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Implicit partial differentiation

The excitement of learning separates youth from old age. As long as you’re learning, you’re not old, Rosalyn S.Yalow.

Recall

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x0, y0) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y0. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x0

$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.

The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.

Implicit Differentiation: An Overview

In mathematics, some equations involving two variables, x and y, do not explicitly define y as a function of x. This means that even though y can be expressed as a function of x implicitly, the equation might be too complex or impossible to rearrange in a way that solves for y explicitly in terms of x. These equations are said to be defined implicitly.

For example, consider an equation like x2 + y2 = 1. This equation implicitly defines a relationship between x and y, but it does not express y directly as a function of x (such as y = f(x)). However, there might still exist a function y=f(x) that satisfies this equation for each x.

The technique of implicit differentiation allows us to find the derivative of y with respect to x (denoted as $\frac{dy}{dx}$) without needing to solve the equation explicitly for y.

Examples:

  1. Implicit Differentiation of a Circle. Consider the equation of a circle with a radius of 1 centered at the origin: x2 + y2 = 1. To find the derivative $\frac{dy}{dx}$ we can differentiate both sides of the equation with respect to x, applying the rules of differentiation: $\frac{d}{dx}(x^2+y^2=1)~⇨[\text{For the term y², we use the chain rule}]~2x+2y\frac{dy}{dx}=0~⇨~\frac{dy}{dx}=\frac{-x}{y}.$ It gives the slope of the tangent line to the circle at any point (x,y) on the circle.
  2. Differentiating the Inverse Sine Function. Consider the function y = sin-1x. To differentiate this implicitly, we start by writing the equation in a form that does not involve the inverse function: sin(y) = x.

Now, differentiate both sides with respect to x: $\frac{d}{dx}sin(y)=1⇨~ cos(y)\frac{dy}{dx}=1⇨~\frac{dy}{dx} = \frac{1}{cos(y)}$

$y’= \frac{1}{cos(y)}=~ \frac{1}{\sqrt{1-sin^{2}y}}=~ \frac{1}{\sqrt{1-x^{2}}}$

$(sin^{-1}x)’=\frac{d}{dx}sin^{-1}x=~ \frac{1}{\sqrt{1-x^{2}}}$

Calculating $\frac{∂z}{∂x}, \frac{∂z}{∂y}$ for implicit function

Rewrite the Implicit Equation. First, let’s rewrite the equation in a form that’s easier to differentiate: f(x, y, z) = $x^2+y^4-z^3+3xy^2-8 = 0$. It defines z implicitly as a function of x and y.

Compute the Partial derivatives of f(x, y, z)

Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:

$f_x = \frac{∂f}{∂x} = 2x + 3y^2, f_y = \frac{∂f}{∂y} = 4y^3+6xy, f_z = \frac{∂f}{∂z} = -3z^2$

Implicit Differentiation.

To compute $\frac{∂z}{∂x}, \frac{∂z}{∂y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating z as a function of x and y, z = z(x, y).

Partial derivative $\frac{∂z}{∂x}$

Differentiate f(x, y, z) = 0 with respect to x: $\frac{∂f}{∂x}\frac{∂x}{∂x} + \frac{∂f}{∂y}\frac{∂y}{∂x}+ \frac{∂f}{∂z}\frac{∂z}{∂x}=$

Since y is fixed, we have: $\frac{∂x}{∂x}=1, \frac{∂y}{∂x} = 0$.

So, the previous equation simplifies to: $\frac{∂f}{∂x}+\frac{∂f}{∂z}\frac{∂z}{∂x} = f_x+f_z\frac{∂z}{∂x}=0⇒[\text{Solve for} \frac{∂z}{∂x}] \frac{∂z}{∂x}=z_x = \frac{-f_x}{f_z}$

$\frac{∂z}{∂x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values}] \frac{-(2x+3y^2)}{-3z^2} = \frac{2x+3y^2}{3z^2}$.

Partial derivative $\frac{∂z}{∂y}$

Similarly, to find $\frac{∂z}{∂y}$, we differentiate f(x, y, z) = 0 with respect to y:

$\frac{∂f}{∂x}\frac{∂x}{∂y} + \frac{∂f}{∂y}\frac{∂y}{∂y}+ \frac{∂f}{∂z}\frac{∂z}{∂y}=$

Since x is fixed, we have: $\frac{∂x}{∂y}=0, \frac{∂y}{∂y} = 1$

So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{∂z}{∂y} = 0↭ f_y + f_z·\frac{∂z}{∂y} = 0$. Solve for $\frac{∂z}{∂y}$: $\frac{∂z}{∂y} = \frac{-f_y}{f_z}$

$\frac{∂z}{∂y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values}] \frac{-( 4y^3+6xy)}{-3z^2} = \frac{4y^3+6xy}{3z^2}$·

Rewrite the Implicit Equation. First, let’s rewrite the equation in a form that’s easier to differentiate: f(x, y, z) = x2sin(2y -5z) - 1 - ycos(6zx) = 0$. It defines z implicitly as a function of x and y.

Compute the Partial derivatives of f(x, y, z)

Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:

$f_x = \frac{∂f}{∂x} = 2xsin(2y -5z) +6yzsin(6zx), f_y = \frac{∂f}{∂y} = 2x^2cos(2y -5z) -cos(6zx) , f_z = \frac{∂f}{∂z} = -5x^2cos(2y -5z) +6yxsin(6zx)$

Implicit Differentiation.

To compute $\frac{∂z}{∂x}, \frac{∂z}{∂y}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating z as a function of x and y, z = z(x, y).

Partial derivative $\frac{∂z}{∂x}$

Differentiate f(x, y, z) = 0 with respect to x: $\frac{∂f}{∂x}\frac{∂x}{∂x} + \frac{∂f}{∂y}\frac{∂y}{∂x}+ \frac{∂f}{∂z}\frac{∂z}{∂x}=$

Since y is fixed, we have: $\frac{∂x}{∂x}=1, \frac{∂y}{∂x} = 0$.

So, the previous equation simplifies to: $\frac{∂f}{∂x}+\frac{∂f}{∂z}\frac{∂z}{∂x} = f_x+f_z\frac{∂z}{∂x}=0⇒[\text{Solve for} \frac{∂z}{∂x}] \frac{∂z}{∂x}=z_x = \frac{-f_x}{f_z}$

$\frac{∂z}{∂x}= z_x = \frac{-f_x}{f_z} =[\text{Plug in the values}] \frac{-(2xsin(2y -5z) +6yzsin(6zx))}{-5x^2cos(2y -5z) +6yxsin(6zx)} = \frac{2xsin(2y -5z) +6yzsin(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$.

Partial derivative $\frac{∂z}{∂y}$

Similarly, to find $\frac{∂z}{∂y}$, we differentiate f(x, y, z) = 0 with respect to y:

$\frac{∂f}{∂x}\frac{∂x}{∂y} + \frac{∂f}{∂y}\frac{∂y}{∂y}+ \frac{∂f}{∂z}\frac{∂z}{∂y}=$

Since x is fixed, we have: $\frac{∂x}{∂y}=0, \frac{∂y}{∂y} = 1$

So, the equation simplifies to: $f_x·0 + f_y·1 + f_z·\frac{∂z}{∂y} = 0↭ f_y + f_z·\frac{∂z}{∂y} = 0$. Solve for $\frac{∂z}{∂y}$: $\frac{∂z}{∂y} = \frac{-f_y}{f_z}$

$\frac{∂z}{∂y}= z_y = \frac{-f_y}{f_z} =[\text{Plug in the values}] \frac{-(2x^2cos(2y -5z) -cos(6zx))}{-5x^2cos(2y -5z) +6yxsin(6zx)} = \frac{2x^2cos(2y -5z)-cos(6zx)}{5x^2cos(2y -5z) -6yxsin(6zx)}$·

Rewrite the Implicit Equation. It is not necessarily.

Compute the Partial derivatives of f(x, y, z)

Next, we need to calculate the partial derivatives of f(x, y, z) with respect to x, y, and z respectively:

$f_x = \frac{∂f}{∂x} = 4x^3+2xy^2z^2, f_y = \frac{∂f}{∂y} = 4y^3 + 2yx^2z^2, f_z = \frac{∂f}{∂z} = 4z^3+2zx^2y^2$

Implicit Differentiation.

To compute $\frac{∂y}{∂z}$, we use implicit differentiation. The key idea is to differentiate the implicit equation f(x, y, z) = 0 with respect to the desired variable while treating y as a function of x and x, z = y(x, z).

Partial derivative $\frac{∂y}{∂z}$

Differentiate f(x, y, z) = 0 with respect to z: $\frac{∂f}{∂x}\frac{∂x}{∂z} + \frac{∂f}{∂y}\frac{∂y}{∂z}+ \frac{∂f}{∂z}\frac{∂z}{∂z}=$

Since x is fixed, we have: $\frac{∂x}{∂z}=0, \frac{∂z}{∂z} = 1$.

So, the previous equation simplifies to: $\frac{∂f}{∂y}\frac{∂y}{∂z}+ \frac{∂f}{∂z} = f_y\frac{∂y}{∂z}+f_z=0⇒[\text{Solve for} \frac{∂y}{∂z}] \frac{∂y}{∂z}=y_z = \frac{-f_z}{f_y}$

$y_z = \frac{-f_z}{f_y} =[\text{Plug in the values}] \frac{-(4z^3+2zx^2y^2)}{4y^3 + 2yx^2z^2} = -\frac{z(2z^2+x^2y^2)}{y(2y^2 + x^2z^2)}$.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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