proof of multivariable chain rule

\end{equation} By the chain rule \begin{equation} \frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\cos \theta +\frac{\partial v}{\partial y}\sin \theta\end{equation} and \begin{equation} \frac{\partial u}{\partial \theta }=-\frac{\partial u}{\partial x}(r \sin \theta )+\frac{\partial u}{\partial y}(r \cos \theta ).\end{equation} Substituting \begin{equation} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\qquad \text{and} \qquad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}, \end{equation} we obtain \begin{equation} \frac{\partial v}{\partial r}=-\frac{\partial u}{\partial y}\cos \theta -\frac{\partial u}{\partial x} \sin \theta \end{equation} and also \begin{equation*} \frac{\partial u}{\partial r}=-\frac{1}{r}\left[\frac{\partial u}{\partial y}(r \cos \theta )-\frac{\partial u}{\partial x}(r \sin \theta )\right]=-\frac{1}{r}\frac{\partial u}{\partial \theta }. be defined by g(t)=(t3,t4)f(x,y)=x2y. State the chain rules for one or two independent variables. \(\displaystyle \dfrac{∂z}{∂u}=0,\dfrac{∂z}{∂v}=\dfrac{−21}{(3\sin 3v+\cos 3v)^2}\). If z is only continuous, the partial derivative, much less the second derivatives, may not even exist. 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. $$, Exercise. Neural networks are one of the most popular and successful conceptual structures in machine learning. $(1) \quad f(x,y)=\left(1+x^2+y^2\right)^{1/2}$ where $x(t)=\cos 5 t$ and $y(t)=\sin 5t $$(2) \quad g(x,y)=x y^2$ where $x(t)=\cos 3t$ and $y(t)=\tan 3t.$, Exercise. Example. What is the equation of the tangent line to the graph of this curve at point \(\displaystyle (2,1)\)? Example \(\PageIndex{1}\): Using the Multivariable Chain Rule Proof of the Chain Rule •If we define ε to be 0 when Δx = 0, the ε becomes a continuous function of Δx. \end{align*}\]. We’ll start with the chain rule that you already know from ordinary functions of one variable. To get the formula for \(\displaystyle dz/dt,\) add all the terms that appear on the rightmost side of the diagram. \[\dfrac { d y } { d x } = \left. There are several versions of the chain rule for functions of more than one variable, each of them giving a rule for differentiating a composite function. All rights reserved. Theorem. \frac { 2 x + y + 7 } { 2 y - x + 3 } \right| _ { ( 3 , - 2 ) } = \dfrac { 2 ( 3 ) + ( - 2 ) + 7 } { 2 ( - 2 ) - ( 3 ) + 3 } = - \dfrac { 11 } { 4 } \nonumber\], Equation of the tangent line: \(\displaystyle y=−\dfrac{11}{4}x+\dfrac{25}{4}\), \(\displaystyle \dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}\), \(\displaystyle \dfrac{dz}{du}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u}\dfrac{dz}{dv}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂v}\), \(\displaystyle \dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_1}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\). Chain Rule (Multivariable Calculus) Chain rule. Calculate \(\displaystyle ∂f/dx\) and \(\displaystyle ∂f/dy\), then use Equation \ref{implicitdiff1}. Then, \[\dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_2}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\]. The upper branch corresponds to the variable \(\displaystyle x\) and the lower branch corresponds to the variable \(\displaystyle y\). \end{align}, Example. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. The single-variable chain rule. 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. Active 5 days ago. \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. The chain rule, part 1 Math 131 Multivariate Calculus D Joyce, Spring 2014 The chain rule. 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 . Then \(\displaystyle f(x,y)=x^2+3y^2+4y−4.\) The ellipse \(\displaystyle x^2+3y^2+4y−4=0\) can then be described by the equation \(\displaystyle f(x,y)=0\). Multivariable Chain Rule SUGGESTED REFERENCE MATERIAL: As you work through the problems listed below, you should reference Chapter 13.5 of the rec-ommended textbook (or the equivalent chapter in your alternative textbook/online resource) and your lecture notes. In other words, we want to compute lim. 1. If $F(u,v,w)$ is differentiable where $u=x-y,$ $v=y-z,$ and $w=z-x,$ then find $$ \frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}. Therefore, there are nine different partial derivatives that need to be calculated and substituted. (You can think of this as the mountain climbing example where f(x,y) isheight of mountain at point (x,y) and the path g(t) givesyour position at time t.)Let h(t) be the composition of f with g (which would giveyour height at time t):h(t)=(f∘g)(t)=f(g(t)).Calculate the derivative h′(t)=dhdt(t)(i.e.,the change in height) via the chain rule. David Smith (Dave) has a B.S. In the section we extend the idea of the chain rule to functions of several variables. In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables. Multivariate chain rule and its applications Having seen that multivariate calculus is really no more complicated than the univariate case, we now focus on applications of the chain rule. How would we calculate the derivative in these cases? The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. James Stewart @http://www.prepanywhere.comA detailed proof of chain rule. However, it may not always be this easy to differentiate in this form. The following theorem gives us the answer for the case of one independent variable. 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). Now that we’ve see how to extend the original chain rule to functions of two variables, it is natural to ask: Can we extend the rule to more than two variables? The Cauchy-Riemann equations are \begin{equation} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \qquad \text{and} \qquad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\end{equation} where $u=u(x,y)$ and $v=v(x,y).$ Show that if $x$ and $y$ are expressed in terms of polar coordinates, the Cauchy-Riemann equations become \begin{equation} \frac{\partial u}{\partial r}=\frac{1}{r}\frac{\partial v}{\partial \theta } \qquad \text{and} \qquad \frac{\partial v}{\partial r}=\frac{-1}{r}\frac{\partial u}{\partial \theta }. The variables \(\displaystyle x\) and \(\displaystyle y\) that disappear in this simplification are often called intermediate variables: they are independent variables for the function \(\displaystyle f\), but are dependent variables for the variable \(\displaystyle t\). As for your second question, one doesn't- what you have written is not true. Example \(\PageIndex{2}\): Using the Chain Rule for Two Variables. \nonumber\]. \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). 6. \\ & \hspace{2cm} \left. \\ & \hspace{2cm} \left. Suppose \(\displaystyle x=g(u,v)\) and \(\displaystyle y=h(u,v)\) are differentiable functions of \(\displaystyle u\) and \(\displaystyle v\), and \(\displaystyle z=f(x,y)\) is a differentiable function of \(\displaystyle x\) and \(\displaystyle y\). +\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\right] \\ & =e^{-2s}\left[\frac{\partial u}{\partial x}e^s \cos t +\left[\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)\right]e^s \cos t\right. Find Textbook Solutions for Calculus 7th Ed. \end{align} Similarly, \begin{align} \frac{\partial F}{\partial y}& =\frac{\partial F}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial y} \\ & =\frac{\partial F}{\partial u}(-1)+\frac{\partial F}{\partial v}(1)+\frac{\partial F}{\partial w}(0) \\ & =-\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v} \end{align} and \begin{align} \frac{\partial F}{\partial z}& =\frac{\partial F}{\partial u}\frac{\partial u}{\partial z}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial z}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial z} \\ & =\frac{\partial F}{\partial u}(0)+\frac{\partial F}{\partial v}(-1)+\frac{\partial F}{\partial w}(1) \\ & =-\frac{\partial F}{\partial v}+\frac{\partial F}{\partial v} \end{align} Therefore the required expression is \begin{equation} \frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z} \left[\frac{\partial F}{\partial u}-\frac{\partial F}{\partial w}\right]+\left[-\frac{\partial F}{\partial u}+\frac{\partial F}{\partial v}\right]+\left[-\frac{\partial F}{\partial v}+\frac{\partial F}{\partial v}\right] =0. \end{align*}\], Then we substitute \(\displaystyle x(u,v)=3u+2v\) and \(\displaystyle y(u,v)=4u−v:\), \[\begin{align*} \dfrac{∂z}{∂v} =14x−6y \\[4pt] =14(3u+2v)−6(4u−v) \\[4pt] =18u+34v \end{align*}\]. Calculate \(\displaystyle dz/dt\) for each of the following functions: a. 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. Let g:R→R2 and f:R2→R (confused?) \end{equation}, Solution. Statements Statement of product rule for differentiation (that we want to prove) uppose and are functions of one variable. This branch is labeled \(\displaystyle (∂z/∂y)×(dy/dt)\). \\ & \hspace{2cm} \left. Find the following higher order partial derivatives: $\displaystyle \frac{ \partial ^2z}{\partial x\partial y}$, $\displaystyle \frac{ \partial ^2z}{\partial x^2}$, and $\displaystyle \frac{\partial ^2z}{\partial y^2}$ for each of the following. Chain rule for functions of 2, 3 variables (Sect. An alternative proof for the chain rule for multivariable functions Raymond Jensen Northern State University 2. Calculate nine partial derivatives, then use the same formulas from Example \(\PageIndex{3}\). Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side): . 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. +\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. 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 \(\PageIndex{1}\)). Calculate \(\displaystyle ∂z/∂u\) and \(\displaystyle ∂z/∂v\) given the following functions: \[ z=f(x,y)=\dfrac{2x−y}{x+3y},\; x(u,v)=e^{2u}\cos 3v,\; y(u,v)=e^{2u}\sin 3v. We wish to prove that \(\displaystyle z=f(x(t),y(t))\) is differentiable at \(\displaystyle t=t_0\) and that Equation \ref{chain1} holds at that point as well. 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}. and write out the formulas for the three partial derivatives of \(\displaystyle w\). Also related to the tangent approximation formula is the gradient of a function. Ask Question Asked 5 days ago. \end{align}, Example. This equation implicitly defines \(\displaystyle y\) as a function of \(\displaystyle x\). Here we see what that looks like in the relatively simple case where the composition is a single-variable function. Example \(\PageIndex{1}\): Using the Chain Rule. \label{chian2b}\]. \end{equation}. 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. It is often useful to create a visual representation of Equation for the chain rule. However, it is simpler to write in the case of functions of the form In particular, we will see that there are multiple variants to the chain rule here all depending on how many variables our function is dependent on and how each of those variables can, in turn, be written in terms of different variables. Not all of them will be proved here and some will only be proved for special cases, but at least you’ll see that some of them aren’t just pulled out of the air. The proof of this chain rule is motivated by appealing to a previously proven chain rule with one independent variable. \end{align*}\]. 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 \(\displaystyle t_0\). Theorem (Chain rule) Assume that \( x,y:\mathbb R\to\mathbb R \) are differentiable at point \( t_0 \). Perform implicit differentiation of a function of two or more variables. Last, each of the branches on the far right has a label that represents the path traveled to reach that branch. And it might have been considered a little bit hand-wavy by some. This field is for validation purposes and should be left unchanged. Find $\frac{\partial w}{\partial s}$ if $w=4x+y^2+z^3$, where $x=e^{r s^2},$ $y=\ln \left(\frac{r+s}{t}\right),$ and $z=r s t^2.$, Solution. When u = u(x,y), for guidance in working out the chain rule… Let $w=\ln(x+y)$, $x=uv$, $ y=\frac uv.$ What is $ \frac {\partial w}{\partial v}$? Example 12.5.3 Using the Multivariable Chain Rule \end{align*}\]. Use the chain rule to find $\frac{dw}{dt}.$ Leave your answer in mixed form $(x,y,z,t).$ $(1) \quad w=\ln \left(x+2y-z^2\right) ,$ $x=2t-1,$ $y=\frac{1}{t},$ and $z=\sqrt{t}.$$(2) \quad w=\sin x y z ,$ $x=1-3t ,$ $y=e^{1-t} ,$ and $z=4t.$ $(3) \quad w=z e^{x y ^2} ,$ $x=\sin t ,$ $y=\cos t ,$ and $z=\tan 2t.$$(4) \quad w=e^{x^3+y z} ,$ $x=\frac{2}{t}$, $y=\ln (2t-3) ,$ and $z=t^2.$$(5) \quad w=\frac{x+y}{2-z} ,$ $x=2 r s$, $y=\sin r t ,$ and $z=s t^2.$, Exercise. 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. 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}}. David is the founder and CEO of Dave4Math. To use the chain rule, we again need four quantities—\(\displaystyle ∂z/∂x,∂z/dy,dx/dt,\) and \(\displaystyle dy/dt:\). This gives us Equation. The method involves differentiating both sides of the equation defining the function with respect to \(\displaystyle x\), then solving for \(\displaystyle dy/dx.\) Partial derivatives provide an alternative to this method. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. \end{equation} Similarly the chain rule is to be used \begin{equation} \frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial v}{\partial y}\frac{\partial y}{\partial r} \qquad \text{and} \qquad \frac{\partial u}{\partial \theta }=\frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta }. Example \(\displaystyle \PageIndex{5}\): Implicit Differentiation by Partial Derivatives, a. x, y, and x, y differentiable wrt. and M.S. Okay, so you know the chain rule from calculus 1, which takes the derivative of a composition of functions. 14.4) I Review: Chain rule for f : D ⊂ R → R. I Chain rule for change of coordinates in a line. Answer: treating everything other than t as a constant, by either the chain rule or the quotient rule you get xq(eq 1)/(1 + xtq)2. We compute, \begin{align} \frac{\partial F}{\partial x}& =\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial F}{\partial w}\frac{\partial w}{\partial x} \\ & =\frac{\partial F}{\partial u}(1)+\frac{\partial F}{\partial v}(0)+\frac{\partial F}{\partial w}(-1) \\ & =\frac{\partial F}{\partial u}-\frac{\partial F}{\partial w}. Solution. $$, Solution. \end{align}, Example. Exercise. First the one you know. 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. 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. \[\begin{align*} \dfrac{∂w}{∂t} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂t}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂t}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂t} \\[4pt] \dfrac{∂w}{∂u} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂u}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂u}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂u} \\[4pt] \dfrac{∂w}{∂v} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂v}+\dfrac{∂w}{∂z}\dfrac{∂z}{∂v}. Using $x=r \cos \theta $ and $y=r \sin \theta $ we can state the chain rule to be used: \begin{equation} \frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \qquad \text{and} \qquad \frac{\partial v}{\partial \theta }=\frac{\partial v}{\partial x}\frac{\partial x}{\partial \theta }+\frac{\partial v}{\partial y}\frac{\partial y}{\partial \theta }. If $u=f(x,y),$ where $x=e^s \cos t$ and $y=e^s \sin t,$ show that \begin{equation} \frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=e^{-2s}\left[\frac{\partial ^2u}{\partial s^2}+\frac{\partial ^2u}{\partial t^2}\right]. \\ & \hspace{2cm} \left. As in single variable calculus, there is a multivariable chain rule. \end{align*} \], \[ \begin{align*} \dfrac{dz}{dt} = \dfrac{1}{2} (e^{4t}−e^{−2t})^{−1/2} \left(4e^{4t}+2e^{−2t} \right) \\[4pt] =\dfrac{2e^{4t}+e^{−2t}}{\sqrt{e^{4t}−e^{−2t}}}. In particular, if we assume that \(\displaystyle y\) is defined implicitly as a function of \(\displaystyle x\) via the equation \(\displaystyle f(x,y)=0\), we can apply the chain rule to find \(\displaystyle dy/dx:\), \[\begin{align*} \dfrac{d}{dx}f(x,y) =\dfrac{d}{dx}(0) \\[4pt] \dfrac{∂f}{∂x}⋅\dfrac{dx}{dx}+\dfrac{∂f}{∂y}⋅\dfrac{dy}{dx} =0 \\[4pt]\dfrac{∂f}{∂x}+\dfrac{∂f}{∂y}⋅\dfrac{dy}{dx} =0. Two terms appear on the right-hand side of the formula, and \(\displaystyle f\) is a function of two variables. Using this function and the following theorem gives us an alternative approach to calculating \(\displaystyle dy/dx.\), Theorem: Implicit Differentiation of a Function of Two or More Variables, Suppose the function \(\displaystyle z=f(x,y)\) defines \(\displaystyle y\) implicitly as a function \(\displaystyle y=g(x)\) of \(\displaystyle x\) via the equation \(\displaystyle f(x,y)=0.\) Then, \[\dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y} \label{implicitdiff1}\], If the equation \(\displaystyle f(x,y,z)=0\) defines \(\displaystyle z\) implicitly as a differentiable function of \(\displaystyle x\) and \(\displaystyle y\), then, \[\dfrac{dz}{dx}=−\dfrac{∂f/∂x}{∂f/∂z} \;\text{and}\; \dfrac{dz}{dy}=−\dfrac{∂f/∂y}{∂f/∂z}\label{implicitdiff2}\], as long as \(\displaystyle f_z(x,y,z)≠0.\), Equation \ref{implicitdiff1} is a direct consequence of Equation \ref{chain2a}. \(\displaystyle z=f(x,y)=4x^2+3y^2,x=x(t)=\sin t,y=y(t)=\cos t\), \(\displaystyle z=f(x,y)=\sqrt{x^2−y^2},x=x(t)=e^{2t},y=y(t)=e^{−t}\), \(\displaystyle \dfrac{∂z}{∂x}=\dfrac{x}{\sqrt{x^2−y^2}}\), \(\displaystyle \dfrac{∂z}{∂y}=\dfrac{−y}{\sqrt{x^2−y^2}}\), \(\displaystyle \dfrac{dx}{dt}=−e^{−t}.\). Next we work through an example which illustrates how to find partial derivatives of two variable functions whose variables are also two variable functions. Set \(\displaystyle f(x,y)=3x^2−2xy+y^2+4x−6y−11=0,\) then calculate \(\displaystyle f_x\) and \(\displaystyle f_y: f_x=6x−2y+4\) \(\displaystyle f_y=−2x+2y−6.\), \[\displaystyle \dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y}=\dfrac{6x−2y+4}{−2x+2y−6}=\dfrac{3x−y+2}{x−y+3}. The chain rule for the case when $n=4$ and $m=2$ yields the following the partial derivatives: \begin{equation} \frac{\partial w}{\partial u}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial u}+\frac{\partial w}{\partial t}\frac{\partial t}{\partial u} \end{equation} and \begin{equation} \frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial v}+\frac{\partial w}{\partial t}\frac{\partial t}{\partial v}. 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. Recall that the chain rule for functions of a single variable gives the rule for differentiating a composite function: if $y=f(x)$ and $x=g(t),$ where $f$ and $g$ are differentiable functions, then $y$ is a a differentiable function of $t$ and \begin{equation} \frac{dy}{d t}=\frac{dy}{dx}\frac{dx}{dt}. \end{align*}\]. 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. \end{align*}\], \[\dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y}=−\dfrac{2x}{6y+4}=−\dfrac{x}{3y+2},\]. The answer is yes, as the generalized chain rule states. \end{align} Finally \begin{align} & e^{-2s}\left[\frac{\partial ^2u}{\partial s^2} +\frac{\partial ^2u}{\partial t^2}\right] \\ & =e^{-2s}\left[\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 \right. 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