Partial Derivatives

This chapter covers multivariable calculus fundamentals: functions of several variables, limits, continuity, partial derivatives, chain rule, directional derivatives, gradients, optimization, and Lagrange multipliers. The concepts build systematically from basic definitions to complex optimization techniques.

A function of several variables is a rule that assigns a single value in the range to each point in the domain. For a function $f(x,y)$, we write:

where $x$ and $y$ are independent variables and $z$ is the dependent variable.

Intuitive Understanding: Just as $y=f(x)$ describes a curve in 2D, $z=f(x,y)$ describes a surface in 3D space. Each point $(x,y)$ in the domain maps to exactly one height $z$.

• Domain: The set of all possible input values $(x,y)$ for which $f(x,y)$ is defined
• Range: The set of all possible output values $f(x,y)$ as $(x,y)$ varies throughout the domain

Example: For $f(x,y)=\sqrt{9-x^2-y^2}$, the domain is the disk $x^2+y^2\le 9$ and the range is $[0,3]$.

For a region $R$ in the xy-plane:

• Interior point: A point $P$ in $R$ such that some disk centered at $P$ lies entirely in $R$
• Boundary point: A point $P$ such that every disk centered at $P$ contains both points in $R$ and points not in $R$
• Limit point: A point $P$ such that every deleted neighborhood of $P$ contains points of $R$

Visualization: Think of a filled circle. Interior points are "safely inside," boundary points are on the edge, and limit points include both interior and boundary points.

• Open region: A region consisting entirely of interior points
• Closed region: A region that contains all its boundary points
• Closure of a region: The union of the region and all its boundary points
• Bounded region: A region that lies inside some disk
• Unbounded region: A region that is not bounded

Key Insight: Open regions are like "intervals without endpoints" - they don't include their boundaries. Closed regions include their boundaries.

• Level curve: For $f(x,y)$, the curve $f(x,y)=c$ for constant $c$
• Level surface: For $f(x,y,z)$, the surface $f(x,y,z)=c$ for constant $c$

Physical Interpretation: Level curves are like contour lines on a topographic map - they connect points of equal elevation. For temperature functions, they're isotherms.

We say $\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=L$ if for every $\epsilon >0$, there exists a $\delta >0$ such that whenever

we have $|f(x,y)-L|<\epsilon$.

Key Difference from Single Variable: We must approach the point from ALL possible directions in the plane, not just left and right.

$\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=L$ if $f(x,y)$ approaches $L$ along every possible path to $(x_0,y_0)$.

Critical Insight: If different paths give different limits, the limit does not exist. This is the basis of the two-path test.

If $\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)=L$ and $\underset{(x,y)\to (x_0,y_0)}{lim}g(x,y)=M$, then:

1. $\underset{(x,y)\to (x_0,y_0)}{lim}[f(x,y)+g(x,y)]=L+M$
2. $\underset{(x,y)\to (x_0,y_0)}{lim}[f(x,y)\cdot g(x,y)]=L\cdot M$
3. $\underset{(x,y)\to (x_0,y_0)}{lim}\frac{f(x,y)}{g(x,y)}=\frac{L}{M}$ if $M\ne 0$
4. Root Rule: $\underset{(x,y)\to (x_0,y_0)}{lim}\sqrt[n]{f(x,y)}=\sqrt[n]{L}$ if $n$ is odd, or if $n$ is even and $L>0$

If $f(x,y)$ approaches different values along two different paths to $(x_0,y_0)$, then $\underset{(x,y)\to (x_0,y_0)}{lim}f(x,y)$ does not exist.

Strategy: Try paths like $y=mx$ (straight lines), $y=m{x}^2$ (parabolas), and $x=0,y=0$ (coordinate axes).

A function $f(x,y)$ is continuous at $(x_0,y_0)$ if:

Practical Test: All polynomial and rational functions are continuous where they're defined. Compositions of continuous functions are continuous.

The partial derivative of $f(x,y)$ with respect to $x$ at $(x_0,y_0)$ is:

Similarly, the partial derivative with respect to $y$ is:

Geometric Interpretation: $\frac{\partial f}{\partial x}$ is the slope of the curve formed by intersecting the surface $z=f(x,y)$ with the plane $y=y_0$. It measures how steeply the surface rises in the $x$-direction while keeping $y$ fixed.

To find $\frac{\partial f}{\partial x}$: Treat all other variables as constants and differentiate normally with respect to $x$.

Example: If $f(x,y)=x^3{y}^2+5xy+\sin (x)$, then:

• $\frac{\partial f}{\partial x}=3x^2{y}^2+5y+\cos (x)$ (treat $y$ as constant)
• $\frac{\partial f}{\partial y}=2x^{3}y+5x$ (treat $x$ as constant)

For $z=f(x,y)$:

• $f_{xx}=\frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)$
• $f_{yy}=\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)$
• $f_{xy}=\frac{{\partial }^{2}f}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)$ (mixed partial)
• $f_{yx}=\frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)$ (mixed partial)

Order of Operations: In $f_{xy}$, differentiate first with respect to $x$, then with respect to $y$.

If $f(x,y)$ and its partial derivatives $f_x$, $f_y$, $f_{xy}$, and $f_{yx}$ are all continuous in an open region containing $(a,b)$, then:

Practical Impact: For "nice" functions, the order of mixed partial differentiation doesn't matter. This dramatically simplifies calculations.

A function $f(x,y)$ is differentiable at $(x_0,y_0)$ if:

where ${\epsilon }_1,{\epsilon }_2\to 0$ as $(\mathrm{\Delta }x,\mathrm{\Delta }y)\to (0,0)$.

Geometric Meaning: The surface has a well-defined tangent plane at the point, and the function is well-approximated by this plane near the point.
If the partial derivatives of a function $f(x,y)$ exist at $(x_0,y_0)$ and are continuous there, then the function is differentiable there. But the converse is not true!!

If $f(x,y)$ is differentiable at $(x_0,y_0)$, then $f$ is continuous at $(x_0,y_0)$.

Warning: The converse is false! A function can be continuous but not differentiable (like $f(x,y)=\sqrt{x^2+y^2}$ at the origin).

If $z=f(x,y)$ is differentiable and $x=g(t)$, $y=h(t)$ are differentiable, then:

Conceptual Understanding: The rate of change of $z$ with respect to $t$ equals the sum of:

• How $z$ changes with $x$ times how $x$ changes with $t$
• How $z$ changes with $y$ times how $y$ changes with $t$

If $z=f(x,y)$ where $x=g(u,v)$, $y=h(u,v)$, then:

If $w=f(x,y,z)$ where $x=g(u,v)$, $y=h(u,v)$, $z=k(u,v)$, then:

Tree diagrams show the dependency relationships between variables. Each path from the top variable to a bottom variable represents a term in the chain rule.

How to Use:

1. Draw the dependent variable at the top
2. Draw intermediate variables in the middle
3. Draw independent variables at the bottom
4. Connect variables with arrows showing dependencies
5. Each complete path gives one term in the chain rule

If $F(x,y)=0$ defines $y$ implicitly as a function of $x$, and $F_y\ne 0$, then:

Extension to More Variables: If $F(x,y,z)=0$ defines $z$ implicitly, then:

Problem: If $w=f(x^2+y^2,xy)$ where $x=r\cos \theta$ and $y=r\sin \theta$, find $\frac{\partial w}{\partial r}$ and $\frac{\partial w}{\partial \theta }$.

Strategy: Let $u=x^2+y^2$, $v=xy$. Then $w=f(u,v)$.

• $\frac{\partial w}{\partial r}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial r}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial r}$
• Need: $\frac{\partial u}{\partial r}=\frac{\partial }{\partial r}(r^2)=2r$
• Need: $\frac{\partial v}{\partial r}=\frac{\partial }{\partial r}(r^2\cos \theta \sin \theta )=2r\cos \theta \sin \theta$

The directional derivative of $f(x,y)$ at point $P_0(x_0,y_0)$ in the direction of unit vector $\mathbf{u}=⟨a,b⟩$ is:

Physical Interpretation: If you're standing at point $(x_0,y_0)$ on a hill represented by $z=f(x,y)$, the directional derivative tells you how steeply the hill rises if you walk in direction $\mathbf{u}$.

The gradient of $f(x,y)$ is the vector:

For functions of three variables:

Fundamental Insight: The gradient is a vector field that points in the direction of steepest increase of the function at each point.

If $f$ is differentiable at $P_0$, then the directional derivative in the direction of unit vector $\mathbf{u}$ is:

This is the most important formula in the chapter! It connects directional derivatives to gradients through the dot product.

1. Maximum rate of increase: $\mathrm{\nabla }f$ points in the direction of maximum rate of increase
2. Magnitude: $|\mathrm{\nabla }f|$ gives the maximum directional derivative
3. Minimum rate: $-\mathrm{\nabla }f$ points in direction of maximum rate of decrease (steepest descent)
4. Zero directional derivative: Directions perpendicular to $\mathrm{\nabla }f$ have zero directional derivative
5. Level curve relationship: $\mathrm{\nabla }f\perp$ level curves

At any point, the gradient vector $\mathrm{\nabla }f$ is perpendicular to the level curve $f(x,y)=c$ passing through that point.

Application: This is why water flows perpendicular to contour lines on topographic maps - it follows the negative gradient (steepest descent).

1. $\mathrm{\nabla }(af+bg)=a\mathrm{\nabla }f+b\mathrm{\nabla }g$ (linearity)
2. $\mathrm{\nabla }(fg)=f\mathrm{\nabla }g+g\mathrm{\nabla }f$ (product rule)
3. $\mathrm{\nabla }\left(\frac{f}{g}\right)=\frac{g\mathrm{\nabla }f-f\mathrm{\nabla }g}{g^2}$ (quotient rule)
4. $\mathrm{\nabla }(f^n)=n{f}^{n-1}\mathrm{\nabla }f$ (power rule)
5. $\mathrm{\nabla }(f(g))=f^{\prime }(g)\mathrm{\nabla }g$ (chain rule)

Problem: Find the directional derivative of $f(x,y,z)=x{e}^{yz}+\ln (x^2+y^2)$ at point $(1,0,2)$ in the direction from $(1,0,2)$ to $(3,-1,1)$.

Solution Strategy:

1. Find $\mathrm{\nabla }f=⟨e^{yz}+\frac{2x}{x^2+y^2},x{e}^{yz}z+\frac{2y}{x^2+y^2},x{e}^{yz}y⟩$
2. Evaluate at $(1,0,2)$: $\mathrm{\nabla }f(1,0,2)=⟨1+2,0,0⟩=⟨3,0,0⟩$
3. Direction vector: $\mathbf{v}=(3,-1,1)-(1,0,2)=⟨2,-1,-1⟩$
4. Unit vector: $\mathbf{u}=\frac{⟨2,-1,-1⟩}{\sqrt{6}}$
5. $D_{\mathbf{u}}f=\mathrm{\nabla }f\cdot \mathbf{u}=⟨3,0,0⟩\cdot \frac{⟨2,-1,-1⟩}{\sqrt{6}}=\frac{6}{\sqrt{6}}=\sqrt{6}$

The tangent plane to the surface $z=f(x,y)$ at point $(x_0,y_0,z_0)$ has equation:

$$z - z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$

Geometric Understanding: The tangent plane is the "best linear approximation" to the surface at the given point. It's the 3D analog of a tangent line.

The normal vector to the surface $z=f(x,y)$ at $(x_0,y_0,z_0)$ is:

Alternative Form: For surfaces defined by $F(x,y,z)=0$, the normal vector is $\mathrm{\nabla }F$.

The linearization of $f(x,y)$ at $(x_0,y_0)$ is:

Approximation: $f(x,y)\approx L(x,y)$ for $(x,y)$ near $(x_0,y_0)$

For a function dependant on three or more variables, it's linearization can be represented as $$L(x,y,z, \dots ) = f(P_{0}) + f_{x}(P_{0})\dd x + f_{y}(P_{0})\dd y + f_{z}(P_{0})\dd z + \dots $$

The total differential of $z=f(x,y)$ is:

Error Estimation: For small changes,

Problem: The radius and height of a cylinder are measured as $r=5\pm 0.1$ cm and $h=12\pm 0.2$ cm. Estimate the maximum error in the calculated volume $V=\pi r^{2}h$.

Solution:

• $\frac{\partial V}{\partial r}=2\pi rh$, $\frac{\partial V}{\partial h}=\pi r^2$
• $dV=2\pi rhdr+\pi r^{2}dh$
• At $(r,h)=(5,12)$ with $dr=\pm 0.1$, $dh=\pm 0.2$:
• $dV=2\pi (5)(12)(0.1)+\pi (5{)}^2(0.2)=12\pi +5\pi =17\pi$ cm³

A point $(a,b)$ is a critical point of $f(x,y)$ if either:

1. $f_x(a,b)=0$ and $f_y(a,b)=0$, or
2. At least one partial derivative does not exist at $(a,b)$

Geometric Interpretation: At interior critical points, the tangent plane is horizontal (if it exists).

• Local maximum: $f(a,b)\ge f(x,y)$ for all $(x,y)$ near $(a,b)$
• Local minimum: $f(a,b)\le f(x,y)$ for all $(x,y)$ near $(a,b)$
• Saddle point: Neither a local max nor min - the surface "saddles" at this point

Let $(a,b)$ be a critical point where $f_x(a,b)=f_y(a,b)=0$, and suppose $f$ has continuous second partial derivatives. Define the discriminant:

Then:

1. $D>0$ and $f_{xx}(a,b)>0$: Local minimum
2. $D>0$ and $f_{xx}(a,b)<0$: Local maximum
3. $D<0$: Saddle point
4. $D=0$: Test inconclusive

The Hessian matrix of $f(x,y)$ is:

The discriminant is: $D=det(H)=f_{xx}{f}_{yy}-(f_{xy}{)}^2$

Connection to Linear Algebra: The second derivative test is really about the eigenvalues of the Hessian matrix. $D>0$ means both eigenvalues have the same sign.

1. Find interior critical points: Solve $\mathrm{\nabla }f=\mathbf{0}$
2. Find boundary extrema: Use techniques like parameterization or Lagrange multipliers
3. Evaluate $f$ at all candidates: Compare values to find absolute max/min

Extreme Value Theorem: Continuous functions on closed, bounded regions always attain their maximum and minimum values.

Problem: Classify all critical points of $f(x,y)=x^4+y^4-4x^{2}y+2y^2$.

Solution Process:

1. $f_x=4x^3-8xy=4x(x^2-2y)=0$
2. $f_y=4y^3-4x^2+4y=4y(y^2+1)-4x^2=0$
3. From first: $x=0$ or $x^2=2y$
4. Case analysis leads to critical points requiring careful algebraic manipulation
5. Second derivative test with $D=f_{xx}{f}_{yy}-(f_{xy}{)}^2$ at each point

To find the extreme values of $f(x,y,z)$ subject to constraint $g(x,y,z)=k$, solve the system:

where $\lambda$ is the Lagrange multiplier.

Geometric Principle: At an extremum, the gradient of the objective function must be parallel to the gradient of the constraint. This happens because both gradients are perpendicular to the constraint surface.

Suppose that $f(x,y,z)$ is differentiable in a region whose interior consists of the smooth curve $C$ given by $$ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} $$

If $P_0$ is a point on $C$ where $f$ has an extremum relative to it's values on $C$, then $\mathrm{\nabla }f$ is orthogonal to $C$ at $P_0$

To find extrema of $f(x,y,z)$ subject to $g(x,y,z)=0$ and $h(x,y,z)=0$:

Geometric Picture: The constraints define a curve (intersection of two surfaces), and we want the extrema of $f$ along this curve.
Invoking Orthogonal Gradient theorem, we know $\mathrm{\nabla }f$ must be perpendicular to this curve $C$, but since this curve itself is an intersection of both $g$ and $h$, both $\mathrm{\nabla }g$ and $\mathrm{\nabla }h$ are perpendicular to it, meaning, $\mathrm{\nabla }f$ lies in the same plane as $\mathrm{\nabla }g$ and $\mathrm{\nabla }h$

Problem: Find the minimum distance from the origin to the curve of intersection of the cylinder $x^2+y^2=1$ and the plane $x+y+z=2$.

Setup: Minimize $f(x,y,z)=x^2+y^2+z^2$ subject to:

• $g(x,y,z)=x^2+y^2-1=0$
• $h(x,y,z)=x+y+z-2=0$

Solution Strategy:

1. $\mathrm{\nabla }f=⟨2x,2y,2z⟩$
2. $\mathrm{\nabla }g=⟨2x,2y,0⟩$
3. $\mathrm{\nabla }h=⟨1,1,1⟩$
4. System: $⟨2x,2y,2z⟩=\lambda ⟨2x,2y,0⟩+\mu ⟨1,1,1⟩$
5. This gives: $2x=2\lambda x+\mu$, $2y=2\lambda y+\mu$, $2z=\mu$
6. Combined with constraints, solve for critical points

Problem: A company produces two goods with production function $f(L,K)=L^{0.6}{K}^{0.4}$ where $L$ is labor and $K$ is capital. If labor costs $w$ per unit and capital costs $r$ per unit with total budget $B$, find the optimal allocation.

Mathematical Setup: Maximize $f(L,K)=L^{0.6}{K}^{0.4}$ subject to $wL+rK=B$

Key Insight: The Lagrange multiplier $\lambda$ represents the marginal utility of money - how much additional output we get per additional dollar of budget.

For a function $f(x,y)$ with continuous partial derivatives up to order 2:

Applications:

• Error analysis for linear approximations
• Theoretical foundation for the second derivative test
• Quadratic approximations of functions

When variables are related by constraints, we must account for these relationships when computing partial derivatives.

Example: If $w=f(x,y,z)$ and $g(x,y,z)=0$, then when we compute $\frac{\partial w}{\partial x}$, we can't hold $y$ and $z$ independent - they're related by the constraint.

1. Method 1: Use the constraint to eliminate one variable, then differentiate
2. Method 2: Use implicit differentiation on the constraint combined with chain rule
3. Method 3: Use the formula involving gradients of the constraint

1. Chain Rule with Multiple Paths: Students often miss terms or get confused about dependency relationships
2. Second Derivative Test Calculations: Computing the discriminant correctly, especially with mixed partials
3. Lagrange Multipliers Setup: Identifying constraints correctly and setting up the gradient equations
4. Directional Derivatives: Forgetting to use unit vectors or miscomputing gradients

If $w=f(u,v)$ where $u=x^2-y^2$, $v=2xy$, and $f$ is differentiable, show that:

Find the points on the surface $xyz=1$ that are closest to the origin. (This requires Lagrange multipliers and careful analysis of the constraint.)

Find the direction in which $f(x,y)=x{e}^y-y{e}^x$ increases most rapidly at $(1,1)$, and find the rate of increase in that direction.

• Partial derivatives extend calculus to multivariable functions by examining rates of change in coordinate directions
• The gradient $\mathrm{\nabla }f$ is the most important concept - it points toward maximum increase and is perpendicular to level curves
• Chain rule becomes more complex but follows systematic patterns using dependency trees
• Optimization uses critical points and the second derivative test, with the Hessian matrix providing geometric insight
• Lagrange multipliers handle constrained optimization by ensuring gradients are parallel at extrema
• Applications span physics, engineering, economics, and all fields involving optimization and rates of change