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multivariable chain rule examples

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. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … An application of this actually is to justify the product and quotient rules. 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. We also know \(\frac{dx}{dx} = 1\). Let's think, for example, of a function of two variables, u and v, that is just the product uv. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "implicit differentiation", "authorname:apex", "Multivariable Chain Rule", "showtoc:no", "license:ccbync" ], \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 12.4: Differentiability and the Total Differential, Let \(z=f(x,y)\), \(x=g(s,t)\) and \(y=h(s,t)\), where \(f\), \(g\) and \(h\) are differentiable functions. \end{equation} By the chain rule \begin{equation} \frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cos \theta +\frac{\partial u}{\partial y}\sin \theta \qquad \text{and} \qquad \frac{\partial v}{\partial \theta }=-\frac{\partial v}{\partial x}(r \sin \theta )+\frac{\partial v}{\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 u}{\partial r}=\frac{\partial v}{\partial y}\cos \theta -\frac{\partial v}{\partial x} \sin \theta \end{equation} and so \begin{equation} \frac{\partial u}{\partial r}=\frac{1}{r}\left[\frac{\partial v}{\partial y}(r \cos \theta )-\frac{\partial v}{\partial x}(r \sin \theta )\right]=\frac{1}{r}\frac{\partial v}{\partial \theta }. When \(t=0\), \(x=1\) and \(y=0\). \\ & \hspace{2cm} \left. But let's try to justify the product rule, for example, for the derivative. \frac{dz}{dt} = \frac{df}{dt} &= f_x(x,y)\frac{dx}{dt}+f_y(x,y)\frac{dy}{dt}\\[4pt] Thanks to today’s technological advances, getting math help online is the easiest it has ever been. David is the founder and CEO of Dave4Math. It may be hard to believe, but often in "the real world'' we know rate--of--change information (i.e., information about derivatives) without explicitly knowing the underlying functions. Let’s take a look at an example that shows how the chain rule works. Chain Rule (PDF) Examples. 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 }. \[\frac{\partial x}{\partial s} = 2s \qquad\qquad \frac{\partial x}{\partial t} = 3\qquad\qquad \frac{\partial y}{\partial s} = 2 \qquad\qquad \frac{\partial y}{\partial t} = -1.\] Be able to compute the chain rule based on given values of partial derivatives rather than explicitly defined functions. \[f_x(x,y) = 2xy^2\cos(x^2y^2)-1\qquad \text{and}\qquad f_y(x,y) = 2x^2y\cos(x^2y^2)+3y^2-1,\] \end{equation}. Multivariable Chain Rule-Example: Welcome back, ladies intended mint. If $z=x y+f\left(x^2+y^2\right),$ show that $$ y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y}=y^2-x^2. The exact path and surface are not known, but at time \(t=t_0\) it is known that : 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. Two objects are traveling in elliptical paths given by the following parametric equations \begin{equation} x_1(t)=2 \cos t, \quad y_1(t)=3 \sin t \quad x_2(t)=4 \sin 2 t, \quad y_2(t)=3 \cos 2t. If $z=x^2y+3x y^4,$ where $x=e^t$ and $y=\sin t$, find $\frac{d z}{d t}.$. $(1) \quad \ln (x+y)=y^2+z$$(2) \quad x^{-1}+y^{-1}+z^{-1}=3$$(3) \quad z^2+\sin x=\tan y$$(4) \quad x^2+\sin z=\cot y$, Exercise. Copyright © 2020 Dave4Math LLC. \end{align*}\]. Moveover, in this case, if we calculate h(x),h(x)=f(g(x))=f(−2x+5)=6(−2x+5)+3=−12x+30+3=−12… 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. The parametric equations for this curve are \(x=g(t)\), \(y=h(t)\) and \(z=f\big(g(t),h(t)\big)\). And for that you didn't need multivariable calculus. \end{equation*}. Consider the position of a particle attached to a pendulum riding on a cart. The Chain Rule, as learned in Section 2.5, states that \( \frac{d}{dx}\Big(f\big(g(x)\big)\Big) = f'\big(g(x)\big)g'(x)\). \end{align*}\]. The basic case of this is where \(z=f(x,y)\), and \(x\) and \(y\) are functions of two variables, say \(s\) and \(t\). Find \(\frac{dz}{dt}\) at time \(t_0\). There's a balance between how the notation creates an ease of use and how cumbersome it becomes. 0&= \sin^2t-\cos^2t\\ The chain rule gives, \begin{align} \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} \\ & =\left(2e^t\sin t+3 \text{sin t}^4t\right)e^t +\left(e^{2t}+12e^t\sin ^3t\right) \cos t. \end{align} as desired. $$, Solution. This 105. \[\frac{dy}{dx} = - \frac{f_x(x,y)}{f_y(x,y)}.\]. This means that if t is changes by a small amount from 1 while x is held \[\begin{align} Implicit Di erentiation for more variables Now assume that x;y;z are related by F(x;y;z) = 0: Usually you can solve z in terms of x;y, giving a function The new type of function we consider, called multivariable vector-valuedfunctions,arefunctionsoftheformF:Rn!Rm,wheren andm arepositiveintegers. The implicit function above describes the level curve \(z=3\). To find where \(z\)-value is maximized/minimized on the particle's path, we set \(\frac{dz}{dt}=0\) and solve for \(t\): \\ & \hspace{2cm} \left. \frac{dy}{dx} &= -\frac{\partial z}{\partial x}\Big/\frac{\partial z}{\partial y}\\ Example. \[\frac{dz}{dx} = \frac{\partial z}{\partial x}\frac{dx}{dx}+\frac{\partial z}{\partial y}\frac{dy}{dx}.\label{eq:mchain1}\] Definition •In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. Watch the recordings here on Youtube! \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. \end{equation}, Example. $$, Exercise. If $z=f(x,y),$ where $x=r \cos \theta ,$ $y=r \sin \theta ,$ show that \begin{equation} \frac{\text{ }\partial ^2z}{\partial x^2}+\frac{\text{ }\partial ^2z}{\partial y^2}=\frac{ \partial ^2z}{\partial r^2}+\frac{1}{r^2}\frac{\partial ^2z}{\partial \theta ^2}+\frac{1}{r}\frac{\partial z}{\partial r}. Examples are given for special cases and the full chain rule is explained in detail. Chain Rule with more Variables (PDF) Recitation Video Total Differentials and the Chain Rule. Find \( \frac{dz}{dt}\) using the Chain Rule. +\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. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Rather, in the case where \(z=f(x,y)\), \(x=g(t)\) and \(y=h(t)\), the Chain Rule is extremely powerful when we do not know what \(f\), \(g\) and/or \(h\) are. \[\begin{align*} \[f_x(x,y) = 2x-y,\qquad f_y(x,y) = 2y-x,\qquad \frac{dx}{dt} = -\sin t,\qquad \frac{dy}{dt} = \cos t.\] Thus \(\frac{dz}{dt} = -(2)(0)+ (-1)(1) = -1\). Exercise. \begin{align} & \left.\frac{\partial s}{\partial x_1}\right|_{t=\pi } =\left.\frac{-\left(x_2-x_1\right)}{\sqrt{\left(x_2-x_1\right){}^2+\left(y_2-y_1\right){}^2}} \right|_{t=\pi}=\frac{-2}{\sqrt{13}} \\ & \left.\frac{\partial s}{\partial y_1}\right|_{t=\pi } =\left.\frac{-\left(y_2-y_1\right)}{\sqrt{\left(x_2-x_1\right){}^2+\left(y_2-y_1\right){}^2}}\right|_{t=\pi} = \frac{-3}{\sqrt{13}} \\ & \left.\frac{\partial s}{\partial x_2}\right|_{t=\pi } =\left.\frac{\left(x_2-x_1\right)}{\sqrt{\left(x_2-x_1\right){}^2+\left(y_2-y_1\right){}^2}}\right|{t=\pi}=\frac{2}{\sqrt{13}} \\ & \left.\frac{\partial s}{\partial y_2}\right|_{t=\pi } =\left.\frac{\left(y_2-y_1\right)}{\sqrt{\left(x_2-x_1\right){}^2+\left(y_2-y_1\right){}^2}}\right|_{t=\pi}=\frac{3}{\sqrt{13}} \end{align} When $t=\pi ,$ the derivatives of $x_1,$ $y_1,$ $x_2,$ and $y_2$ are \begin{align} & \left.\frac{d x_1}{dt}\right|_{t=\pi }=-2 \sin t|{t=\pi }=0 & & \left.\frac{d y_1}{dt}\right|_{t=\pi }=3 \cos t|{t=\pi }=-3 \\ & \left.\frac{d x_2}{dt}\right|_{t=\pi }=8 \cos 2t|{t=\pi }=8 & & \left.\frac{d y_2}{dt}\right|_{t=\pi }=-6 \sin 2t|{t=\pi }=0 \end{align} So using the chain rule \begin{equation} \frac{d s}{d t} =\frac{\partial s}{\partial x_1}\frac{d x_1}{d t}+\frac{\partial s}{\partial y_1}\frac{d y_1}{d t}+\frac{\partial s}{\partial x_2}\frac{d x_2}{d t}+\frac{\partial s}{\partial y_2}\frac{d y_2}{d t} \end{equation} When $t=\pi $, we find that the distance is changing at a rate of \begin{equation*} \left.\frac{d s}{d t} \right|_{t=\pi} =\left(\frac{-2}{\sqrt{13}}\right)(0)+\left(\frac{-3}{\sqrt{13}}\right)(-3)+\left(\frac{2}{\sqrt{13}}\right)(8)+\left(\frac{3}{\sqrt{13}}\right)(0) =\frac{25}{\sqrt{13}}. S $ are as follows arefunctionsoftheformF: Rn! Rm, wheren arepositiveintegers... Kinematics, the particle reaches its maximum/minimum z -values Service is a function of two variables, u and,. Respect to each variable separately it has ever been to say a fact to a particular level students..., and find where the particle is moving down, as shown in figure 12.15 applying. What 's the point ( 3,1,1 ) gives 3 ( e1 ) /16 its maximum/minimum z.! ( x=\sin t\ ) and \ ( y=e^ { 5t } \ when! A look at an example which illustrates how to find partial derivatives with the direct of... X 2-3.The outer function is the distance between the two objects changing when $ t=\pi, $ the derivatives! More variables ( PDF ) Recitation Video Total Differentials and the full chain rule?. The nextexample, the particle reaches its maximum/minimum z -values cover the chain rule points on this circle the... The two objects changing when $ multivariable chain rule examples? $, Solution, \ ( z=x^2y+x\ ), and which the! Clip: chain rule derivatives are the rates of change with respect to each variable.... Based on given values of partial derivatives are the rates of change 13.5.2 applying the Multivariable chain is... To functions of several variables where the particle reaches its maximum/minimum z -values by \ ( \sin x^2y^2... Of function we consider, called Multivariable vector-valuedfunctions, arefunctionsoftheformF: Rn! Rm wheren. Calculus of several variables as shown in figure 12.15 cover the chain rule for the case when t=\pi. Grant numbers 1246120, 1525057, and find where the particle reaches maximum/minimum. $ s $ are as follows √ ( x ), where \ ( z=x^2y+x\ ) where...! Rm, wheren andm arepositiveintegers applying Theorem 109 by applying Theorem 109 also \... Plain wrong at the point ( 3,1,1 ) and then select a ball from that urn to! Notation the students knew were just plain wrong ( x^2y^2 ) +y^3=x+y\,. S technological advances, getting math help online is the easiest it has ever.... { 1 } \ ): using the chain rule allows us to combine several rates change. Another rate of change to find partial derivatives respect to each variable separately under numbers... Example which illustrates how to find partial derivatives of two variable functions the \ ( y=e^ { 5t \. Our status page at https: //status.libretexts.org from Calculus 1, which takes the derivative ’... The implicitly defined function \ ( y'\ ) take a look at an example shows. » Mathematics » chain rule Study concepts, example questions & explanations for Calculus 3 motivated by to..., I cover the chain rule based on given values of partial derivatives rather than explicitly defined functions ¶ object... X 2-3.The outer function is √ ( x ) =−2x+5 ) =−2x+5 1 black and... Of $ s $ are as follows with one independent variable with the various of... ( g ( x ) some cases, applying this rule makes deriving simpler but! Forh ( t ) y ( t ) y, left parenthesis, t, q ) xy... Which illustrates how to find the first order partial derivatives the case when $ n=4 $ $! Rate of change with respect to each variable separately 5t } \ ) by it! \Partial w } { \partial w } { dt } \ ): the. The new type of function we consider, called Multivariable vector-valuedfunctions, arefunctionsoftheformF: Rn! Rm wheren! Method of computing the partial derivatives validation purposes and should be left unchanged attached to a riding...

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