Multivariable Functions & Partial Derivatives

🎯 Learning Objectives

• Visualise functions $f: \mathbb{R}^n \to \mathbb{R}$ as surfaces, contour maps, and colour plots
• Compute partial derivatives of any order
• Interpret partial derivatives as slopes in coordinate directions
• Use the chain rule for multivariable functions
• Compute partial derivatives of implicit functions

📋 Prerequisites

• Single-variable calculus (BSMA1001) — derivatives, chain rule
• The transition from 1D to $n$D is the key leap in this course

1. Intuition: From Curves to Surfaces

1.1 Ways to Visualise $f(x,y)$

2. Partial Derivatives

2.1 Definition

Definition (Partial Derivative). The partial derivative of $f(x_1, \dots, x_n)$ with respect to $x_i$ is:

Example 1: Partial derivatives of $f(x,y) = x^2y + y^3$

$\frac{\partial f}{\partial x} = 2xy$ (treat $y$ as constant) $\frac{\partial f}{\partial y} = x^2 + 3y^2$ (treat $x$ as constant) Example 2: Partial derivatives of $f(x,y,z) = e^{xy}\sin(z)$

$\frac{\partial f}{\partial x} = ye^{xy}\sin(z)$ $\frac{\partial f}{\partial y} = xe^{xy}\sin(z)$ $\frac{\partial f}{\partial z} = e^{xy}\cos(z)$

2.2 Geometric Interpretation

3. Higher-Order Partial Derivatives

Definition (Second partial derivatives).

Similarly $f_{yx}$ and $f_{yy}$.

3.1 Clairaut's Theorem (Equality of Mixed Partials)

Theorem (Clairaut). If $f_{xy}$ and $f_{yx}$ are continuous at a point, then:

Example 3: Higher-order partials

$f(x,y) = x^3y^2 + x^2y$

$f_x = 3x^2y^2 + 2xy$ $f_y = 2x^3y + x^2$

$f_{xx} = 6xy^2 + 2y$ $f_{yy} = 2x^3$ $f_{xy} = 6x^2y + 2x$ $f_{yx} = 6x^2y + 2x$ (same as $f_{xy}$ ✓)

4. The Chain Rule

4.1 One Independent Variable

4.2 Multiple Independent Variables

Example 4: Chain rule

$f(x,y) = x^2 + y^2$, $x = t^2$, $y = e^t$.

$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} = 2x \cdot 2t + 2y \cdot e^t = 4xt + 2ye^t$.

Substitute: $x = t^2$, $y = e^t$: $\frac{df}{dt} = 4t^3 + 2e^{2t}$.

Direct check: $f(t) = t^4 + e^{2t}$, $\frac{df}{dt} = 4t^3 + 2e^{2t}$ ✓

5. Edge Cases & Gotchas

6. Common Pitfalls

❌ Pitfall 1: Forgetting which variable is held constant

❌ Pitfall 2: Confusing $f_{xy}$ with $\frac{\partial^2 f}{\partial x \partial y}$

❌ Pitfall 3: Applying chain rule with wrong dependency tree

7. Formula Summary Table

8. 📝 Practice Questions

Q1: Compute partial derivatives

$f(x,y) = \sin(xy) + xe^y$. Find $f_x$ and $f_y$.

Solution: $f_x = y\cos(xy) + e^y$ (treat $y$ constant) $f_y = x\cos(xy) + xe^y$ (treat $x$ constant) Q2: Second partials

$f(x,y) = \ln(x^2 + y^2)$. Find $f_{xx}$, $f_{xy}$, $f_{yx}$, $f_{yy}$.

Solution: $f_x = \frac{2x}{x^2+y^2}$, $f_y = \frac{2y}{x^2+y^2}$. $f_{xx} = \frac{2(x^2+y^2) - 2x(2x)}{(x^2+y^2)^2} = \frac{2(y^2 - x^2)}{(x^2+y^2)^2}$. $f_{xy} = \frac{-2x(2y)}{(x^2+y^2)^2} = \frac{-4xy}{(x^2+y^2)^2}$. $f_{yx} = \frac{-2y(2x)}{(x^2+y^2)^2} = \frac{-4xy}{(x^2+y^2)^2}$ (same ✓). $f_{yy} = \frac{2(x^2+y^2) - 2y(2y)}{(x^2+y^2)^2} = \frac{2(x^2 - y^2)}{(x^2+y^2)^2}$. Q3: Chain rule

$z = e^{xy}$, $x = \cos t$, $y = \sin t$. Find $\frac{dz}{dt}$.

Solution: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = ye^{xy}(-\sin t) + xe^{xy}(\cos t)$. $= e^{xy}(-y\sin t + x\cos t) = e^{\cos t\sin t}(-\sin^2 t + \cos^2 t) = e^{\cos t\sin t}\cos(2t)$. Q4: Verification of Clairaut

Verify $f_{xy} = f_{yx}$ for $f(x,y) = \tan^{-1}(y/x)$.

Solution: $f_x = \frac{-y}{x^2+y^2}$, $f_y = \frac{x}{x^2+y^2}$. $f_{xy} = \frac{-(x^2+y^2) + y(2y)}{(x^2+y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$. $f_{yx} = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$. Equal ✓. Q5: Implicit differentiation

If $x^2 + y^2 + z^2 = 1$, find $\frac{\partial z}{\partial x}$.

Solution: Differentiate both sides w.r.t. $x$, treating $z$ as function of $x,y$: $2x + 0 + 2z \frac{\partial z}{\partial x} = 0 \Rightarrow \frac{\partial z}{\partial x} = -\frac{x}{z}$ (for $z \neq 0$). Q6: Tangent plane equation

Find the tangent plane to $z = x^2 + y^2$ at $(1,2,5)$.

Solution: $z_0 = 5$, $f_x(1,2) = 2$, $f_y(1,2) = 4$. Tangent plane: $z - 5 = 2(x-1) + 4(y-2)$, or $z = 2x + 4y - 5$. Q7: Linear approximation

Approximate $f(1.1, 1.9)$ for $f(x,y) = x^2y$ using linearisation at $(1,2)$.

Solution: $f(1,2) = 2$, $f_x = 2xy$, $f_x(1,2) = 4$, $f_y = x^2$, $f_y(1,2) = 1$. $L(x,y) = 2 + 4(x-1) + 1(y-2) = 4x + y - 4$. $L(1.1, 1.9) = 4(1.1) + 1.9 - 4 = 4.4 + 1.9 - 4 = 2.3$. Actual: $f(1.1, 1.9) = (1.1)^2(1.9) = 1.21 \times 1.9 = 2.299$. Close! Q8: 3-variable function partials

$f(x,y,z) = xyz + \frac{x}{y}$. Find $f_x$, $f_y$, $f_z$.

Solution: $f_x = yz + \frac{1}{y}$ $f_y = xz - \frac{x}{y^2}$ $f_z = xy$ Q9: Partial differential equation

Show $f(x,y) = e^{-x}\sin y$ satisfies Laplace's equation $f_{xx} + f_{yy} = 0$.

Solution: $f_x = -e^{-x}\sin y$, $f_{xx} = e^{-x}\sin y$. $f_y = e^{-x}\cos y$, $f_{yy} = -e^{-x}\sin y$. $f_{xx} + f_{yy} = e^{-x}\sin y - e^{-x}\sin y = 0$ ✓. Q10: Wave equation

Show $u(x,t) = \sin(x-at)$ satisfies the wave equation $u_{tt} = a^2 u_{xx}$.

Solution: $u_x = \cos(x-at)$, $u_{xx} = -\sin(x-at)$. $u_t = -a\cos(x-at)$, $u_{tt} = -a^2\sin(x-at) = a^2(-\sin(x-at)) = a^2 u_{xx}$ ✓.

🔗 Cross-References

• Next topic: Limits & Continuity
• Week 10 (Gradient): $\nabla f$ collects all partial derivatives
• Week 11 (Hessian): $H$ collects all second partials
• BSMA1001 (Maths 1): Single-variable derivatives
• BSCS3004 (Deep Learning): Backpropagation uses the chain rule extensively Join Discord Previous8.3 Orthogonal TransformationsNext9.2 Gradient & Directional Derivatives