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Total Differential. Gradient. Linearization in Two Variables. Part 1 - nerjik.com

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Total Differential. Gradient. Linearization in Two Variables. Part 1

Total Differential. Gradient. Linearization in Two Variables. Part 1

Almaz Khusnutdinov

Almaz Khusnutdinov•Math enthusiast, love coding and mathematics

Published 9/10/2025

Total Differential in Two Variables. Gradient Map

Problem 0.1

\text{Let }f:\mathbb{R}^2\to\mathbb{R}\text{ have partial derivatives continuous at }(x_0,y_0).\\ \text{Prove that }f\text{ is differentiable at }(x_0,y_0)\text{ and the total differential is }\\ dz=f_x(x_0,y_0)\,dx+f_y(x_0,y_0)\,dy.\\[6pt] \textbf{Outline and detailed explanation: }\\ \text{Set }\Delta z=f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0).\\ \text{Add and subtract }f(x_0+\Delta x,y_0):\\ \Delta z=\big[f(x_0+\Delta x,y_0)-f(x_0,y_0)\big]+\big[f(x_0+\Delta x,y_0+\Delta y)-f(x_0+\Delta x,y_0)\big].\\ \text{By the 1D mean value theorem in }x,\ \exists\,\xi\text{ between }x_0\text{ and }x_0+\Delta x:\\ f(x_0+\Delta x,y_0)-f(x_0,y_0)=f_x(\xi,y_0)\,\Delta x.\\ \text{By the 1D mean value theorem in }y,\ \exists\,\eta\text{ between }y_0\text{ and }y_0+\Delta y:\\ f(x_0+\Delta x,y_0+\Delta y)-f(x_0+\Delta x,y_0)=f_y(x_0+\Delta x,\eta)\,\Delta y.\\ \text{Hence } \Delta z=f_x(\xi,y_0)\,\Delta x+f_y(x_0+\Delta x,\eta)\,\Delta y.\\ \text{Continuity of }f_x,f_y\text{ at }(x_0,y_0)\Rightarrow f_x(\xi,y_0)\to f_x(x_0,y_0),\ f_y(x_0+\Delta x,\eta)\to f_y(x_0,y_0)\\ \text{as }(\Delta x,\Delta y)\to(0,0). \text{ Write }\\ \Delta z=f_x(x_0,y_0)\,\Delta x+f_y(x_0,y_0)\,\Delta y+r(\Delta x,\Delta y),\\ \text{where }r=\big(f_x(\xi,y_0)-f_x(x_0,y_0)\big)\Delta x+\big(f_y(x_0+\Delta x,\eta)-f_y(x_0,y_0)\big)\Delta y.\\ \text{Then } \displaystyle \frac{|r|}{\sqrt{(\Delta x)^2+(\Delta y)^2}}\to 0,\ \text{so }f\text{ is differentiable and }\\ dz:=f_x(x_0,y_0)\,dx+f_y(x_0,y_0)\,dy\ \text{is the total differential.}\\[6pt] \textbf{Answer: }dz=f_x(x_0,y_0)\,dx+f_y(x_0,y_0)\,dy.

Problem 0.2

\text{Let }f(x,y)=x^2y+e^{xy}. \text{ Find }dz\text{ at }(1,0),\ \text{the linearization }L(x,y)\text{ there, and estimate }f(1.02,-0.01).\\[4pt] \textbf{Answer: }dz=\big(0\big)\,dx+\big(2\big)\,dy,\ L(x,y)=1+2y,\ f(1.02,-0.01)\approx 0.98.\\[6pt] \textit{Note: }L(x,y)\ \text{denotes the affine linearization at }a=(1,0):\ L(x,y)=f(a)+Df(a)[(x-a,y-b)].\ \text{It is not the linear map on a displacement.}\\ \text{Common split: }T_a(x)=f(a)+Df(a)[x-a]\ \text{(affine value)},\quad L(h)=Df(a)[h]\ \text{(linear change)}.

Problem 0.3

\text{Let }f(x,y)=\ln(1+x+2y). \text{ Along }x(t)=t^2,\ y(t)=\sin t,\ \text{use }dz\text{ to find }\left.\frac{dz}{dt}\right|_{t=0}\\ \text{and a first-order approximation to }f(x(t),y(t))\text{ for small }t.\\[4pt] \textbf{Answer: }\left.\frac{dz}{dt}\right|_{t=0}=2,\ \ f(x(t),y(t))\approx 0+2t.

Problem 0.4

\text{Let }f(x,y)=\sqrt{x^2+4y^2}. \text{ At }(3,4)\text{ use }dz\text{ to predict the change for }dx=-0.03,\ dy=0.02.\\[4pt] \textbf{Answer: }dz=\frac{3}{\sqrt{73}}\,dx+\frac{16}{\sqrt{73}}\,dy,\ \ \Delta z\approx \frac{-0.09+0.32}{\sqrt{73}}\approx 0.0269.

Problem 0.5

\text{Let }f(x,y)=x^3+y^3\text{ and take a step }(dx,dy)\text{ from }(x,y). \text{ Compute }dz\text{ and the exact remainder }R=\Delta f-dz.\\[4pt] \textbf{Answer: }dz=3x^2\,dx+3y^2\,dy,\ \ R=3x\,dx^2+dx^3+3y\,dy^2+dy^3.

Linearization in Two variables.

Problem 1.1

\text{Let } f(x,y)=\sqrt{\,1+x+2y\,}.\ \text{Use differentials to estimate the increase in } f \text{ when } (x,y):(0,0)\to(0.04,0.02).\\ \text{Also estimate the relative change } \dfrac{\Delta f}{f(0,0)}.\\[6pt] \textbf{Answer: }\Delta f\approx 0.04,\quad \dfrac{\Delta f}{f}\approx 0.04\ (4\%).

Problem 1.2

\text{Let } f(x,y)=x^{2}e^{y}.\ \text{Use differentials to estimate the increase in } f \text{ when } (x,y):(1,0)\to(1.10,-0.05).\\ \text{Also estimate the relative change } \dfrac{\Delta f}{f(1,0)}.\\[6pt] \textbf{Answer: }\Delta f\approx 0.15,\quad \dfrac{\Delta f}{f}\approx 0.15\ (15\%).

Problem 1.3

\text{Let } f(x,y)=e^{x}\cos y.\ \text{Use differentials to estimate the increase in } f \text{ when } (x,y):(0,0)\to(0.06,0.02).\\ \text{Also estimate the relative change } \dfrac{\Delta f}{f(0,0)}.\\[6pt] \textbf{Answer: }\Delta f\approx 0.06,\quad \dfrac{\Delta f}{f}\approx 0.06\ (6\%).

Approximations

Problem 2.1

\text{Let } f(x,y)=e^{xy}\sin(x-y).\ \text{Use differentials to estimate the change in } f \text{ when } (x,y):(1,0)\to(0.97,0.04).\\ \text{Also estimate the relative change } \dfrac{\Delta f}{f(1,0)}.\\[6pt] \textbf{Answer: }\Delta f\approx -0.00416,\quad \dfrac{\Delta f}{f}\approx -0.00495\ (\!-0.495\%).

Problem 2.2

\text{Let } f(x,y)=\ln\!\big(1+x^{2}+2y^{2}\big).\ \text{From } (1,1),\ \text{estimate (via differentials) the maximal possible increase in } f \text{ over any step of length } s=0.05.\\ \text{Also report a unit direction achieving it (angle from the } +x\text{-axis}).\\[6pt] \textbf{Answer: }\Delta f_{\max}\approx 0.05590,\ \ u\approx(0.4472136,\,0.8944272)\ \ (\theta\approx 1.1071\ \text{rad}).

Problem 2.3

\text{Let } F(x,y)=x e^{y}+y^{3}-3=0 \text{ define } y=y(x) \text{ near } (3,0).\ \text{Use differentials to estimate the new } y \text{ when } x:3\to 3.06.\\ \text{Also report the predicted } \Delta y.\\[6pt] \textbf{Answer: } y\approx -0.02,\quad \Delta y\approx -0.02.

Problem 2.4

\text{Let } f(x,y)=\arctan(xy)+e^{x-y}.\ \text{Move along the line } y=2x \text{ starting at } (1,2) \text{ with } \Delta x=0.03.\ \text{Use differentials to estimate } \Delta f.\\ \text{Also estimate the relative change } \dfrac{\Delta f}{f(1,2)}.\\[6pt] \textbf{Answer: }\Delta f\approx 0.01296,\quad \dfrac{\Delta f}{f}\approx 0.00879\ (0.879\%).

Problem 2.5

\text{Let } f(x,y)=x^{2}y+\ln(1+x-y).\ \text{From } (1,0),\ \text{suppose } |\Delta x|\le 0.02,\ |\Delta y|\le 0.03.\ \text{Use differentials to estimate the maximal increase in } f.\\ \text{Also give an increment } (\Delta x,\Delta y) \text{ that attains it.}\\[6pt] \textbf{Answer: }\Delta f_{\max}\approx 0.025,\ \ \text{attained by }\Delta x=+0.02,\ \Delta y=+0.03.

Problem 2.6

\text{Let }f(x,y)=e^{\,x-y^2}\cos(2x+y).\ \text{At }(0,0)\text{ find }dz,\ \text{the linearization }T(x,y),\ \text{and estimate }f(0.02,-0.01).\\[6pt] \textbf{Answer: }f(0,0)=1,\ f_x(0,0)=1,\ f_y(0,0)=0;\ dz=1\,dx+0\,dy;\ T(x,y)=1+(x-0)+0\cdot(y-0)=1+x;\\ f(0.02,-0.01)\approx T(0.02,-0.01)=1.02.\\[6pt]

Problem 2.7

\text{Let }f:\mathbb{R}^2\to\mathbb{R}\text{ have partial derivatives continuous at }(a,b).\ \text{Here we fix the notation: }\\ a=(a,b)\in\mathbb{R}^2,\quad h=(h_1,h_2)\in\mathbb{R}^2,\quad a+h:=(a+h_1,\ b+h_2).\\[6pt] \text{Prove that }f\text{ is differentiable at }a\text{ and }f(a+h)-f(a)=f_x(a,b)\,h_1+f_y(a,b)\,h_2+o(\|h\|).\\[6pt] \textbf{Answer (outline): }\\ \text{Define the increment } \Delta f:=f(a+h)-f(a)=f(a+h_1,b+h_2)-f(a,b).\\ \text{Insert and subtract the middle term }f(a+h_1,b)\ \text{(purely algebraic identity }X-Z=(X-Y)+(Y-Z)\text{):}\\ \Delta f=\underbrace{\big[f(a+h_1,b)-f(a,b)\big]}_{\text{move in }x\text{ only}}+\underbrace{\big[f(a+h_1,b+h_2)-f(a+h_1,b)\big]}_{\text{then move in }y}.\\ \text{Apply 1D MVT to each bracket: } \exists\,\xi\text{ between }a\text{ and }a+h_1,\ \exists\,\eta\text{ between }b\text{ and }b+h_2\ \text{such that}\\ f(a+h_1,b)-f(a,b)=f_x(\xi,b)\,h_1,\qquad f(a+h_1,b+h_2)-f(a+h_1,b)=f_y(a+h_1,\eta)\,h_2.\\ \Rightarrow\ \Delta f=f_x(\xi,b)\,h_1+f_y(a+h_1,\eta)\,h_2.\\ \text{Continuity of }f_x,f_y\text{ at }(a,b)\text{ gives }f_x(\xi,b)\to f_x(a,b),\ f_y(a+h_1,\eta)\to f_y(a,b)\ \text{as }h\to 0.\\ \text{Write } \Delta f=f_x(a,b)\,h_1+f_y(a,b)\,h_2+\underbrace{\Big([f_x(\xi,b)-f_x(a,b)]\,h_1+[f_y(a+h_1,\eta)-f_y(a,b)]\,h_2\Big)}_{=:r(h)}.\\ \text{Then } \displaystyle \frac{|r(h)|}{\|h\|}\to 0,\ \text{so }f\text{ is differentiable at }a,\ \text{and }dz=f_x(a,b)\,dx+f_y(a,b)\,dy.\\[6pt] \textit{Notation reminders: }a\text{ is the fixed base point; }h\text{ is the displacement; }a+h=(a+h_1,b+h_2).\ \Delta f\text{ is split by adding and subtracting }f(a+h_1,b).

Problem 2.8

\text{Let }f(x,y)=\ln(1+x+2y).\ \text{On the level set }f(x,y)=0\text{ near }(0,0),\ \text{use }dz=0\text{ to find a linear approximation }y(x).\\[6pt] \textbf{Answer: }f_x(0,0)=1,\ f_y(0,0)=2.\ \ dz=f_x\,dx+f_y\,dy=dx+2\,dy=0\Rightarrow dy/dx=-\tfrac{1}{2}.\\ y(x)\approx -\tfrac{1}{2}x\ \text{for }x\text{ near }0.

Problem 2.9

\text{Uncertainty propagation.}\ z=\dfrac{x^2y}{\ 1+x\ }.\ \text{At }(x,y)=(2,3)\ \text{with }|dx|\le 0.01,\ |dy|\le 0.02,\ \text{estimate }|\,\Delta z\,| \text{ via }dz.\\[6pt] \textbf{Answer: }f_x=\dfrac{(2xy)(1+x)-x^2y}{(1+x)^2}=\dfrac{xy(2+ x)}{(1+x)^2},\ f_y=\dfrac{x^2}{1+x}.\\ f_x(2,3)=\dfrac{3\cdot 2\cdot(2+2)}{(1+2)^2}=\dfrac{24}{9}=\tfrac{8}{3},\ \ f_y(2,3)=\dfrac{4}{3}.\\ |dz|\le |f_x||dx|+|f_y||dy|\le \tfrac{8}{3}\cdot 0.01+\tfrac{4}{3}\cdot 0.02=\tfrac{8+8}{300}=\tfrac{16}{300}\approx 0.0533.\\ z(2,3)=\dfrac{4\cdot 3}{3}=4\ \Rightarrow\ z\approx 4\pm 0.0533.

Problem 2.10

\text{Let }f(x,y)=\sqrt{x^2+4y^2},\ x(t)=1+0.1t^2,\ y(t)=2-0.05t.\ \text{Find }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}\ \text{and the first-order }t\text{-approximation.}\\[6pt] \textbf{Answer: }f_x=\dfrac{x}{\sqrt{x^2+4y^2}},\ f_y=\dfrac{4y}{\sqrt{x^2+4y^2}}.\ (x,y)=(1,2)\Rightarrow f_x=\tfrac{1}{\sqrt{17}},\ f_y=\tfrac{8}{\sqrt{17}}.\\ x'(0)=0,\ y'(0)=-0.05\ \Rightarrow\ \dfrac{d}{dt}f(x(t),y(t))\big|_{0}=f_x\cdot 0+f_y\cdot(-0.05)=-\dfrac{0.4}{\sqrt{17}}.\\ f(x(t),y(t))\approx f(1,2)+\Big(-\dfrac{0.4}{\sqrt{17}}\Big)t=\sqrt{17}-\dfrac{0.4}{\sqrt{17}}\,t.

Problem 2.11

\text{Let }f(x,y)=x^2y.\ \text{At }(x,y)=(3,1)\text{ and step }(dx,dy)=(-0.03,\,0.02),\ \text{compute }dz\text{ and compare with the exact }\Delta f.\\[6pt] \textbf{Answer: }f_x=2xy,\ f_y=x^2.\ \ dz=2xy\,dx+x^2\,dy=2\cdot3\cdot1\cdot(-0.03)+9\cdot0.02=-0.18+0.18=0.\\ \Delta f=f(2.97,1.02)-f(3,1)=(2.97)^2\cdot1.02-9=8.8209\cdot1.02-9=8.997318-9=-0.002682.\\ \text{Exact remainder: }\Delta f-dz=\underbrace{2x\,dx\,dy+y\,dx^2+dx^2dy}_{\text{for }f=x^2y}.\ \text{At }(3,1):\ -0.0036+0.0009+0.000018=-0.002682.

Problem 2.12

\text{Let }f(x,y)=e^{xy},\ x(t)=2-0.1t,\ y(t)=-1+0.2t^2.\ \text{Find }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}\ \text{and the first-order }t\text{-approximation.}\\[6pt] \textbf{Answer: }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}=0.1e^{-2},\quad f(x(t),y(t))\approx e^{-2}+0.1e^{-2}\,t.

Problem 2.13

\text{Let }f(x,y)=\ln\!\big(1+x^{2}+3y\big),\ x(t)=-1+t,\ y(t)=1-t^{2}.\ \text{Find }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}\ \text{and the first-order }t\text{-approximation.}\\[6pt] \textbf{Answer: }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}=-\tfrac{2}{5},\quad f(x(t),y(t))\approx \ln 5-\tfrac{2}{5}\,t.

Problem 2.14

\text{Let }f(x,y)=\sqrt{\,1+xy^{2}\,},\ x(t)=4+0.5t,\ y(t)=1-0.1t.\ \text{Find }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}\ \text{and the first-order }t\text{-approximation.}\\[6pt] \textbf{Answer: }\left.\dfrac{d}{dt}f(x(t),y(t))\right|_{t=0}=-\dfrac{3}{20\sqrt{5}},\quad f(x(t),y(t))\approx \sqrt{5}-\dfrac{3}{20\sqrt{5}}\,t.

General Approximation in case of N-variables

Problem 3.1

\text{Let }f(x,y,z)=\sqrt{\,1+2x-y+3z^2\,}\ \text{and set }w=f(x,y,z).\ \text{At }a=(0,0,0)\text{ find }dw,\ \text{the linearization }T(x,y,z),\ \text{and estimate }f(0.02,0.01,-0.01).\\[6pt] \textbf{Answer: }f_x(a)=1,\ f_y(a)=-\tfrac12,\ f_z(a)=0.\quad dw=1\,dx-\tfrac12\,dy+0\cdot dz,\quad T(x,y,z)=1+(x-0)-\tfrac12(y-0)=1+x-\tfrac12 y,\\ f(0.02,0.01,-0.01)\approx T(0.02,0.01,-0.01)=1.015.\\[6pt] \textit{Note: }w=f(x,y,z)\ \text{is the output; }dw\ \text{is its differential. Inputs have }dx,dy,dz.

Problem 3.2

\text{Quadratic model in }\mathbb{R}^4.\ f(x)=\tfrac12 x^\top A x+b\cdot x,\ A=\mathrm{diag}(2,1,3,4),\ b=(1,-2,0,3).\\ \text{At }a=(1,1,1,1)\text{ find }dz\text{ and the linearization }T(x).\\[4pt] \textbf{Answer: }\nabla f(x)=Ax+b\ \Rightarrow\ \nabla f(a)=(3,-1,3,7).\\ dz=\nabla f(a)\cdot h=(3,-1,3,7)\cdot h,\quad f(a)=7,\quad T(x)=7+(3,-1,3,7)\cdot(x-a).

Problem 3.3

\text{Norm of a linear image in }\mathbb{R}^3.\ f(x)=\|Mx\|,\ M=\mathrm{diag}(1,2,3).\ \text{At }a=(1,1,1)\\ \text{find }dz\text{ and the directional derivative along }v=(1,-2,2).\\[4pt] \textbf{Answer: }Mx=(x_1,2x_2,3x_3),\ \nabla f(x)=\dfrac{M^\top Mx}{\|Mx\|}=\dfrac{(x_1,4x_2,9x_3)}{\sqrt{x_1^2+4x_2^2+9x_3^2}}.\\ \nabla f(a)=\dfrac{(1,4,9)}{\sqrt{14}},\ dz=\nabla f(a)\cdot h.\\ D_v f(a)=\nabla f(a)\cdot \dfrac{v}{\|v\|}=\dfrac{1\cdot1+4(-2)+9\cdot2}{3\sqrt{14}}=\dfrac{11}{3\sqrt{14}}.

Problem 3.4

\text{Error bound in }\mathbb{R}^4.\ f(x)=\sin(2x_1-x_2+3x_4). \text{ At }a=0, \text{ with }|h_1|\le 0.01,\ |h_2|\le 0.02,\\ |h_3|\le 0.01,\ |h_4|\le 0.005,\ \text{give a first-order bound on }|\Delta f|.\\[4pt] \textbf{Answer: }dz=\nabla f(a)\cdot h=(2,-1,0,3)\cdot h\ \Rightarrow\ |dz|\le 2|h_1|+|h_2|+3|h_4|\le 0.055.\\ \text{So }|\Delta f|\approx |dz|\le 0.055\ \text{(first-order bound).}

Problem 3.5

\text{Linearized constraint in }\mathbb{R}^3.\ G(x_1,x_2,x_3)=x_1^2+2x_2-x_3-5=0.\\ \text{Near }a=(2,1,1)\text{ find the linearized formula for }x_3\text{ as a function of }(x_1,x_2).\\[4pt] \textbf{Answer: }dG=2x_1\,dx_1+2\,dx_2-dx_3.\ \text{At }a:\ 2\cdot2\,dx_1+2\,dx_2-dx_3=0\Rightarrow dx_3=4\,dx_1+2\,dx_2.\\ x_3\approx 1+4(x_1-2)+2(x_2-1).

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