Multivariable Calculus: Partial Differentiation, Jacobians, Extrema, and Leibniz Rule

Functions of Several Variables and Partial Differentiation

• Functions of One vs. Two Variables:

  • A function of one variable is written as $f(x)$, for example: $f(x) = 3x + 2$.
  • A function of two variables is written as $f(x,y)$, for example: $f(x,y) = x^2 + 3ay$.

• Formal Definitions of Partial Derivatives:

  • The partial derivative of $f(x,y)$ with respect to $x$ is defined as the limit:     $\frac{\partial f}{\partial x} = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x, y) - f(x,y)}{\Delta x}$
  • The partial derivative of $f(x,y)$ with respect to $y$ is defined as the limit:     $\frac{\partial f}{\partial y} = \lim_{\Delta y \rightarrow 0} \frac{f(x, y + \Delta y) - f(x,y)}{\Delta y}$

• Example 1: First and Second Order Partial Derivatives:

  • Given the function:     $z = x^3 + y^3 - 3axy$
  • First-order partial derivative with respect to $x$:     $\frac{\partial z}{\partial x} = 3x^2 - 3ay$
  • First-order partial derivative with respect to $y$:     $\frac{\partial z}{\partial y} = 3y^2 - 3ax$
  • Second-order partial derivative with respect to $x$:     $\frac{\partial^2 z}{\partial x^2} = 6x$
  • Second-order partial derivative with respect to $y$:     $\frac{\partial^2 z}{\partial y^2} = 6y$
  • Mixed partial derivatives:     $\frac{\partial^2 z}{\partial x \partial y} = -3a$$\frac{\partial^2 z}{\partial y \partial x} = -3a$
  • Equivalence of mixed partial derivatives:     $\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}$

• Example 2: Verification of Inverse Tangent Partial Derivatives:

  • Given:     $u = \tan^{-1}\left(\frac{x^2+y^2}{x-y}\right)$
  • Differentiating partially with respect to $x$:     $\frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{x^2+y^2}{x-y}\right)^2} \times \frac{(x-y)(2x) - (x^2+y^2)(1)}{(x-y)^2} = \frac{2x^2 - 2xy - x^2 - y^2}{(x-y)^2 + (x^2+y^2)^2}$
  • Differentiating partially with respect to $y$:     $\frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{x^2+y^2}{x-y}\right)^2} \times \frac{(x-y)(2y) - (x^2+y^2)(-1)}{(x-y)^2} = \frac{2xy - 2y^2 + x^2 + y^2}{(x-y)^2 + (x^2+y^2)^2}$

• Example 3: Three-Variable Partial Derivative Identity:

  • Given:     $u = (x^2 + y^2 + z^2)^{-1/2}$
  • Prove that:     $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = -u$
  • Proof Steps:
  • Differentiating $u$ partially with respect to $x$:       $\frac{\partial u}{\partial x} = -\frac{1}{2}(x^2 + y^2 + z^2)^{-3/2} (2x) = -x(x^2 + y^2 + z^2)^{-3/2}$
  • Multiplying by $x$:       $x \frac{\partial u}{\partial x} = -x^2 (x^2 + y^2 + z^2)^{-3/2}$
  • Similarly, for $y$ and $z$:       $y \frac{\partial u}{\partial y} = -y^2 (x^2 + y^2 + z^2)^{-3/2}$$z \frac{\partial u}{\partial z} = -z^2 (x^2 + y^2 + z^2)^{-3/2}$
  • Adding the three expressions:       $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = -(x^2 + y^2 + z^2)(x^2 + y^2 + z^2)^{-3/2}$$= -(x^2 + y^2 + z^2)^{-1/2} = -u$

• Practice Problems on Partial Differentiation:

  • Problem 1: If $u = x^2(y-z) + y^2(z-x) + z^2(x-y)$, show that:     $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$
  • Problem 2: If $V = (1 - 2xy + y^2)^{-1/2}$, prove that:     $x \frac{\partial V}{\partial x} - y \frac{\partial V}{\partial y} = y^2 V^3$
  • Problem 3: If $u = \log(x^3 + y^3 + z^3 - 3xyz)$, show that:
  • (i) $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3}{x+y+z}$
    • Proof:         $\frac{\partial u}{\partial x} = \frac{3x^2 - 3yz}{x^3 + y^3 + z^3 - 3xyz}$$\frac{\partial u}{\partial y} = \frac{3y^2 - 3xz}{x^3 + y^3 + z^3 - 3xyz}$$\frac{\partial u}{\partial z} = \frac{3z^2 - 3xy}{x^3 + y^3 + z^3 - 3xyz}$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \frac{3(x^2 + y^2 + z^2 - xy - yz - zx)}{(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)} = \frac{3}{x+y+z}$
  • (ii) $\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 u = \frac{-9}{(x+y+z)^2}$
    • Proof:         $\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right)^2 u = \left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}\right) \left(\frac{3}{x+y+z}\right)$$= \frac{\partial}{\partial x}\left(\frac{3}{x+y+z}\right) + \frac{\partial}{\partial y}\left(\frac{3}{x+y+z}\right) + \frac{\partial}{\partial z}\left(\frac{3}{x+y+z}\right)$$= \frac{-3}{(x+y+z)^2} + \frac{-3}{(x+y+z)^2} + \frac{-3}{(x+y+z)^2} = \frac{-9}{(x+y+z)^2}$

• Example 4: Heat/Diffusion Differential Equation:

  • Given:     $\theta = t^n e^{-\frac{r^2}{4t}}$
  • Find the value of $n$ which satisfies:     $\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = \frac{\partial \theta}{\partial t}$
  • Solution Steps:
  • Compute first derivative with respect to $r$:       $\frac{\partial \theta}{\partial r} = t^n e^{-\frac{r^2}{4t}} \left(-\frac{2r}{4t}\right) = -\frac{r}{2} t^{n-1} e^{-\frac{r^2}{4t}}$
  • Multiply by $r^2$:       $r^2 \frac{\partial \theta}{\partial r} = -\frac{r^3}{2} t^{n-1} e^{-\frac{r^2}{4t}}$
  • Compute derivative of $r^2 \frac{\partial \theta}{\partial r}$ with respect to $r$:       $\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = -\frac{3r^2}{2} t^{n-1} e^{-\frac{r^2}{4t}} - \frac{r^3}{2} t^{n-1} e^{-\frac{r^2}{4t}} \left(-\frac{2r}{4t}\right)$$= \left(-\frac{3r^2}{2} t^{n-1} + \frac{r^4}{4} t^{n-2}\right) e^{-\frac{r^2}{4t}}$
  • Divide by $r^2$:       $\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r}\right) = \left(-\frac{3}{2} t^{n-1} + \frac{r^2}{4} t^{n-2}\right) e^{-\frac{r^2}{4t}}$
  • Compute derivative of $\theta$ with respect to $t$:       $\frac{\partial \theta}{\partial t} = n t^{n-1} e^{-\frac{r^2}{4t}} + t^n e^{-\frac{r^2}{4t}} \left(\frac{r^2}{4t^2}\right) = \left(n t^{n-1} + \frac{r^2}{4} t^{n-2}\right) e^{-\frac{r^2}{4t}}$
  • Equating the two expressions:       $n t^{n-1} = -\frac{3}{2} t^{n-1} \implies n = -\frac{3}{2}$

• Example 5: One-Dimensional Wave Equation Solution:

  • If $\theta = \phi(y + ax) + \psi(y - ax)$, show that:     $\frac{\partial^2 \theta}{\partial x^2} = a^2 \frac{\partial^2 \theta}{\partial y^2}$
  • Proof:     $\frac{\partial \theta}{\partial x} = a \phi'(y + ax) - a \psi'(y - ax)$$\frac{\partial^2 \theta}{\partial x^2} = a^2 \phi''(y + ax) + a^2 \psi''(y - ax) = a^2 [\phi''(y + ax) + \psi''(y - ax)]$$\frac{\partial \theta}{\partial y} = \phi'(y + ax) + \psi'(y - ax)$$\frac{\partial^2 \theta}{\partial y^2} = \phi''(y + ax) + \psi''(y - ax)$$\frac{\partial^2 \theta}{\partial x^2} = a^2 \frac{\partial^2 \theta}{\partial y^2}$

Euler's Theorem on Homogeneous Functions

• Definition of a Homogeneous Function:

  • A function $f(x,y)$ is said to be homogeneous of degree $n$ in $x$ and $y$ if it can be expressed in the form:     $z = x^n f\left(\frac{y}{x}\right)$
  • Example: For $z = \frac{x^3 - y^3}{x + y}$, factoring $x^3$ from numerator and $x$ from denominator:     $z = \frac{x^3 \left(1 - \frac{y^3}{x^3}\right)}{x \left(1 + \frac{y}{x}\right)} = x^2 f\left(\frac{y}{x}\right)$     Therefore, $z$ is a homogeneous function of degree $n = 2$

• Euler's Theorem Statement:

  • If $z = f(x,y)$ is a homogeneous function of $x$ and $y$ of order $n$, then:     $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z$     for all $(x,y)$ in the domain of the function.

• Complete Proof of Euler's Theorem:

  • Since $z$ is a homogeneous function of $x$ and $y$ of order $n$:     $z = x^n f\left(\frac{y}{x}\right)$
  • Differentiating partially with respect to $x$:     $\frac{\partial z}{\partial x} = n x^{n-1} f\left(\frac{y}{x}\right) + x^n f'\left(\frac{y}{x}\right) \left(-\frac{y}{x^2}\right) = n x^{n-1} f\left(\frac{y}{x}\right) - y x^{n-2} f'\left(\frac{y}{x}\right)$
  • Multiplying both sides by $x$:     $x \frac{\partial z}{\partial x} = n x^n f\left(\frac{y}{x}\right) - y x^{n-1} f'\left(\frac{y}{x}\right)$$x \frac{\partial z}{\partial x} = n z - y x^{n-1} f'\left(\frac{y}{x}\right)$
  • Differentiating $z$ partially with respect to $y$:     $\frac{\partial z}{\partial y} = x^n f'\left(\frac{y}{x}\right) \left(\frac{1}{x}\right) = x^{n-1} f'\left(\frac{y}{x}\right)$
  • Multiplying both sides by $y$:     $y \frac{\partial z}{\partial y} = y x^{n-1} f'\left(\frac{y}{x}\right)$
  • Adding the equations for $x \frac{\partial z}{\partial x}$ and $y \frac{\partial z}{\partial y}$:     $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n z - y x^{n-1} f'\left(\frac{y}{x}\right) + y x^{n-1} f'\left(\frac{y}{x}\right) = n z$

• Second-Order Extension of Euler's Theorem:

  • Statement: If $u$ is a homogeneous function of degree $n$, then:     $x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = n(n-1) u$
  • Proof:
  • From Euler's theorem:       $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$
  • Differentiating partially with respect to $x$:       $\frac{\partial u}{\partial x} + x \frac{\partial^2 u}{\partial x^2} + y \frac{\partial^2 u}{\partial x \partial y} = n \frac{\partial u}{\partial x}$$x \frac{\partial^2 u}{\partial x^2} + y \frac{\partial^2 u}{\partial x \partial y} = (n-1) \frac{\partial u}{\partial x}$
  • Differentiating partially with respect to $y$:       $x \frac{\partial^2 u}{\partial y \partial x} + \frac{\partial u}{\partial y} + y \frac{\partial^2 u}{\partial y^2} = n \frac{\partial u}{\partial y}$$x \frac{\partial^2 u}{\partial x \partial y} + y \frac{\partial^2 u}{\partial y^2} = (n-1) \frac{\partial u}{\partial y}$
  • Multiplying the first derivative equation by $x$ and the second by $y$, and adding:       $x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = (n-1)\left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right) = n(n-1)u$

• Applications & Worked Examples of Euler's Theorem:

  • Example 1: If $u = \sin^{-1}\left(\frac{x^2+y^2}{x+y}\right)$, show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \tan(u)$.
  • Solution:       Let $z = \sin(u) = \frac{x^2+y^2}{x+y} = \frac{x^2\left(1 + \frac{y^2}{x^2}\right)}{x\left(1 + \frac{y}{x}\right)} = x^1 f\left(\frac{y}{x}\right)$.       $z$ is a homogeneous function of degree $n = 1$.       By Euler's Theorem:       $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 1 \times z$       Since $z = \sin(u)$, $\frac{\partial z}{\partial x} = \cos(u) \frac{\partial u}{\partial x}$ and $\frac{\partial z}{\partial y} = \cos(u) \frac{\partial u}{\partial y}$.       $x \cos(u) \frac{\partial u}{\partial x} + y \cos(u) \frac{\partial u}{\partial y} = \sin(u)$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{\sin(u)}{\cos(u)} = \tan(u)$
  • Example 2: If $u = \log\left(\frac{x^2+y^2}{x+y}\right)$, show that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 1$
  • Solution:       Let $z = e^u = \frac{x^2+y^2}{x+y}$, which is homogeneous of degree $n = 1$.       By Euler's Theorem:       $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 1 \times z$$x e^u \frac{\partial u}{\partial x} + y e^u \frac{\partial u}{\partial y} = e^u$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 1$

• Practice Exercises on Euler's Theorem:

  • Problem 1: If $u = \log\left(\frac{x^5 + y^5 + z^5}{x^2 + y^2 + z^2}\right)$, show that:     $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 3$
  • Problem 2: If $u = \sin^{-1}\left(\frac{x^3 + y^3 + z^3}{ax + by + cz}\right)$, prove that:     $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 2 \tan(u)$
  • Problem 3: If $u = f\left(\frac{y}{x}\right) + g\left(\frac{x}{y}\right)$, prove that:     $x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = 0$
  • Problem 4: If $u = \tan^{-1}\left(\frac{y}{x}\right)$, prove that:     $x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = -\sin(2u) \sin^2(u)$

Total Differentials, Chain Rule, and Composite Functions

• Definitions & Concepts:

  • Let $z = f(x,y)$ be a function of variables $x$ and $y$, where $x$ and $y$ are themselves functions of another variable $t$ ($x = h(t)$, $y = g(t)$). Then $z$ is called a composite function of single variable $t$.
  • Theorem on Total Differentials: If $z = f(x,y)$ possesses continuous first-order partial derivatives, the total differential of $z$ is given by:     $dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$
  • Total Derivative (Chain Rule for 1 Variable $t$):     $\frac{dz}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$

• Example 1: Total Differential Calculation:

  • Find total differential of $f(x,y) = y e^{x+y}$.     $df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$\frac{\partial f}{\partial x} = y e^{x+y}$$\frac{\partial f}{\partial y} = y e^{x+y} + e^{x+y} = (y+1) e^{x+y}$$df = y e^{x+y} dx + (y+1) e^{x+y} dy$

• Example 2: Total Derivative Calculation:

  • Find total derivative of $f(x,y) = x^2 + 3xy$ with respect to $u$, given $x = \sin(u)$, $y = \cos(u)$.     $\frac{df}{du} = \frac{\partial f}{\partial x} \frac{dx}{du} + \frac{\partial f}{\partial y} \frac{dy}{du}$$\frac{\partial f}{\partial x} = 2x + 3y, \quad \frac{\partial f}{\partial y} = 3x$$\frac{dx}{du} = \cos(u), \quad \frac{dy}{du} = -\sin(u)$$\frac{df}{du} = (2x + 3y) \cos(u) + (3x) (-\sin(u)) = (2 \sin(u) + 3 \cos(u)) \cos(u) - 3 \sin(u) \sin(u)$

• Change of Variables (Chain Rule for Two Independent Variables):

  • Let $u = f(x,y)$ where $x = \phi(s,t)$ and $y = \psi(s,t)$. Then:     $\frac{\partial u}{\partial s} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s}$$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t}$

• Example 3: Partial Derivatives in Polar Coordinates:

  • Given $z = x^3 - xy + y^3$ where $x = r \cos(\theta)$ and $y = r \sin(\theta)$, find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial \theta}$.
  • Solution:     $\frac{\partial z}{\partial x} = 3x^2 - y, \quad \frac{\partial z}{\partial y} = -x + 3y^2$$\frac{\partial x}{\partial r} = \cos(\theta), \quad \frac{\partial y}{\partial r} = \sin(\theta)$$\frac{\partial z}{\partial r} = (3x^2 - y) \cos(\theta) + (-x + 3y^2) \sin(\theta)$     Substitute $x = r \cos(\theta)$ and $y = r \sin(\theta)$:     $= [3(r \cos(\theta))^2 - r \sin(\theta)] \cos(\theta) + [-r \cos(\theta) + 3(r \sin(\theta))^2] \sin(\theta)$$= 3r^2 (\cos^3(\theta) + \sin^3(\theta)) - 2r \sin(\theta) \cos(\theta)$$\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} (-r \sin(\theta)) + \frac{\partial z}{\partial y} (r \cos(\theta))$$= (3x^2 - y) (-r \sin(\theta)) + (-x + 3y^2) (r \cos(\theta))$$= r [3r^2 \sin^2(\theta) \cos(\theta) - 3r^2 \cos^2(\theta) \sin(\theta) + r \cos(\theta) \sin(\theta) - r \sin(\theta) \cos(\theta)]$$= r^2 (\sin(\theta) - \cos(\theta)) [3r \sin(\theta) \cos(\theta) + 1]$

• Problem 4: Proof for Composite Exponential Function:

  • If $z = e^{ax+by} f(ax-by)$, show that:     $b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = 2ab z$
  • Proof:     Let $u = ax + by$ and $v = ax - by$. Then $z = e^u f(v)$.     $\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = [e^u f(v)] a + [e^u f'(v)] a = a e^u [f(v) + f'(v)]$$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = [e^u f(v)] b + [e^u f'(v)] (-b) = b e^u [f(v) - f'(v)]$$b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = ab e^u [f(v) + f'(v)] + ab e^u [f(v) - f'(v)] = 2ab e^u f(v) = 2ab z$

• Problem 5: Proof for Cyclic Variable Function:

  • If $u = f(y-z, z-x, x-y)$, prove that:     $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$
  • Proof:     Let $X = y-z$, $Y = z-x$, $Z = x-y$.     $\frac{\partial u}{\partial x} = \frac{\partial f}{\partial X}(0) + \frac{\partial f}{\partial Y}(-1) + \frac{\partial f}{\partial Z}(1) = -\frac{\partial f}{\partial Y} + \frac{\partial f}{\partial Z}$$\frac{\partial u}{\partial y} = \frac{\partial f}{\partial X}(1) + \frac{\partial f}{\partial Y}(0) + \frac{\partial f}{\partial Z}(-1) = \frac{\partial f}{\partial X} - \frac{\partial f}{\partial Z}$$\frac{\partial u}{\partial z} = \frac{\partial f}{\partial X}(-1) + \frac{\partial f}{\partial Y}(1) + \frac{\partial f}{\partial Z}(0) = -\frac{\partial f}{\partial X} + \frac{\partial f}{\partial Y}$     Adding all three equations yields zero.

Transformation of Coordinates and Differential Operators

• Transformation of Laplace Equation into Polar Coordinates:
  • Transform the equation:     $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$     into polar coordinates $(r, \theta)$.
  • Coordinate Relations:     $x = r \cos(\theta), \quad y = r \sin(\theta)$$r^2 = x^2 + y^2 \implies r = \sqrt{x^2+y^2}$$\tan(\theta) = \frac{y}{x} \implies \theta = \tan^{-1}\left(\frac{y}{x}\right)$
  • Derivative Relations:     $\frac{\partial r}{\partial x} = \frac{x}{r} = \cos(\theta), \quad \frac{\partial r}{\partial y} = \frac{y}{r} = \sin(\theta)$$\frac{\partial \theta}{\partial x} = -\frac{y}{r^2} = -\frac{\sin(\theta)}{r}, \quad \frac{\partial \theta}{\partial y} = \frac{x}{r^2} = \frac{\cos(\theta)}{r}$
  • First Derivatives:     $\frac{\partial u}{\partial x} = \cos(\theta) \frac{\partial u}{\partial r} - \frac{\sin(\theta)}{r} \frac{\partial u}{\partial \theta}$$\frac{\partial u}{\partial y} = \sin(\theta) \frac{\partial u}{\partial r} + \frac{\cos(\theta)}{r} \frac{\partial u}{\partial \theta}$
  • Operator Equivalences:     $\frac{\partial}{\partial x} = \cos(\theta) \frac{\partial}{\partial r} - \frac{\sin(\theta)}{r} \frac{\partial}{\partial \theta}$$\frac{\partial}{\partial y} = \sin(\theta) \frac{\partial}{\partial r} + \frac{\cos(\theta)}{r} \frac{\partial}{\partial \theta}$
  • Second Derivatives:     $\frac{\partial^2 u}{\partial x^2} = \left(\cos(\theta) \frac{\partial}{\partial r} - \frac{\sin(\theta)}{r} \frac{\partial}{\partial \theta}\right) \left(\cos(\theta) \frac{\partial u}{\partial r} - \frac{\sin(\theta)}{r} \frac{\partial u}{\partial \theta}\right)$$\frac{\partial^2 u}{\partial y^2} = \left(\sin(\theta) \frac{\partial}{\partial r} + \frac{\cos(\theta)}{r} \frac{\partial}{\partial \theta}\right) \left(\sin(\theta) \frac{\partial u}{\partial r} + \frac{\cos(\theta)}{r} \frac{\partial u}{\partial \theta}\right)$
  • Summing the two expressions yields the polar form of the Laplacian:     $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0$

Differentiation of Implicit Functions

• Definitions:

  • Implicit Function: A relation of the form $f(x,y) = c$ in which $x$ and $y$ cannot be separated out explicitly.
  • Example: The relation $x^3 + y^3 - 3axy = 0$ is an implicit function.
  • Explicit Function: A relation in which one variable can be isolated completely in terms of the other.
  • Example: The relation $y^2 = x^2 + 2x$ is an explicit function.

• Formulas for Differentiation of Implicit Functions:

  • Let $f(x,y) = c$ be an implicit function. Then:
  • (i) First derivative formula:       $\frac{dy}{dx} = -\frac{f_x}{f_y}, \quad f_y \neq 0$
  • (ii) Second derivative formula:       $\frac{d^2 y}{dx^2} = -\frac{f_{xx}(f_y)^2 - 2 f_x f_y f_{xy} + f_{yy}(f_x)^2}{(f_y)^3}, \quad f_y \neq 0 \quad \text{provided } f_{xy} = f_{yx}$

• Example 1: Exponential Variable Power:

  • Find $\frac{dy}{dx}$ if $x^y + y^x = c$
  • Solution:     Let $f(x,y) = x^y + y^x - c = 0$     Partial derivative with respect to $x$:     $f_x = y x^{y-1} + y^x \log(y)$     Partial derivative with respect to $y$:     $f_y = x^y \log(x) + x y^{x-1}$     Applying formula $\frac{dy}{dx} = -\frac{f_x}{f_y}$:     $\frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log(y)}{x^y \log(x) + x y^{x-1}}$

• Example 2: Folium of Descartes Derivative:

  • If $x^3 + y^3 - 3axy = 0$, find $\frac{dy}{dx}$ at $x = a, y = a$:     $f_x = 3x^2 - 3ay, \quad f_y = 3y^2 - 3ax$$\frac{dy}{dx} = -\frac{3x^2 - 3ay}{3y^2 - 3ax} = -\frac{x^2 - ay}{y^2 - ax}$

Exact and Inexact Differentials

• Definitions:

  • Exact Differential: Differentials of the form:     $M(x,y) dx + N(x,y) dy = d(f(x,y))$     which can be integrated directly, are called exact differentials.
  • Inexact Differential: Differentials that cannot be written directly as an exact differential of a single function without an integrating factor.

• Condition for Exactness:

  • A differential $M dx + N dy$ is exact if and only if:     $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

• Examples:

  • Example of Exact Differential: $x dy + y dx = d(xy) \implies f(x,y) = xy + c$.     Here $M = y$ and $N = x$, so $\frac{\partial M}{\partial y} = 1$ and $\frac{\partial N}{\partial x} = 1$.
  • Example of Inexact Differential: Show that $x dy + 3y dx$ is inexact.     Here $M = 3y$ and $N = x$. $\frac{\partial M}{\partial y} = 3$ whereas $\frac{\partial N}{\partial x} = 1$. Since $3 \neq 1$, the differential is inexact.

Jacobians (Functional Determinants)

• Definition of Jacobian for Two Variables:

  • If $u$ and $v$ are functions of two independent variables $x$ and $y$, then the Jacobian of $u, v$ with respect to $x, y$ is defined as the determinant:     $J = \frac{\partial(u,v)}{\partial(x,y)} = J\left(\frac{u,v}{x,y}\right) = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}$

• Definition of Jacobian for Three Variables:

  • Similarly, the Jacobian of $u, v, w$ with respect to $x, y, z$ is:     $\frac{\partial(u,v,w)}{\partial(x,y,z)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} & \frac{\partial w}{\partial z} \end{vmatrix}$

• Property 1: Reciprocal Property of Jacobians:

  • Statement: If $J = \frac{\partial(u,v)}{\partial(x,y)}$ and $J' = \frac{\partial(x,y)}{\partial(u,v)}$, then:     $J J' = 1$
  • Proof:     Let $u = f(x,y)$ and $v = g(x,y)$. Solving for $x$ and $y$ gives $x = \phi(u,v)$ and $y = \psi(u,v)$.     Total differentials:     $du = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy$$dv = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y} dy$     Substituting $dx = \frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv$ and $dy = \frac{\partial y}{\partial u} du + \frac{\partial y}{\partial v} dv$ yields identity matrix product of partial derivatives:     $\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$     Taking determinants on both sides gives $J J' = 1$

• Property 2: Chain Rule for Jacobians:

  • If $u, v$ are functions of $r, s$, and $r, s$ are functions of $x, y$, then:     $\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial(u,v)}{\partial(r,s)} \times \frac{\partial(r,s)}{\partial(x,y)}$

• Jacobian of Implicit Functions:

  • If $u_1, u_2$ are functions of $x, y$ defined implicitly by two equations:     $f_1(u_1, u_2, x, y) = 0$$f_2(u_1, u_2, x, y) = 0$     Then:     $\frac{\partial(u_1, u_2)}{\partial(x,y)} = (-1)^2 \frac{\frac{\partial(f_1, f_2)}{\partial(x,y)}}{\frac{\partial(f_1, f_2)}{\partial(u_1, u_2)}}$
  • Similarly, if $u_1, u_2, u_3$ are functions of $x, y, z$ defined implicitly by three equations $f_1 = 0, f_2 = 0, f_3 = 0$:     $\frac{\partial(u_1, u_2, u_3)}{\partial(x,y,z)} = (-1)^3 \frac{\frac{\partial(f_1, f_2, f_3)}{\partial(x,y,z)}}{\frac{\partial(f_1, f_2, f_3)}{\partial(u_1, u_2, u_3)}}$

• Worked Examples on Jacobians:

  • Example 1: Given $u = x^2 + xy$ and $v = xy$, find $\frac{\partial(u,v)}{\partial(x,y)}$.     $\frac{\partial u}{\partial x} = 2x + y, \quad \frac{\partial u}{\partial y} = x$$\frac{\partial v}{\partial x} = y, \quad \frac{\partial v}{\partial y} = x$$\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 2x+y & x \\ y & x \end{vmatrix} = x(2x+y) - xy = 2x^2 + xy - xy = 2x^2$
  • Example 2: Verification of $J J' = 1$ for Polar Coordinates:     $x = r \cos(\theta), \quad y = r \sin(\theta)$$J = \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos(\theta) & -r \sin(\theta) \\ \sin(\theta) & r \cos(\theta) \end{vmatrix} = r \cos^2(\theta) + r \sin^2(\theta) = r$     For inverse functions $r = \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$:     $J' = \frac{\partial(r,\theta)}{\partial(x,y)} = \begin{vmatrix} \frac{x}{r} & \frac{y}{r} \\ -\frac{y}{r^2} & \frac{x}{r^2} \end{vmatrix} = \frac{x^2}{r^3} + \frac{y^2}{r^3} = \frac{r^2}{r^3} = \frac{1}{r}$$J J' = r \times \frac{1}{r} = 1$
  • Example 3: Implicit Jacobian System:     If $x^2 + y^2 + u^2 - v^2 = 0$ and $uv + xy = 0$, prove that:     $\frac{\partial(u,v)}{\partial(x,y)} = \frac{x^2 - y^2}{u^2 + v^2}$
  • Solution:       Let $f_1 = x^2 + y^2 + u^2 - v^2 = 0$ and $f_2 = uv + xy = 0$.       $\frac{\partial(f_1, f_2)}{\partial(x,y)} = \begin{vmatrix} 2x & 2y \\ y & x \end{vmatrix} = 2x^2 - 2y^2 = 2(x^2 - y^2)$$\frac{\partial(f_1, f_2)}{\partial(u,v)} = \begin{vmatrix} 2u & -2v \\ v & u \end{vmatrix} = 2u^2 + 2v^2 = 2(u^2 + v^2)$$\frac{\partial(u,v)}{\partial(x,y)} = (-1)^2 \frac{2(x^2 - y^2)}{2(u^2 + v^2)} = \frac{x^2 - y^2}{u^2 + v^2}$
  • Example 4: Three-Variable Symmetric Determinant:     If $u = xyz$, $v = x^2 + y^2 + z^2$, $w = x + y + z$, find $\frac{\partial(u,v,w)}{\partial(x,y,z)}$.
  • Solution:       $\frac{\partial(u,v,w)}{\partial(x,y,z)} = \begin{vmatrix} yz & zx & xy \\ 2x & 2y & 2z \\ 1 & 1 & 1 \end{vmatrix} = 2 \begin{vmatrix} yz & zx & xy \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix}$       Applying column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:       $= 2 (x-y)(y-z)(z-x)$