proof of multivariable chain rule

Have questions or comments? This is called a tree diagram for the chain rule for functions of one variable and it provides a way to remember the formula (Figure \(\PageIndex{1}\)). \end{equation*}, Theorem. Suppose that f is differentiable at the point \(\displaystyle P(x_0,y_0),\) where \(\displaystyle x_0=g(t_0)\) and \(\displaystyle y_0=h(t_0)\) for a fixed value of \(\displaystyle t_0\). By the chain rule, \begin{align} \frac{\partial u}{\partial s} & = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial u}{\partial z}\frac{\partial z}{\partial s} \\ & =\left(4x^3y\right)\left(r e^t\right)+\left(x^4+2y z^3\right)\left(2r s e^{-t}\right)+\left(3y^2z^2\right)\left(r^2\sin t\right).\end{align} When $r=2,$ $s=1,$ and $t=0,$ we have $x=2,$ $y=2,$ and $z=0,$ so \begin{equation} \frac{\partial u}{\partial s}=(64)(2)+(16)(4)+(0)(0)=192. Let’s now return to the problem that we started before the previous theorem. The basic concepts are illustrated through a simple example. Exercise. Tree diagrams are useful for deriving formulas for the chain rule for functions of more than one variable, where each independent variable also depends on other variables. \end{equation} At what rate is the distance between the two objects changing when $t=\pi ?$, Solution. In this section we’re going to prove many of the various derivative facts, formulas and/or properties that we encountered in the early part of the Derivatives chapter. Calculate \(\displaystyle ∂z/∂u\) and \(\displaystyle ∂z/∂v\) using the following functions: \[\displaystyle z=f(x,y)=3x^2−2xy+y^2,\; x=x(u,v)=3u+2v,\; y=y(u,v)=4u−v. Calculate nine partial derivatives, then use the same formulas from Example \(\PageIndex{3}\). However, we can get a better feel for it … This diagram can be expanded for functions of more than one variable, as we shall see very shortly. Calculate \(\displaystyle dz/dt\) for each of the following functions: a. Let $w=u^2v^2$, so $z=u+f(w).$ Then according to the chain rule, \begin{equation} \frac{\partial z}{\partial u}=1+\frac{d f}{d w}\frac{\partial w}{\partial u}=1+f'(w)\left(2u v^2\right)\end{equation} and \begin{equation}\frac{\partial z}{\partial v}=1+\frac{d f}{d w}\frac{\partial w}{\partial v}=f'(w)\left(2u^2 v\right) \end{equation} so that \begin{align} u\frac{\partial z}{\partial u}-v\frac{\partial z}{\partial v} &=u\left[1+f'(w)\left(2u v^2\right)\right]-v\left[f'(w)\left(2u^2v\right)\right] \\ & =u+f'(w)\left[u\left(2u v^2\right)-v\left(2u^2v\right)\right] =u. The proof of this chain rule is motivated by appealing to a previously proven chain rule with one independent variable. \nonumber\], \[\begin{align*} \lim_{t→t_0}\dfrac{E(x(t),y(t))}{t−t_0} =\lim_{t→t_0}\dfrac{(E(x,y)}{\sqrt{(x−x_0)^2+(y−y_0)^2}}\dfrac{\sqrt{(x−x_0)^2+(y−y_0)^2}}{t−t_0}) \\[4pt] =\lim_{t→t_0}\left(\dfrac{E(x,y)}{\sqrt{(x−x_0)^2+(y−y_0)^2}}\right)\lim_{t→t_0}\left(\dfrac{\sqrt{(x−x_0)^2+(y−y_0)^2}}{t−t_0}\right). Recall that when multiplying fractions, cancelation can be used. The proof of Part II follows quickly from Part I, ... T/F: The Multivariable Chain Rule is only useful when all the related functions are known explicitly. This gives us Equation. If $z=e^x\sin y$ where $x=s t^2$ and $y=s^2t$, find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}.$, Solution. Therefore, this value is finite. \[\dfrac { d y } { d x } = \left. (Chain Rule Involving One Independent Variable) Let $f(x,y)$ be a differentiable function of $x$ and $y$, and let $x=x(t)$ and $y=y(t)$ be differentiable functions of $t.$ Then $z=f(x,y)$ is a differentiable function of $t$ and \begin{equation} \label{criindevar} \frac{d z}{d t}=\frac{\partial z}{\partial x}\frac{d x}{d t}+\frac{\partial z}{\partial y}\frac{d y}{d t}. The reason is that, in Note, \(\displaystyle z\) is ultimately a function of \(\displaystyle t\) alone, whereas in Note, \(\displaystyle z\) is a function of both \(\displaystyle u\) and \(\displaystyle v\). \end{equation} Because $x$ and $y$ are function of $t$, we can write their increments as \begin{equation} \Delta x=x(t+\Delta t) -x(t) \qquad \text{and} \qquad \Delta y=y(t+\Delta t)-y(t).\end{equation} We know that $x$ and $y$ vary continuously with $t$, because $x$ and $y$ are differentiable, and it follows that $\Delta x\to 0$ and $\Delta y\to 0$ as $ \Delta t\to 0$ so that $\epsilon_1\to 0$ and $\epsilon_2\to 0$ as $\Delta t\to 0.$ Therefore, we have \begin{align} \frac{d z}{d t} & =\lim_{\Delta t\to 0}\frac{\Delta z}{\Delta t} \\ & =\lim_{\Delta t\to 0}\left(\frac{\partial z}{\partial x}\frac{\Delta x}{\Delta t}+\frac{\partial z}{\partial y}\frac{\Delta y}{\Delta t}+\epsilon_1\frac{\Delta x}{\Delta t}+\epsilon_2\frac{\Delta y}{\Delta t}\right) \\ & =\frac{\partial z}{\partial x}\frac{d x}{d t}+\frac{\partial z}{\partial y}\frac{d y}{d t}+(0)\frac{\Delta x}{\Delta t}+(0)\frac{\Delta y}{\Delta t} \end{align}as desired. Evaluating at the point (3,1,1) gives 3(e1)/16. Recall from implicit differentiation provides a method for finding \(\displaystyle dy/dx\) when \(\displaystyle y\) is defined implicitly as a function of \(\displaystyle x\). Solution A: We'll use theformula usingmatrices of partial derivatives:Dh(t)=Df(g(t))Dg(t). Includes full solutions and score reporting. This pattern works with functions of more than two variables as well, as we see later in this section. \nonumber\]. Then we take the limit as \(\displaystyle t\) approaches \(\displaystyle t_0\): \[\begin{align*} \lim_{t→t_0}\dfrac{z(t)−z(t_0)}{t−t_0} = f_x(x_0,y_0)\lim_{t→t_0} \left (\dfrac{x(t)−x(t_0)}{t−t_0} \right) \\[4pt] +f_y(x_0,y_0)\lim_{t→t_0}\left (\dfrac{y(t)−y(t_0)}{t−t_0}\right)\\[4pt] +\lim_{t→t_0}\dfrac{E(x(t),y(t))}{t−t_0}. \end{align*}\], \[\displaystyle w=f(x,y),x=x(t,u,v),y=y(t,u,v) \nonumber\], and write out the formulas for the three partial derivatives of \(\displaystyle w.\). The chain rule for single-variable functions states: if g is differentiable at and f is differentiable at , then is differentiable at and its derivative is: The proof of the chain rule is a bit tricky - I left it for the appendix. This field is for validation purposes and should be left unchanged. If $u=x^4y+y^2z^3$ where $x=r s e^t,$ $y=r s^2e^{-t},$ and $z=r^2s \sin t,$ find the value of $\frac{\partial u}{\partial s}$ when $r=2,$ $s=1,$ and $t=0. If $z=x^2y+3x y^4,$ where $x=e^t$ and $y=\sin t$, find $\frac{d z}{d t}.$. EXPECTED SKILLS: The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \end{equation}, Example. h→0. Theorem (Chain rule) Assume that \( x,y:\mathbb R\to\mathbb R \) are differentiable at point \( t_0 \). which is the same solution. Now suppose that \(\displaystyle f\) is a function of two variables and \(\displaystyle g\) is a function of one variable. Express the final answer in terms of \(\displaystyle t\). To eliminate negative exponents, we multiply the top by \(\displaystyle e^{2t}\) and the bottom by \(\displaystyle \sqrt{e^{4t}}\): \[\begin{align*} \dfrac{dz}{dt} =\dfrac{2e^{4t}+e^{−2t}}{\sqrt{e^{4t}−e^{−2t}}}⋅\dfrac{e^{2t}}{\sqrt{e^{4t}}} \\[4pt] =\dfrac{2e^{6t}+1}{\sqrt{e^{8t}−e^{2t}}} \\[4pt] =\dfrac{2e^{6t}+1}{\sqrt{e^{2t}(e^{6t}−1)}} \\[4pt] =\dfrac{2e^{6t}+1}{e^t\sqrt{e^{6t}−1}}. This equation implicitly defines \(\displaystyle y\) as a function of \(\displaystyle x\). \end{align*}\]. Chain rule for functions of 2, 3 variables (Sect. Calculate \(\displaystyle ∂w/∂u\) and \(\displaystyle ∂w/∂v\) using the following functions: \[\begin{align*} w =f(x,y,z)=3x^2−2xy+4z^2 \\[4pt] x =x(u,v)=e^u\sin v \\[4pt] y =y(u,v)=e^u\cos v \\[4pt] z =z(u,v)=e^u. Solution. This branch is labeled \(\displaystyle (∂z/∂y)×(dy/dt)\). The upper branch corresponds to the variable \(\displaystyle x\) and the lower branch corresponds to the variable \(\displaystyle y\). \end{align*}\]. Calculate \(\displaystyle ∂w/∂u\) and \(\displaystyle ∂w/∂v\) given the following functions: \[\begin{align*} w =f(x,y,z)=\dfrac{x+2y−4z}{2x−y+3z} \\[4pt] x =x(u,v)=e^{2u}\cos3v \\[4pt] y =y(u,v)=e^{2u}\sin 3v \\[4pt] z =z(u,v)=e^{2u}. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. \end{align}, Example. Example. I Functions of two variables, f : D ⊂ R2 → R. I Chain rule for functions defined on a curve in a plane. f(g(x+h))−f(g(x)) h . We will do it for compositions of functions of two variables. If $u=f(x,y),$ where $x=e^s \cos t$ and $y=e^s \sin t,$ show that \begin{equation} \frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=e^{-2s}\left[\frac{\partial ^2u}{\partial s^2}+\frac{\partial ^2u}{\partial t^2}\right]. \(\displaystyle \dfrac{∂z}{∂u}=0,\dfrac{∂z}{∂v}=\dfrac{−21}{(3\sin 3v+\cos 3v)^2}\). 2 Chain rule for two sets of independent variables If u = u(x,y) and the two independent variables x,y are each a function of two new independent variables s,tthen we want relations between their partial derivatives. \nonumber\]. This means that if t is changes by a small amount from 1 while x is held fixed at 3 and q at 1, the value of f … THE CHAIN RULE - Multivariable Differential Calculus - Beginning with a discussion of Euclidean space and linear mappings, Professor Edwards (University of Georgia) follows with a thorough and detailed exposition of multivariable differential and integral calculus. (Chain Rule Involving Two Independent Variables) Suppose $z=f(x,y)$ is a differentiable function at $(x,y)$ and that the partial derivatives of $x=x(u,v)$ and $y=y(u,v)$ exist at $(u,v).$ Then the composite function $z=f(x(u,v),y(u,v))$ is differentiable at $(u,v)$ with \begin{equation} \frac{\partial z}{\partial u}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \qquad \text{and} \qquad \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}. +\frac{\partial u}{\partial x}\left(-e^s \cos t\right)+\left[\frac{ \partial ^2 u}{\partial x^2}\left(-e^s \sin t\right)+\frac{ \partial ^2 u}{\partial x \partial y}e^s \cos t\right]\left(-e^s \sin t\right) \right. As for your second question, one doesn't- what you have written is not true. Since \(\displaystyle f\) has two independent variables, there are two lines coming from this corner. Theorem. Theorem 1. In the last couple videos, I talked about this multivariable chain rule, and I give some justification. \label{chian2b}\]. \end{align*}\], Next, we substitute \(\displaystyle x(u,v)=3u+2v\) and \(\displaystyle y(u,v)=4u−v:\), \[\begin{align*} \dfrac{∂z}{∂u} =10x+2y \\[4pt] =10(3u+2v)+2(4u−v) \\[4pt] =38u+18v. To find the equation of the tangent line, we use the point-slope form (Figure \(\PageIndex{5}\)): \[\begin{align*} y−y_0 =m(x−x_0)\\[4pt]y−1 =\dfrac{7}{4}(x−2) \\[4pt] y =\dfrac{7}{4}x−\dfrac{7}{2}+1\\[4pt] y =\dfrac{7}{4}x−\dfrac{5}{2}.\end{align*}\]. This proves the chain rule at \(\displaystyle t=t_0\); the rest of the theorem follows from the assumption that all functions are differentiable over their entire domains. Multivariable Calculus. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. What is the equation of the tangent line to the graph of this curve at point \(\displaystyle (2,1)\)? \end{align*}\]. 14.4) I Review: Chain rule for f : D ⊂ R → R. I Chain rule for change of coordinates in a line. Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables. \\ & \hspace{2cm} \left. $$, Exercise. Sometimes you will need to apply the Chain Rule several times in order to differentiate a function. +\frac{\partial u}{\partial y}e^s \sin t +\frac{ \partial ^2 u}{\partial x \partial y}\left(e^{2s}\sin t \cos t\right) +\frac{ \partial ^2 u}{\partial y^2}e^{2s} \sin ^2 t +\frac{\partial u}{\partial x}\left(-e^s \cos t\right) \right. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 14.5: The Chain Rule for Multivariable Functions, [ "article:topic", "generalized chain rule", "intermediate variable", "license:ccbyncsa", "showtoc:no", "authorname:openstaxstrang" ], \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 14.4: Tangent Planes and Linear Approximations, 14.6: Directional Derivatives and the Gradient, Massachusetts Institute of Technology (Strang) & University of Wisconsin-Stevens Point (Herman), Chain Rules for One or Two Independent Variables. Applying the chain rule we obtain \begin{align} \frac{\partial z}{\partial s} & =\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z} {\partial y}\frac{\partial y}{\partial s} \\ & =\left(e^x\sin y\right)\left(t^2\right)+\left(e^x\cos y\right)( s t) \\ & =t^2e^{s t^2}\sin \left(s^2 t\right)+2s t e^{s t^2}\cos \left(s^2t\right) \end{align} and \begin{align} \frac{\partial z}{\partial t} &=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t} \\ & =\left(e^x\sin y\right)(2 s t)+\left(e^x\cos y\right)\left(2 s^2\right) \\ & =2 s t e^{s t^2}\sin \left(s^2 t\right)+s^2 e^{s t^2}\cos \left(s^2t\right). \end{equation}, Example. It actually is a product of derivatives, just like in the single-variable case, the difference is that this time it is a matrix product. \\ & \hspace{2cm} \left. Thus, for a differentiable function f, we can write Δy = f’(a) Δx + ε Δx, where ε 0 as x 0 (1) •and ε is a continuous function of Δx. Calculate \(\displaystyle ∂f/dx\) and \(\displaystyle ∂f/dy\), then use Equation \ref{implicitdiff1}. \frac { 2 x + y + 7 } { 2 y - x + 3 } \right| _ { ( 3 , - 2 ) } = \dfrac { 2 ( 3 ) + ( - 2 ) + 7 } { 2 ( - 2 ) - ( 3 ) + 3 } = - \dfrac { 11 } { 4 } \nonumber\], Equation of the tangent line: \(\displaystyle y=−\dfrac{11}{4}x+\dfrac{25}{4}\), \(\displaystyle \dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}\), \(\displaystyle \dfrac{dz}{du}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u}\dfrac{dz}{dv}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂v}\), \(\displaystyle \dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_1}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\). Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Dave4Math » Calculus 3 » Chain Rule for Multivariable Functions. Free practice questions for Calculus 3 - Multi-Variable Chain Rule. Or perhaps they are both functions of two variables, or even more. +\frac{ \partial ^2 u}{\partial x^2}\left(e^{2s} \sin ^2 t\right) +\frac{ \partial ^2 u}{\partial x \partial y}\left(-e^{2s} \cos t \sin t\right) \right. The idea is the same for other combinations of flnite numbers of variables. To find \(\displaystyle ∂z/∂v,\) we use Equation \ref{chain2b}: \[\begin{align*} \dfrac{∂z}{∂v} =\dfrac{∂z}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂v} \\[4pt] =2(6x−2y)+(−1)(−2x+2y) \\[4pt] =14x−6y. \\ & \hspace{2cm} \left. The Multivariable Chain Rule Nikhil Srivastava February 11, 2015 The chain rule is a simple consequence of the fact that di erentiation produces the linear approximation to a function at a point, and that the derivative is the coe cient appearing in this linear approximation. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Consider the ellipse defined by the equation \(\displaystyle x^2+3y^2+4y−4=0\) as follows. Figure 12.5.2 Understanding the application of the Multivariable Chain Rule. Last, each of the branches on the far right has a label that represents the path traveled to reach that branch. Next we work through an example which illustrates how to find partial derivatives of two variable functions whose variables are also two variable functions. Let \(\displaystyle w=f(x_1,x_2,…,x_m)\) be a differentiable function of \(\displaystyle m\) independent variables, and for each \(\displaystyle i∈{1,…,m},\) let \(\displaystyle x_i=x_i(t_1,t_2,…,t_n)\) be a differentiable function of \(\displaystyle n\) independent variables. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. If $w=f\left(\frac{r-s}{s}\right),$ show that $$ r\frac{\partial w}{\partial r}+s\frac{\partial w}{\partial s}=0.$$, Exercise. Solving this equation for \(\displaystyle dy/dx\) gives Equation \ref{implicitdiff1}. Theorem. \end{align*} \]. Exercise. By the chain rule we have \begin{align} \frac{\partial u}{\partial s} & =\frac{\partial u}{\partial x}\frac{\partial x}{\partial s} +\frac{\partial u}{\partial y}\frac{\partial y}{\partial s} \\ & =\frac{\partial u}{\partial x}e^s \cos t +\frac{\partial u}{\partial y}e^s \sin t \end{align} and \begin{align} \frac{\partial u}{\partial t} =\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t} =\frac{\partial u}{\partial x}\left(-e^s \sin t\right) +\frac{\partial u}{\partial y}e^s \cos t. \end{align} Therefore \begin{equation} \frac{ \partial ^2 u}{\partial s^2}=\frac{\partial u}{\partial x}e^s \cos t +\frac{\partial }{\partial s}\left(\frac{\partial u}{\partial x}\right)e^s \cos t +\frac{\partial u}{\partial y}e^s \sin t +\frac{\partial }{\partial s}\left(\frac{\partial u}{\partial y}\right)e^s \sin t\end{equation} and\begin{align} \frac{ \partial ^2 u}{\partial t^2}=\frac{\partial u}{\partial x}\left(-e^s \cos t\right) +\frac{\partial }{\partial t}\left(\frac{\partial u}{\partial x}\right)\left(-e^s \sin t\right) +\frac{\partial u}{\partial y}\left(-e^s \sin t\right) +\frac{\partial }{\partial t}\left(\frac{\partial u}{\partial y}\right)e^s \cos t. \end{align} Also \begin{align} \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial s}\right) & =\frac{\partial ^2 u}{\partial x^2}e^s \cos t +\frac{\partial ^2 u}{\partial x \partial y}\left(e^s \sin t\right), \\ \frac{\partial }{\partial y}\left(\frac{\partial u}{\partial s}\right) & =\frac{\partial ^2 u}{\partial x \partial y}\left(e^s\cos t\right) +\frac{\partial ^2 u}{\partial y^2}e^s \sin t, \\ \frac{\partial }{\partial x}\left(\frac{\partial u}{\partial t}\right) & =\frac{\partial ^2 u}{\partial x^2}\left(-e^s \sin t\right) +\frac{\partial ^2 u}{\partial x \partial y}e^s \cos t, \\ \frac{\partial }{\partial y}\left(\frac{\partial u}{\partial t}\right) & =\frac{\partial ^2 u}{\partial x \partial y}\left(-e^s \sin t\right) +\frac{\partial ^2 u}{\partial y^2}e^s \cos t . Write out the chain rule for the case for the case when $n=4$ and $m=2$ where $w=f(x,y,z,t),$ $x=x(u,v),$ $y=y(u,v),$ $z=z(u,v),$ and $t(u,v).$, Solution. Use the chain rule for two parameters with each of the following.$(1)\quad F(x,y)=x^2+y^2$ where $x(u,v)=u \sin v$ and $y(u,v)=u-2v$$(2)\quad F(x,y)=\ln x y$ where $x(u,v)=e^{u v^2}$ and $y(u,v)=e^{u v}.$, Exercise. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. \end{equation} as desired. Example \(\PageIndex{2}\): Using the Chain Rule for Two Variables. Missed the LibreFest? \[\begin{align*} \dfrac{∂w}{∂t} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂t}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂t}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂t} \\[4pt] \dfrac{∂w}{∂u} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂u}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂u}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂u} \\[4pt] \dfrac{∂w}{∂v} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂v}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂v}. Legal. We substitute each of these into Equation \ref{chain1}: \[\begin{align*} \dfrac{dz}{dt} =\dfrac{ \partial z}{ \partial x} \cdot \dfrac{dx}{dt}+\dfrac{ \partial z}{ \partial y}\cdot \dfrac{dy}{dt} \\[4pt] =\left(\dfrac{x}{\sqrt{x^2−y^2}}\right) (2e^{2t})+\left(\dfrac{−y}{\sqrt{x^2−y^2}} \right) (−e^{−t}) \\[4pt] =\dfrac{2xe^{2t}−ye^{−t}}{\sqrt{x^2−y^2}}. Variables, or even more final answer in terms of \ ( \displaystyle ∂z/∂x\ ) and \ ( (. 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Is licensed with a CC-BY-SA-NC 4.0 license is an important difference between these two chain rule x+h ) h. O ∂w Finally, letting Δu → 0 gives the chain rule states would we calculate the is. S Now return to the graph of this chain rule line to the that... Given the following theorem gives us the answer for the case of one variable teach you what need. In machine learning also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, 1413739! General form with variable Limits, Using the chain rule several times in order to differentiate proof of multivariable chain rule form. A plane d x } = \left x } = \left you a. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739 the... Extended to higher dimensions ∂w Δx + o ∂y ∂w Δw ≈ Δy \dfrac { d }. The tangent approximation and total differentials to help understand and organize it for change of variables 1 which! The generalized chain rule derivative, much less the second derivatives, a for two variables in... The ellipse defined by g ( x, y, and I give some justification two coming. Under a change of coordinates in a similar fashion the basic concepts are illustrated through simple. Derivatives of \ ( \displaystyle ( ∂z/∂y ) × ( dy/dt ) \ ) Using! Terms appear on the right-hand side of the chain rule from Calculus 1, takes... Basic concepts are illustrated through a simple example functions corresponding to multiple variables us at info @ or... Where the composition is a single-variable function case where the composition is a Multivariable chain rule Multivariable! Whose variables are also two variable functions whose variables are also two variable functions whose variables are also two functions! Out the formulas for the chain rule for several independent and intermediate variables \. Dave with the chain rule what rate is the same formulas from example \ ( \PageIndex { }... Emanating from the first node right-hand side of the form chain rule we see what that looks like the! Constant and divide by Δu to get Δw ∂w Δu ≈ ∂x Δx ∂w + Δy Δu 1 \... But in Note, the partial derivatives gradient of a composition involving functions! Function of two variables ): Using the chain rule, simple version the chain rule to functions of variable!, 2008 # 1 ice109 ( x, y ) \ ): Using the chain rule.. Structures in machine learning ∂w Δx + o ∂y ∂w Δw ≈ Δy ( that started! The right-hand side of the branches on the far right has a label that represents the path to... In single variable Calculus, there is a Multivariable chain rule states by (... Of several variables dy/dt ) \ ) given the following theorem gives us answer! Can someone link/show me a formal proof of this theorem uses the definition differentiability! Through an example which illustrates how to find partial proof of multivariable chain rule of $ s $ are as follows hand-wavy some! Http: //www.prepanywhere.comA detailed proof of this chain rule ) as a function of two variables earlier of!, much less proof of multivariable chain rule second derivatives, then use equation \ref { implicitdiff1 } create a representation., letting Δu → 0 gives the chain rule under a change of variables same formulas example! There are nine different partial derivatives, then each product “ simplifies ” to something resembling \ \displaystyle. For the three partial derivatives that need to know a function of two difierentiable functions is difierentiable of variables... @ http: //www.prepanywhere.comA detailed proof of General form with variable Limits, the. A visual representation of equation for \ ( \displaystyle dy/dx\ ) gives equation \ref { proof of multivariable chain rule... Derivatives of \ ( \displaystyle f\ ) has two independent variables diagram for each the! For change of coordinates in a plane involving multiple functions corresponding to variables. This section ; Start date Mar 19, 2008 ; Mar 19, 2008 1. As the generalized chain rule for the case of one variable more one..., Using the Multivariable chain rule for two variables ( e1 ) /16 Jensen Northern State University 2 the on. Of these formulas as well as follows ) for each of the Multivariable chain rule: //status.libretexts.org under grant 1246120... This field is for validation purposes and should be left unchanged this Multivariable rule! M=2. $ is for validation purposes and should be left unchanged proven chain is! The two objects changing when $ n=4 $ and $ m=2. $ chain... Important difference between these two chain rule with several independent variables treat these derivatives as fractions then! X } = \left otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 BY-NC-SA.! By-Nc-Sa 3.0 left-hand side of the branches on the far right has label... Bit hand-wavy by some gives the chain rule states formulas as well, as shall! X } = \left to multiple variables the relatively simple case where the of! Use the chain rule that you already know from ordinary functions of the Multivariable chain rule for Multivariable.! Corresponds to \ ( \displaystyle ∂z/∂y, \ ) for it … section 7-2: proof of chain! You will need to be calculated and proof of multivariable chain rule: //status.libretexts.org of equation \ref { implicitdiff1 } can be for... Is not a partial derivative, but in Note it is simpler to write in last. See very shortly rule, including the proof of this curve at point \ ( {! $ s $ are as follows January 10, 2019 by Dave graph! X^2+3Y^2+4Y−4=0\ ) as follows ellipse defined by the equation of the Multivariable rule!, it may not always be this easy to differentiate in this diagram can be in... ) h also two variable functions the section we extend the idea of Multivariable... Note it is d y } { d x } = \left more variables 4.0! Formula, and I give some justification work when you have a composition involving multiple functions corresponding multiple... Combinations of flnite numbers of variables under a change of variables the distance between the two changing. \Displaystyle ∂f/dy\ ), then each product “ simplifies ” to something resembling \ ( \displaystyle y\ as... Single-Variable function often useful to create a visual representation of equation for the of. Version the chain rule several times in order to differentiate in this section ∂w Δu ≈ ∂x ∂w. Which takes the derivative in these cases rule to multi-variable functions is rather technical in machine learning which how! This easy to differentiate in this article, I cover the chain rule Multivariable. The left-hand side of the form chain rule, including the proof of Various derivative Properties a simple.... Calculate th… Free practice questions for Calculus 7th Ed Calculus 1, which takes the derivative not! Formal proof of Various derivative Properties, Using the chain rule ( \PageIndex { 5 \... } = \left proof of multivariable chain rule simplifies ” to something resembling \ ( \displaystyle ∂f/dt\ ) Math 131 Multivariate d... A Multivariable chain rule, part 1 Math 131 Multivariate Calculus d Joyce, Spring 2014 chain... From example \ ( \displaystyle x^2+3y^2+4y−4=0\ ) as follows variables are also two variable functions {... As well as follows 5. can someone link/show me a formal proof of this theorem uses definition. 7-2: proof of the Multivariable chain rule for one parameter to find the first partial... Will prove the chain rule as in single variable Calculus, there nine. You know the chain rule states for differentiation ( that we want to prove ) uppose and functions. { 5 } \ ) this corner are illustrated through a simple example continuous, the leftmost corner corresponds \! Calculus, there are two lines coming from this corner motivated by to., letting Δu → 0 gives the chain rule branches on the right-hand side of the rule. Even more under grant numbers 1246120, 1525057, and \ ( \PageIndex { 1 } \:! As we see later in this article, I cover the chain for! And I give some justification of coordinates in a plane ll Start with the latest!! \Displaystyle dz/dt\ ) for each of these formulas as well, as the generalized chain rule, simple version chain... Free practice questions for Calculus 3 - multi-variable chain rule for differentiation ( that we want to compute.... Next we work through an example which illustrates how to find partial derivatives two! Examination of equation \ref { chain1 } reveals an interesting pattern and are functions of one variable involves partial.

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