knitr::opts_chunk$set(echo = TRUE)
library(tinytex)
library(Deriv)
library(symengine)## Warning: package 'symengine' was built under R version 3.6.2## SymEngine Version: 0.6.0## _____ _____ _
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## |___| |___|library(ggplot2)Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/]
In Exercises 23 – 26, find \(\frac{dy}{dx}\) using Implicit Differentiation and Theorem 12.5.3.
\[\text{ln}(x^2 + xy + y^2) = 1\]
Theorem 12.5.3 - Implicit Differentiation
Let \(f\) be a differentiable function of \(x\) and \(y\), where \(f(x, y) = c\) defines \(y\) as an implicit function of \(x\), for some constant \(c\). Then
\[\frac{dy}{dx} = -\frac{f_x(x,y)}{f_y(x,y)}\]
\[f(x, y) = \text{ln}(x^2 + xy + y^2) -1\]
Partial Derivative of \(f(x, y)\) with respect to \(x\):
\[\frac{\partial{}}{\partial{x}}(\text{ln}(x^2 + xy + y^2)) + \frac{\partial{}}{\partial{x}}(-1)\\ \frac{\partial{}}{\partial{x}}(\text{ln}(x^2 + xy + y^2)) + 0\\ \frac{\partial{}}{\partial{x}}(\text{ln}(x^2 + xy + y^2))\]
Using the Chain Rule with: \(u = x^2 +xy + y^2\):
\[\frac{\partial{}}{\partial{x}}(\text{ln}(x^2 + xy + y^2)) = \frac{\partial{}\text{ln}(u)}{\partial{u}}\frac{\partial{u}}{\partial{x}},\\ \text{where }u = x^2 + xy + y^2\text{ and } \frac{\partial{}}{\partial{u}}(ln(u)) = \frac{1}{u}\\ = \frac{\frac{\partial{}}{\partial{x}}(x^2 + xy + y^2)}{x^2 + xy + y^2}\\ = \frac{2x + y}{x^2 + xy + y^2}\\\]
Partial Derivative of \(f(x, y)\) with respect to \(y\):
\[\frac{\partial{}}{\partial{y}}(\text{ln}(x^2 + xy + y^2)) + \frac{\partial{}}{\partial{y}}(-1)\\ \frac{\partial{}}{\partial{y}}(\text{ln}(x^2 + xy + y^2)) + 0\\ \frac{\partial{}}{\partial{y}}(\text{ln}(x^2 + xy + y^2))\]
Using the Chain Rule with: \(u = x^2 +xy + y^2\):
\[\frac{\partial{}}{\partial{y}}(\text{ln}(x^2 + xy + y^2)) = \frac{\partial{}\text{ln}(u)}{\partial{u}}\frac{\partial{u}}{\partial{y}},\\ \text{where }u = x^2 + xy + y^2\text{ and } \frac{\partial{}}{\partial{u}}(ln(u)) = \frac{1}{u}\\ = \frac{\frac{\partial{}}{\partial{y}}(x^2 + xy + y^2)}{x^2 + xy + y^2}\\ = \frac{2y + x}{x^2 + xy + y^2}\]
Substituting into Theorem 12.5.3:
\[\frac{dy}{dx} = -\frac{f_x(x,y)}{f_y(x,y)}\\ \frac{dy}{dx} = -\frac{\frac{2x + y}{x^2 + xy + y^2}}{\frac{2y + x}{x^2 + xy + y^2}}\\ \frac{dy}{dx} = -(\frac{2x + y}{x^2 + xy + y^2})(\frac{x^2 + xy + y^2}{2y + x})\\ \frac{dy}{dx} = -\frac{2x + y}{2y + x}\]
# Using Deriv package
f <- function(x, y){log(x^2 + x*y + y^2) - 1}
(f_partial_x <- Deriv(f, "x"))## function (x, y)
## (2 * x + y)/(x * (x + y) + y^2)(f_partial_y <- Deriv(f, "y"))## function (x, y)
## (2 * y + x)/(x * (x + y) + y^2)This confirms our partial derivatives above are correct. To solve the Theorem 12.5.3 symbolically, I pulled in the SymEngine.R package which can handle symbolic math. Note: SymEngine lets us define our variables and create expressions directly.
# Using SymEngine.R package
old_options <- options(symengine.latex = TRUE, symengine.latex.center = TRUE)
use_vars(x, y, .quiet = TRUE)
f <- (log(x^2 + x*y + y^2) - 1)
(f_partial_x <- D(f, x))\[ \frac{\left(0.0 + y + 2.0 x^{1.0}\right)}{\left(x y + x^{2.0} + y^{2.0}\right)} \]
(f_partial_y <- D(f, y))\[ \frac{\left(0.0 + x + 2.0 y^{1.0}\right)}{\left(x y + x^{2.0} + y^{2.0}\right)} \]
(dx_dy <- -f_partial_x / f_partial_y)\[ \frac{-\left(0.0 + y + 2.0 x^{1.0}\right)}{\left(0.0 + x + 2.0 y^{1.0}\right)} \]
options(old_options)Other than the annoying floating point exponentials and unnecessary 0.0’s, this matches our analytical solution above.