4.3 Multivariable Chain Rule and Implicit Differentiation

4.3 Multivariable Chain Rule and Implicit Differentiation

When variables depend on intermediate variables that in turn depend on independent parameters, computing rates of change requires the multivariable chain rule. Furthermore, because many geometric and physical relations are defined implicitly rather than explicitly, the Implicit Function Theorem provides the analytical engine for computing derivatives and understanding local invertibility.

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The Multivariable Chain Rule

Case 1: One Independent Parameter (Curves)

Let $z = f(x, y)$ be differentiable, where $x = x(t)$ and $y = y(t)$ are differentiable curves. Then: $\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$

Case 2: Several Independent Parameters (Surfaces)

Let $z = f(x, y)$ where $x = g(u, v)$ and $y = h(u, v)$. Then: $\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u}, \quad \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v}$

Dependency Tree Algorithm

To differentiate dependent variable $z$ with respect to independent variable $s$:

1. Construct a tree: root $z$ at top, intermediate variables in middle, leaves at bottom.
2. Trace all distinct paths connecting root $z$ to leaf $s$.
3. Multiply derivatives along each path.
4. Sum products across all paths: $\frac{\partial z}{\partial s} = \sum_{i=1}^n \frac{\partial z}{\partial x_i} \frac{\partial x_i}{\partial s}$.

Matrix Formulation (Jacobians)

For composite map $h = f \circ g$, the derivative is the matrix product of Jacobians: $Dh(\mathbf{x}) = Df(g(\mathbf{x})) \cdot Dg(\mathbf{x})$

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Implicit Differentiation via Gradients

Implicit Curves in $\mathbb{R}^2$

For $F(x, y) = 0$ defining $y = y(x)$, differentiating with respect to $x$ gives: $F_x + F_y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{F_x}{F_y} \quad (F_y \neq 0)$

Implicit Surfaces in $\mathbb{R}^3$

For $F(x, y, z) = 0$ defining $z = z(x, y)$, differentiating with respect to $x$ and $y$ gives: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \quad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} \quad (F_z \neq 0)$

Higher-Order Implicit Differentiation

Differentiating $\frac{dy}{dx} = -\frac{F_x}{F_y}$ with respect to $x$ yields: $\frac{d^2 y}{dx^2} = -\frac{F_{xx} F_y^2 - 2 F_{xy} F_x F_y + F_{yy} F_x^2}{F_y^3}$

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The Implicit Function Theorem (IFT)

Let $F: U \subset \mathbb{R}^{n+m} \to \mathbb{R}^m$ be $C^1$ with coordinates $(\mathbf{x}, \mathbf{y}) \in \mathbb{R}^n \times \mathbb{R}^m$. Suppose $(\mathbf{x}_0, \mathbf{y}_0) \in U$ satisfies:

1. $F(\mathbf{x}_0, \mathbf{y}_0) = \mathbf{0}$.
2. The $m \times m$ Jacobian matrix $D_{\mathbf{y}} F(\mathbf{x}_0, \mathbf{y}0) = \left[ \frac{\partial F_i}{\partial y_j} \right]$ is invertible: $\det(D{\mathbf{y}} F) \neq 0$.

Then there exists an open neighborhood $V \subset \mathbb{R}^n$ of $\mathbf{x}_0$ and a unique $C^1$ function $g: V \to \mathbb{R}^m$ such that $g(\mathbf{x}_0) = \mathbf{y}_0$ and $F(\mathbf{x}, g(\mathbf{x})) = \mathbf{0}$. Its derivative is: $Dg(\mathbf{x}) = -\left[ D_{\mathbf{y}} F(\mathbf{x}, g(\mathbf{x})) \right]^{-1} \cdot D_{\mathbf{x}} F(\mathbf{x}, g(\mathbf{x}))$

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Jacobians and Coordinate Transformations

For transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $x = x(u, v), y = y(u, v)$: $J = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} x_u & x_v \\ y_u & y_v \end{pmatrix} = x_u y_v - x_v y_u$

Inverse Function Theorem

If $F: \mathbb{R}^n \to \mathbb{R}^n$ is $C^1$ and $\det(DF(\mathbf{x}_0)) \neq 0$, then $F$ is locally invertible near $\mathbf{x}0$, and $J{F^{-1}}(F(\mathbf{x}_0)) = \frac{1}{J_F(\mathbf{x}_0)}$.

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Step-by-Step Worked Problems

Problem 1: Chain Rule with Two Intermediate Variables

Let $z = u^2 - v^3$, where $u = x^2 - y^2$ and $v = 2xy$. Compute $\frac{\partial z}{\partial x}$ at $(x, y) = (2, 1)$.

Solution:

1. Intermediate values at $(2, 1)$: $u = 4 - 1 = 3$ and $v = 2(2)(1) = 4$.
2. Outer partials: $\frac{\partial z}{\partial u} = 2u = 6$ and $\frac{\partial z}{\partial v} = -3v^2 = -3(16) = -48$.
3. Inner partials: $\frac{\partial u}{\partial x} = 2x = 4$ and $\frac{\partial v}{\partial x} = 2y = 2$.
4. Sum paths: $\frac{\partial z}{\partial x} = (6)(4) + (-48)(2) = 24 - 96 = -72$.

Problem 2: Implicit Surface Partial Derivative

Find $\frac{\partial z}{\partial x}$ for $x^2 y + y^2 z + z^3 x = 3$ at $(1, 1, 1)$.

Solution:

1. Set $F(x, y, z) = x^2 y + y^2 z + z^3 x - 3$. Verify $F(1, 1, 1) = 0$.
2. Compute $F_x = 2xy + z^3 \implies F_x(1, 1, 1) = 2(1) + 1 = 3$.
3. Compute $F_z = y^2 + 3z^2 x \implies F_z(1, 1, 1) = 1 + 3 = 4$.
4. Apply formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{3}{4}$.

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GRE Exam Traps & Pitfalls

Trap 1: Missing Minus Sign in Implicit Differentiation Always write $\frac{dy}{dx} = -\frac{F_x}{F_y}$ and $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$. The negative sign originates from moving $F_x$ across the equals sign.

Trap 2: Omitting Paths in Chain Rule Trees When an independent variable appears in several intermediate functions, sum over every path connecting the root to the leaf.

Trap 3: Vanishing Denominators in IFT If $F_z(x_0, y_0, z_0) = 0$, the IFT cannot guarantee $z = g(x, y)$; the tangent plane is vertical.