Total differential

And there are never really endings, happy or otherwise. Things keep going on, they overlap and blur, your story is part of your sister’s story is part of many other stories, and there is no telling where any of them may lead. Everything turns in circles and spirals with the cosmic heart until infinity.

Functions, Partials, and Gradients

• A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns to each ordered pair in its domain a unique real number. Examples include geometric formulas like Area = $\frac{1}{2}bh$, or algebraic surfaces like $z = f(x, y) = 2x + 3y$, $f(x, y) = x^2 + y^2$, and $f(x,y) = e^{x+y}$.

• Partial derivatives differentiate a multivariable function $f(x_1, x_2, \dots, x_n)$ with respect to one target variable while treating all other variables as constants. They measure how the function changes in a single specific direction.
  If $f(x, y) = 2x^2y^3$, then $\frac{\partial f}{\partial x} = f_x = 4xy^3$ and $\frac{\partial f}{\partial y} = f_y = 6x^2y^2$.
  If $f(x, y) = x^3y + y^2$, then $f_x = 3x^2y$ and $f_y = x^3+2y$.

• A point $(x_1, x_2, \dots, x_n)$ is considered a critical point if the gradient of $f$ at that point is the zero vector, or if the gradient does not exist.

• The gradient $\nabla f$ (del) is a vector containing all first-order partial derivatives of $f$. It geometrically points in the direction of the function’s steepest ascent. $\nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n} \right\rangle$.

• In 2D space ($n = 2$), finding a critical point $(x_0, y_0)$ requires solving the system: $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$.

Implicit Differentiation & The Total Differential

Implicit differentiation is used to find the derivative of functions defined implicitly ($F(x,y) = 0$) rather than explicitly ($y = f(x)$).

• Single Variable Example: $y = \sin^{-1}(x) \implies x = \sin(y) \implies dx = \cos(y)dy \implies \frac{dy}{dx} = \frac{1}{\cos(y)}$. Using the Pythagorean identity $\cos(y) = \sqrt{1-\sin^2(y)}$ and knowing $\sin(y)=x$, we get $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$.
• Multivariable Shortcut: For an implicitly defined curve $F(x,y) = 0$, the derivative is simply $\frac{dy}{dx} = -\frac{F_x}{F_y}$.
  🎯 Because the curve is defined by the constraint $F(x,y) = 0$, any movement along the curve must keep F equal to zero. Apply the chain rule for multivariable functions $\frac{d}{dx}F(x, y) = F_x(x, y)\cdot 1 + F_y(x, y)\frac{dy}{dx} = 0$.
  Solve for $\frac{dy}{dx}, \frac{dy}{dx} = -\frac{F_x}{F_y}$

The total differential of a function $w = f(x, y, z)$ describes how small changes in each independent variable cumulatively affect the function’s overall value. It provides a linear (tangent plane) approximation of the function for small variations $\Delta x$, $\Delta y$, and $\Delta z$.

It is denoted by $df$ or $dw$ and is defined as follows: $dw = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = f_x dx + f_y dy + f_z dz$ where $f_x = \frac{\partial f}{\partial x}, f_y = \frac{\partial f}{\partial y}$, and $f_z = \frac{\partial f}{\partial z}$ are the partial derivatives of f.

1. The total differential is the unique linear function whose slopes in the x-, y-, and z-directions match the partial derivatives of f.
2. 🧱 A function $w = f(x,y,z)$ defines a surface in 4‑dimensional space. Near a point $(x_0,y_0,z_0)$, the surface behaves almost linearly: $f(x_0+\Delta x, y_0 +\Delta y, z_0 +\Delta z) \approx f(x_0,y_0,z_0) + f_x \Delta x + f_y \Delta y + f_z \Delta z.$
   This is the first‑order Taylor polynomial.
   The differential is simply the increment part: $dw \approx f_x\Delta x + f_y\Delta y + f_z\Delta z.$
   Replace $\Delta x, \Delta y, \Delta z$ with the differential symbols $dx, dy, dz$: $dw = f_xdx + f_ydy + f_zdz.$
3. 🧭 Chain rule viewpoint.
   Think of f as a function of three variables that each depend on some parameter t: $w(t) = f(x(t), y(t), z(t)).$
   Differentiate with respect to t: $\frac{dw}{dt} = f_x \frac{dx}{dt} + f_y\frac{dy}{dt} + f_z \frac{dz}{dt}.$
   Now multiply both sides by $dt: dw = f_xdx + f_ydy + f_zdz.$ This is the total differential.
4. 🧠 Geometric intuition: the gradient is the “slope vector”.
   The gradient $\nabla f = (f_x, f_y, f_z)$ tells you how $f$ changes in each coordinate direction.
   A small displacement $d\mathbf{r} = (dx, dy, dz)$ changes the function by the dot product: $dw = \nabla f \cdot d\mathbf{r}.$
   The total differential tells us how the function w changes as you move along an infinitesimal displacement in each of its variables, hence providing information about the rate of change of the function in the x, y, and z directions.

Alternative Notations

When considering how w changes with respect to another variable t, we can write: $\frac{dw}{dt} = w_x\frac{dx}{dt}+ w_y\frac{dy}{dt}+ w_z\frac{dz}{dt}$.

This can be elegantly expressed as the dot product of the gradient $\nabla w$ and the velocity vector $\frac{d\vec{r}}{dt}$: $\boxed{\frac{dw}{dt} = \nabla w \cdot \frac{d\vec{r}}{dt}} \quad \text{where} \quad \frac{d\vec{r}}{dt} = \left\langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right\rangle$

The Multivariable Chain Rule

The Chain Rule for multivariable functions allows us to compute the derivative of a composite function. It states that if we have a function $f(x, y, z)$ where $x = x(t)$, $y = y(t)$, and $z = z(t)$ are functions of another variable $t$, then: the derivative of f with respect to t is given by $\frac{df}{dt} = f_x\frac{dx}{dt} + f_y\frac{dy}{dt} + f_z\frac{dz}{dt}$

Proof of the Chain Rule (2 versions)

Version 1: Using Total Differential

1. Start with the total differential: df = $f_xdx + f_ydy +f_zdz,$ where $x = x(t)$, $y = y(t)$, and $z = z(t)$ are functions of another variable t.

2. Substitute $dx = x'(t)dt$, $dy = y'(t)dt$, $dz = z'(t)dt$. Then, $df = f_xdx + f_ydy +f_zdz = f_xx'(t)dt + f_yy'(t)dt +f_zz'(t)dt$.

3. Factor out $dt$ and divide: $\frac{df}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}.$

Version 2: Using Incremental Changes

1. Approximate the change in f for small increments: $\Delta f \approx f_x\Delta x + f_y\Delta y + f_z\Delta z$.
   🧩 Understanding the Components: $\Delta f$ is the total change in the function’s output value. $\Delta x, \Delta y, \Delta z$ are the small changes (increments) made to the input variables $x$, $y$, and $z$. $f_x, f_y, f_z$ represent the rate of change of the function along each individual axis.
   💡 Why It Works: Imagine you are standing on a hill. If you walk a tiny step forward $\Delta x$, your elevation changes based on how steep the hill is in that direction $f_{x}$.
   If you move slightly in all three directions $x$, $y$, and $z$ at the same time, your total change in elevation $\Delta f$ is roughly the sum of the changes caused by each individual movement:
   Change from x step (change in the x-direction) = $f_x \cdot \Delta x$. Change from y step = $f_y \cdot \Delta y$. Change from z step = $f_z \cdot \Delta z$
   Summing these up gives you the total approximate change.

2. Divide by $\Delta t$: $\frac{\Delta f}{\Delta t} \approx f_x\frac{\Delta x}{\Delta t} + f_y\frac{\Delta y}{\Delta t} + f_z\frac{\Delta z}{\Delta t}$.

3. As $\Delta t \to 0$, this approximation becomes exact and we get the Chain Rule again, $\frac{df}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}.$

Exercises

Calculate the total differential

• $z = 2x^2y^3 \implies dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = 4xy^3dx + 6x^2y^2dy$
• $z = e^x\sin(y) \implies dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = e^x\sin(y)dx + e^x\cos(y)dy$

Approximate $f(2.5, 1.25)$ for $f(x, y) = ye^x$ using the base point $(2, 1)$.

• Exact values: $f(2, 1) = e^2 \approx 7.3891$. $f(2.5, 1.25) = 1.25e^{2.5} \approx 15.2281$. True $\Delta z = 7.839$.
• Differential approximation: $dz = f_x \Delta x + f_y \Delta y = ye^x \Delta x + e^x \Delta y$. Given $\Delta x = 0.5$ and $\Delta y = 0.25$: $dz = (1)(e^2)(0.5) + (e^2)(0.25) = 0.75e^2 \approx 5.5418$.
• Note: The approximation (5.54) undershoots the true change (7.84) significantly (it is definitely not a good approximation) because $\Delta x = 0.5$ is too large a step for an exponential function, causing the tangent plane to diverge rapidly from the actual surface.

Chain Rule vs. Direct Substitution

Given $w = x^2y+z$ where $x = t$, $y = e^t$, and $z = \sin(t)$.

• Method A (Chain Rule): $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt} = (2xy)(1) + (x^2)(e^t) + (1)\cos(t)$. Substituting the $t$ definitions yields $2te^t + t^2e^t + \cos(t)$.
• Method B (Direct): $w(t) = (t)^2(e^t) + \sin(t) \implies \frac{dw}{dt} = 2te^t + t^2e^t + \cos(t)$. This confirms that both methods yield the same result.

Deriving standard rules from the Multivariable Chain Rule

• Product Rule: Let $f = uv$, where $u = u(t)$ and $v = v(t)$ are differentiable functions of t.
  $\frac{d(uv)}{dt} = f_u\frac{du}{dt} + f_v\frac{dv}{dt} = v\frac{du}{dt} + u\frac{dv}{dt}$. This is the traditional product rule for differentiation.

• Quotient Rule: Let $g = \frac{u}{v} = uv^{-1}$, where $u(t)$ and $v(t)$.
  $\frac{d(u/v)}{dt} = g_u\frac{du}{dt} + g_v\frac{dv}{dt} = \left(\frac{1}{v}\right)\frac{du}{dt} + \left(-\frac{u}{v^2}\right)\frac{dv}{dt} = \frac{u'v - v'u}{v^2}$, that is, the quotient rule.

Multi-parameter Chain Rule

Given $w = f(x, y)$ where $x(u, v)$ and $y(u, v)$. Starting with the total differential $dw = f_xdx + f_ydy$, substitute (since x and y are functions of u and v) $dx = x_udu + x_vdv$ and $dy = y_udu + y_vdv$: $dw = f_x(x_udu + x_vdv) + f_y(y_udu + y_vdv)$

Group by $du$ and $dv$: $dw = (f_xx_u + f_yy_u)du + (f_xx_v + f_yy_v)dv$

Therefore, the partials with respect to the new parameters are: $\boxed{\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}} \quad \text{and} \quad \boxed{\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v}}$

• Polar coordinates. Given the transformations $x = r\cos(\theta)$, $y = r\sin(\theta)$.

We want to find: $\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r} = f_x\cos(\theta) + f_y\sin(\theta)$

• Applied Physics: The Overheating Bird.

The temperature of a bird in flight is $T(x, y, z) = 0.09x^2 + 1.4xy + 95z^2$. Approximate the change in temperature when headwind $x$ increases from 1 meters per second to 2 meters per second, bird heart rate $y$ increases from 50 beats per minute to 55 beats per minute and flapping rate $z$ increases from 3 flaps per second to 4 flaps per second.

1. To approximate the change in temperature T given the changes in the head wind x, bird heart rate y, and flapping rate z, we can use the concept of the total differential in multivariable calculus.
2. Deltas: $\Delta x = 2m/s−1m/s = 1m/s$, $\Delta y = 55beats/min −50beats/min = 5$, $\Delta z = 4flaps/s −3flaps/s = 1flaps/s$.
3. Calculate the partial derivatives: $T_x = 0.18x + 1.4y$, $T_y = 1.4x$, $T_z = 190z$.
4. Evaluate at initial state $(1, 50, 3)$:
   $T_x(1, 50, 3) = 0.18(1) + 1.4(50) = 70.18$, $T_y(1, 50, 3) = 1.4(1) = 1.4$, and $T_z(1, 50, 3) = 190(3) = 570$
5. Total Differential Calculation: $dT \approx T_x\Delta x + T_y\Delta y + T_z\Delta z$
   $dT \approx (70.18)(1) + (1.4)(5) + (570)(1) = 647.18$
6. 🔬Conclusion. Thus, the approximate change in temperature is $647.18 °C$. The math is flawless, but an instantaneous temperature increase of 647 degrees suggests this poor bird has spontaneously combusted mid-flight!
   Why? The reason for this absurd result lies entirely in the unrealistic coefficient for the $z^2$ term (95) and the massive step size ($\Delta z = 1$) for a quadratic variable. As a consequence, the partial derivative for the flapping rate is $T_z = 190z$. Evaluated at $z = 3$, this single component contributes $570$ (or roughly 88%) to the total estimated temperature change.

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