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.

Recall

• A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x² + y², e^x+y, etc.

• Partial derivatives are derivatives of a function of multiple variables, say f(x₁, x₂, ···, x_n) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. Examples: f(x, y) = 2x²y³, $\frac{\partial f}{\partial x} = f_x = 4x·y^3, \frac{\partial f}{\partial y} = f_y = 6x^2·y^2$; f(x, y) = x³y + y², $\frac{\partial f}{\partial x} = f_x = 3x^2·y, \frac{\partial f}{\partial y} = f_y = x^3+2y.$

• Let f(x₁, x₂, ···, x_n) be a function of n variables. A point (x₁, x₂, ···, x_n) in the domain of f 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 ∇f is a vector of partial derivatives of f: ∇f = ($\frac{∂f}{∂x_1}, \frac{∂f}{∂x_2}, ···, \frac{∂f}{∂x_n}$).

In particular, n = 2, this means that a critical point (x₀, y₀), $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$

Total differential

Implicit differentiation is a technique used to find the derivative of functions defined implicitly, rather than explicitly or to find derivatives of functions that cannot be explicitly solved for one variable, y = f(x), dy = f’(x)dx, e.g., y = sin^-1(x) ⇒ x = sin(y) ⇒ dx = cos(y)dy ⇒ $\frac{dy}{dx}=\frac{1}{cos(y)}$=[Using the Pythagorean identity cos(y) = $\sqrt{1-sin^2(y)}$ and since sin(y)=x] $\frac{1}{\sqrt{1-x^2}}$.

The total differential of a multivariable function w = f(x₁, x₂, ···,x_n) describes how small changes in each of its variables affects the function’s value. It is denoted by df and is defined as follows: $\frac{df}{dt} =\frac{\partial f}{\partial x}d_x+\frac{\partial f}{\partial y}d_y+\frac{\partial f}{\partial z}d_z = f_xdx + f_ydy +f_zdz$ 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.

Using Different Notation

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 also be expressed using the gradient ∇w = ⟨w_x, w_y, w_z⟩ and the derivate of the position vector $\vec{r}(t)$: $∇w·\frac{d\vec{r}}{dt}$, where $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$

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. Besides, it provides a linear approximation of the function for small variations Δx, Δy, and Δz: Δw ≈ f_xΔx + f_yΔy + f_zΔz.

The Chain Rule for multivariable functions

The Chain Rule for multivariable functions is a fundamental concept in calculus that 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 derivate 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), dz = z’(t)dt ⇒ df = $f_xdx + f_ydy +f_zdz = f_xx’(t)dt + f_yy’(t)dt +f_zz’(t)dt$.
3. Divide by dt: $\frac{df}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}.$

Version 2: Using Incremental Changes

1. Approximate the change in f for small increments: Δf ≈ $f_xΔx + f_yΔy +f_zΔz$.
2. Divide by Δt: $\frac{Δf}{Δt} = \frac{f_xΔx + f_yΔy +f_zΔz}{Δt} = f_x\frac{Δx}{Δt} + f_y\frac{Δy}{Δt} + f_z\frac{Δz}{Δt}⇒$.
3. As Δt → 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

• Find the total differential of z = 2x²y³ and z = e^xsin(y)

$dz = \frac{∂z}{∂x}dx + \frac{∂z}{∂y}dy = 4xy^3dx + 6x^2y^2dy.$

$dz = \frac{∂z}{∂x}dx + \frac{∂z}{∂y}dy = e^xsin(y)dx + e^xcos(y)dy.$

• Let f(x, y) = ye^x. Approximate f(2.5, 1.25) using f(2, 1).

f(2, 1) = 1·e² ≈ 7.3891, f(2.5, 1.25) = 1·e² = 1.25·e^2.5 ≈ 15.2281. Δz = 15.2281-7.3891 = 7.839.

$Δz = \frac{∂z}{∂x}Δx + \frac{∂z}{∂y}Δy = ye^xΔx+ e^xΔy =$[Δx = 2.5 -2 = 0.5, Δy = 1.25 -1 = 0.25] = 1·e²·0.5 + e²·0.25 ≈ 5.5418. It is definitely not a good approximation.

• Given w = x²y+z where x = t, y = e^t, and z = sin(t).

Let’s apply the Chain Rule, $\frac{dw}{dt} = \frac{∂w}{∂x}\frac{dx}{dt} + \frac{∂w}{∂y}\frac{dy}{dt} + \frac{∂w}{∂z}\frac{dz}{dt} = 2xy·1 + x^2·e^t + 1·cos(t) =[\text{Since x = t and y =}e^t] 2te^t+t^2e^t+cos(t)$.

Alternatively, w(t) = x²y+z = t²e^t + sin(t) ⇒ $\frac{dw}{dt} = 2te^t+t^2e^t+cos(t)$. This confirms that both methods yield the same result.

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

• Quotient Rule. Let g = ^u⁄_v where u = u(t) and v = v(t)⇒[Using the quotient rule]$\frac{d(u/v)}{dt} =f_u\frac{du}{dt}+f_v\frac{dv}{dt} = \frac{1}{v}\frac{du}{dt}+\frac{-u}{v^2}\frac{dv}{dt} = \frac{u’v-v’u}{v^2}$, that is, the quotient rule. This derives from the product rule applied to g = uv^-1.

• Multivariable Chain Rule. Given w = f(x, y) where x = x(u, v) and y = y(u, v)

We start with the total differential: $dw = f_xdx +f_ydy$ =[Since x and y are functions of u and v: $dx = \frac{∂x}{∂u}du + \frac{∂x}{∂v}dv = x_udu+x_vdv, dy = \frac{∂y}{∂u}du + \frac{∂y}{∂v}dv = y_udu+y_vdv$] $f_x(x_udu +x_vdv) +f_y(y_udu +y_vdv) = […]$

Collecting terms involving du and dv: = […] $(f_xx_u+f_yy_u)du + (f_xx_v + f_yy_v)dv$ ⇒ $\frac{\partial f}{\partial u} = f_xx_u+f_yy_u, \frac{\partial f}{\partial v} = f_xx_v + f_yy_v$.

Thus, we have, $\frac{\partial f}{\partial u} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}, \frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial v}$.

• Polar coordinates. Given the transformations x = rcos(θ), y = rsin(θ),

We want to find: $\frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r} = f_xcos(θ) + f_ysin(θ)$

• The temperature for a bird in fight is given by T(x, y, z) = 0.09x² + 1.4xy + 95z². Approximate change in temperature when head wind 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.

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.

Given the changes: Δx = 2m/s−1m/s = 1m/s, Δy = 55beats/min −50beats/min = 5beats/min, Δz = 4flaps/s −3flaps/s = 1flaps/s

Calculate the partial derivatives: $T_x = 0.18x + 1.4y, T_y = 1.4x, T_z = 190z.$

Calculate the total change in temperature:

$ΔT ≈ T_x(x, y, z)Δx + T_y(x, y, z)Δy + T_z(x, y, z)Δz =[\text{Substitute the given changes and values into the total differential}] (0.18x + 1.4y)Δx + 1.4xΔy + 190zΔz = (0.18·1 + 1.4·50)(2-1) + 1.4·1(55-50) + 190·3·1 = 70.18⋅1+1.4⋅5+570⋅1 ≈ 647.18$. Thus, the approximate change in temperature is 647.18 °C.

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