confirmed.
This connects to the single-variable implicit differentiation you learned in Calc I, extended now to surfaces in three dimensions. The gradient and tangent plane reappear immediately in the next sections on optimisation: Lagrange multipliers use the fact that ∇f is parallel to ∇g at a constrained optimum, which is the same gradient-is-normal-to-level-surface idea.
The cyclic identity also appears in thermodynamics courses (CHEM 260 / PHYSICS 240) as a tool for relating partial derivatives of state variables.
Related Terms / Search Tags
implicit differentiation multivariable, cyclic chain rule, cyclic relation partial derivatives, tangent plane to surface, gradient as normal vector, cone tangent plane origin, intercept form tangent plane, √x + √y + √z = √c, MATH 215 HW5, Calculus III §14.5, §14.6, University of Michigan