Linearization in Two Variables: Total Differential and Tangent Plane

Published 9/9/2025

\begin{aligned} &\text{\textbf{Extension to Multiple Variables (Two-Variable Case)}}\\[8pt] &\text{Let } f:\mathbb{R}^2\to\mathbb{R}. \text{ We study the change of } f \text{ when moving from } (x_0,y_0) \text{ by } (h,k).\\[6pt] &\text{If } f \text{ is differentiable at } (x_0,y_0), \text{ there exist constants } A,B \text{ and an error } E(h,k) \text{ with}\\ &\qquad f(x_0+h,\;y_0+k)-f(x_0,y_0)=A\,h+B\,k+E(h,k),\\ &\qquad \lim\limits_{(h,k)\to(0,0)}\dfrac{E(h,k)}{\sqrt{h^2+k^2}}=0 \quad\big(\text{equivalently } E(h,k)=\varepsilon(h,k)\sqrt{h^2+k^2},\ \varepsilon\to0\big).\\[6pt] &\text{These constants are the partials: } A=\dfrac{\partial f}{\partial x}(x_0,y_0),\quad B=\dfrac{\partial f}{\partial y}(x_0,y_0).\\[10pt] &\text{\textbf{Total Differential \& Linearization}}\\[6pt] &df=\dfrac{\partial f}{\partial x}(x_0,y_0)\,dx+\dfrac{\partial f}{\partial y}(x_0,y_0)\,dy,\\ &\text{and the linear approximation (tangent plane) is}\\ &\qquad f(x_0+h,\;y_0+k)\approx f(x_0,y_0)+\dfrac{\partial f}{\partial x}(x_0,y_0)\,h+\dfrac{\partial f}{\partial y}(x_0,y_0)\,k.\\[4pt] &\text{In vector form: } f(x_0+h,y_0+k)\approx f(x_0,y_0)+\nabla f(x_0,y_0)\cdot\begin{bmatrix}h\\[2pt]k\end{bmatrix},\ \text{with error } o\!\big(\sqrt{h^2+k^2}\big).\\[6pt] &\text{For a chosen finite step } (\Delta x,\Delta y),\ \text{set } dx=\Delta x,\ dy=\Delta y \text{ to get } \Delta f\approx df.\\[12pt] &\text{\textbf{Why } A,B \text{ are the partials (Uniqueness Argument)}}\\[6pt] &\text{Fix } k=0:\ f(x_0+h,y_0)-f(x_0,y_0)=A\,h+E(h,0),\ \ \dfrac{E(h,0)}{|h|}\to 0.\\ &\text{Dividing by } h \text{ and letting } h\to 0 \text{ gives } A=\partial f/\partial x(x_0,y_0). \text{ Similarly with } h=0 \text{ for } B.\\[10pt] &\text{\textbf{Differentiability vs.\ Partial Derivatives}}\\[6pt] &\text{Differentiability is stronger than mere existence of partials. If the partials}\\ &\text{are continuous near } (x_0,y_0) \text{ (the usual } C^1 \text{ condition), then } f \text{ is differentiable at } (x_0,y_0).\\ &\text{Equivalently, differentiability means a \emph{single} linear map } L(h,k)=Ah+Bk \text{ fits all directions,}\\ &\text{with a remainder } o\!\big(\sqrt{h^2+k^2}\big).\\[10pt] &\text{\textbf{Directional Derivatives \& Gradient}}\\[6pt] &\text{For a unit vector } u=(u_1,u_2),\ \ D_u f(x_0,y_0)=\lim_{t\to 0}\dfrac{f((x_0,y_0)+tu)-f(x_0,y_0)}{t}.\\ &\text{If } f \text{ is differentiable at } (x_0,y_0),\ \ D_u f(x_0,y_0)=\nabla f(x_0,y_0)\cdot u.\\ &\text{Thus } \nabla f(x_0,y_0) \text{ gives the direction of steepest increase, with magnitude } \lVert\nabla f\rVert.\\[10pt] &\text{\textbf{Why }}\ D_{u} f=\nabla f\cdot u\ \text{ and why }\ \|\nabla f\|\ \text{ is the maximal rate}\\[6pt] &\text{Differentiability at } (x_0,y_0) \text{ gives one linear map } L \text{ with}\\ &\quad f(x_0+h,y_0+k)=f(x_0,y_0)+L(h,k)+r(h,k),\ \ r(h,k)=o\!\big(\sqrt{h^2+k^2}\big).\\ &\text{For } f:\mathbb{R}^2\to\mathbb{R},\ L(h,k)=f_x(x_0,y_0)h+f_y(x_0,y_0)k=\nabla f(x_0,y_0)\cdot(h,k).\\ &\text{Along } \gamma(t)=(x_0,y_0)+tu,\ \|u\|=1:\ f(\gamma(t))-f(x_0,y_0)=t\,\nabla f(x_0,y_0)\cdot u + r(tu).\\ &\text{Divide by } t \text{ and let } t\to 0:\ \ D_u f(x_0,y_0)=\nabla f(x_0,y_0)\cdot u.\\[6pt] &\text{By Cauchy--Schwarz, for unit } u:\ \ D_u f=\nabla f\cdot u\le \|\nabla f\|,\\ &\text{with equality when } u=\nabla f/\|\nabla f\|\ \ (\nabla f\neq 0).\ \text{Thus } \nabla f \text{ points in the }\\ &\text{direction of steepest increase and } \|\nabla f\| \text{ is the maximal directional rate.} \end{aligned}

\begin{aligned} &\text{\textbf{Example.}}\ \text{Let } f(x,y)=x^2+xy \text{ at } (x_0,y_0)=(2,1).\\[4pt] &\frac{\partial f}{\partial x}=2x+y,\quad \frac{\partial f}{\partial y}=x \ \Rightarrow\ \frac{\partial f}{\partial x}(2,1)=5,\ \frac{\partial f}{\partial y}(2,1)=2,\ f(2,1)=6.\\[6pt] &\text{Linearization near } (2,1):\ f(2+h,\,1+k)\approx 6+5h+2k.\\[6pt] &\text{Check for } h=k=0.1:\ f(2.1,1.1)=6.72,\ \text{linear } \approx 6+5(0.1)+2(0.1)=6.7,\ \text{error } 0.02.\\ &\text{Distance of step } \sqrt{h^2+k^2}=\sqrt{0.1^2+0.1^2}\approx 0.141,\ \text{and the error is } o\big(\sqrt{h^2+k^2}\big). \end{aligned}