All rights reserved. To reduce this to one variable, we use the fact that \(\displaystyle x(t)=e^{2t}\) and \(\displaystyle y(t)=e^{−t}\). 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. Dave4Math » Calculus 3 » Chain Rule for Multivariable Functions. 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}}. How does the chain rule work when you have a composition involving multiple functions corresponding to multiple variables? \end{equation}. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. To people who need to learn Calculus but are afraid they can't. df dx = df dt dt dx. Collection of Multivariable Chain Rule exercises and solutions, Suitable for students of all degrees and levels and will help you pass the Calculus test successfully. 10 Multivariable functions and integrals 10.1 Plots: surface, contour, intensity To understand functions of several variables, start by recalling the ways in which you understand a function f of one variable. 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. 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. 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 \(\displaystyle dy/dx:\), \[\begin{align*} \dfrac{d}{dx}f(x,y) =\dfrac{d}{dx}(0) \\[4pt] \dfrac{∂f}{∂x}⋅\dfrac{dx}{dx}+\dfrac{∂f}{∂y}⋅\dfrac{dy}{dx} =0 \\[4pt]\dfrac{∂f}{∂x}+\dfrac{∂f}{∂y}⋅\dfrac{dy}{dx} =0. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. \end{align*}\]. \end{align*}\]. Ask Question Asked 20 days ago. In this section we extend the Chain Rule to functions of more than one variable. Limits for multivariable functions-I; Limits for multivariable functions-II; Continuity of multivariable functions; Partial Derivatives-I; Unit 2. 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 \(\displaystyle t\). Therefore, there are nine different partial derivatives that need to be calculated and substituted. \end{align*} \]. And this is known as the chain rule. Suppose that \(\displaystyle x=g(t)\) and \(\displaystyle y=h(t)\) are differentiable functions of \(\displaystyle t\) and \(\displaystyle z=f(x,y)\) is a differentiable function of \(\displaystyle x\) and \(\displaystyle y\). 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 \(\displaystyle t_0\). Free detailed solution and explanations Multivariable Chain Rule - Proving an equation of partial derivatives - Exercise 6472. and write out the formulas for the three partial derivatives of \(\displaystyle w\). 2. To find the equation of the tangent line, we use the point-slope form (Figure \(\PageIndex{5}\)): \[\begin{align*} y−y_0 =m(x−x_0)\\[4pt]y−1 =\dfrac{7}{4}(x−2) \\[4pt] y =\dfrac{7}{4}x−\dfrac{7}{2}+1\\[4pt] y =\dfrac{7}{4}x−\dfrac{5}{2}.\end{align*}\]. The chain rule, part 1 Math 131 Multivariate Calculus D Joyce, Spring 2014 The chain rule. Then, \[\dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_2}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\]. 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 nine partial derivatives, then use the same formulas from Example \(\PageIndex{3}\). Example \(\PageIndex{3}\): Using the Generalized Chain Rule. 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]. If we treat these derivatives as fractions, then each product “simplifies” to something resembling \(\displaystyle ∂f/dt\). Now, we substitute each of them into the first formula to calculate \(\displaystyle ∂w/∂u\): \[\begin{align*} \dfrac{∂w}{∂u} =\dfrac{∂w}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂w}{∂y}⋅\dfrac{∂y}{∂u}+\dfrac{∂w}{∂z}⋅\dfrac{∂z}{∂u} \\[4pt] =(6x−2y)e^u\sin v−2xe^u\cos v+8ze^u, \end{align*}\]. Therefore, this value is finite. 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. \end{align*}\], \[\dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y}=−\dfrac{2x}{6y+4}=−\dfrac{x}{3y+2},\]. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. 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 }. David Smith (Dave) has a B.S. 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). 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 . which is the same solution. \end{align}, Example. Example \(\PageIndex{2}\): Using the Chain Rule for Two Variables. 1. curve in 3-space (x,y,z)=F(t)=f(g(t)). As Preview Activity 10.3.1 suggests, the following version of the Chain Rule holds in general. Viewed 136 times 5. }\) Find \(\ds \frac{dz}{dt}\) using the Chain Rule. Now suppose that \(\displaystyle f\) is a function of two variables and \(\displaystyle g\) is a function of one variable. This derivative can also be calculated by first substituting \(\displaystyle x(t)\) and \(\displaystyle y(t)\) into \(\displaystyle f(x,y),\) then differentiating with respect to \(\displaystyle t\): \[\displaystyle z=f(x,y)=f(x(t),y(t))=4(x(t))^2+3(y(t))^2=4\sin^2 t+3\cos^2 t. \nonumber\], \[\displaystyle \dfrac{dz}{dt}=2(4\sin t)(\cos t)+2(3\cos t)(−\sin t)=8\sin t\cos t−6\sin t\cos t=2\sin t\cos t, \nonumber\]. Chain Rule Calculator (If you have issues viewing the output make sure that your browser is set to accept third-party cookies. We have \begin{align} \frac{\partial w}{\partial s} & =\frac{\partial w}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial s}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial s} \\ & =\left[\frac{\partial }{\partial x}\left(4x+y^2 +z^3\right)\right] \left[\frac{\partial }{\partial s}\left(e^{r s^2}\right)\right] \\ & \hspace{2cm} +\left[\frac{\partial }{\partial y}\left(4x+y^2+z^3\right)\right] \left[\frac{\partial }{\partial s}\left(\ln \frac{r+s}{t}\right)\right] \\ & \hspace{3cm} +\left[\frac{\partial}{\partial z}\left(4x+y^2+z^3\right)\right] \left[\frac{\partial }{\partial s}\left(r s t^2\right)\right] \\ & =4\left[e^{r s^2}(2r s)\right]+2y\left(\frac{1}{\frac{r+s}{t}}\right)\left(\frac{1}{t}\right)+3z^2\left(r t^2\right) \\ & =8r s e^{r s^2}+2\frac{y}{r+s}+3r t^2z^2. A more general chain rule As you can probably imagine, the multivariable chain rule generalizes the chain rule from single variable calculus. Next we work through an example which illustrates how to find partial derivatives of two variable functions whose variables are also two variable functions. In this diagram, the leftmost corner corresponds to \(\displaystyle z=f(x,y)\). Thanks!) +\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. 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 can draw a tree diagram for each of these formulas as well as follows. We now practice applying the Multivariable Chain Rule. Inconsistent notation in definitions of the multivariable chain rule. Example \(\displaystyle \PageIndex{5}\): Implicit Differentiation by Partial Derivatives, a. And it's not just any old scalar calculus that pops up---you need differential matrix calculus, the shotgun wedding of linea… 12.5: The Multivariable Chain Rule. Untitled; A more elegant form of representing Euler's equation This equation implicitly defines \(\displaystyle y\) as a function of \(\displaystyle x\). For the formula for \(\displaystyle ∂z/∂v\), follow only the branches that end with \(\displaystyle v\) and add the terms that appear at the end of those branches. 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. 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 \(\displaystyle f(x,y)=0\). The proof of this theorem uses the definition of differentiability of a function of two variables. (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}. For instance, consider the implicit function \(x^2y-xy^3=3.\) We learned to use the following steps to find \(\frac{dy}{dx}\): 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. We substitute each of these into Equation \ref{chain1}: \[\begin{align*} \dfrac{dz}{dt} =\dfrac{ \partial z}{ \partial x} \cdot \dfrac{dx}{dt}+\dfrac{ \partial z}{ \partial y}\cdot \dfrac{dy}{dt} \\[4pt] =\left(\dfrac{x}{\sqrt{x^2−y^2}}\right) (2e^{2t})+\left(\dfrac{−y}{\sqrt{x^2−y^2}} \right) (−e^{−t}) \\[4pt] =\dfrac{2xe^{2t}−ye^{−t}}{\sqrt{x^2−y^2}}. Partial Derivative / Multivariable Chain Rule Notation. However, it is simpler to write in the case of functions of the form \end{align*} \], As \(\displaystyle t\) approaches \(\displaystyle t_0, (x(t),y(t))\) approaches \(\displaystyle (x(t_0),y(t_0)),\) so we can rewrite the last product as, \[\displaystyle \lim_{(x,y)→(x_0,y_0)}\dfrac{(E(x,y)}{\sqrt{(x−x_0)^2+(y−y_0)^2}}\lim_{(x,y)→(x_0,y_0)}(\dfrac{\sqrt{(x−x_0)^2+(y−y_0)^2}}{t−t_0}). The Chain rule of derivatives is a direct consequence of differentiation. 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}. Implicit Function Theorem [Understanding theorem in book] 1. Proof of multivariable chain rule. }\) Find \(\ds \frac{dz}{dt}\) using the Chain Rule. Inconsistent notation in definitions of the multivariable chain rule. \end{align}. 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. Partial Derivatives-II ; Differentiability-I; Differentiability-II; Chain rule-I; Chain rule-II; Unit 3. \end{align*}\]. The single variable chain rule tells you how to take the derivative of the composition of two functions: \dfrac {d} {dt}f (g (t)) = \dfrac {df} {dg} \dfrac {dg} {dt} = f' (g (t))g' (t) dtd 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. Our mission is to provide a free, world-class education to anyone, anywhere. \end{equation}. We have \(\displaystyle f(x,y,z)=x^2e^y−yze^x.\) Therefore, \[\begin{align*} \dfrac{∂f}{∂x} =2xe^y−yze^x \\[4pt] \dfrac{∂f}{∂y} =x^2e^y−ze^x \\[4pt] \dfrac{∂f}{∂z} =−ye^x\end{align*}\], \[\begin{align*} \dfrac{∂z}{∂x} =−\dfrac{∂f/∂x}{∂f/∂y} \dfrac{∂z}{∂y} =−\dfrac{∂f/∂y}{∂f/∂z} \\[4pt] =−\dfrac{2xe^y−yze^x}{−ye^x} \text{and} =−\dfrac{x^2e^y−ze^x}{−ye^x} \\[4pt] =\dfrac{2xe^y−yze^x}{ye^x} =\dfrac{x^2e^y−ze^x}{ye^x} \end{align*}\]. The top branch is reached by following the \(\displaystyle x\) branch, then the t branch; therefore, it is labeled \(\displaystyle (∂z/∂x)×(dx/dt).\) The bottom branch is similar: first the \(\displaystyle y\) branch, then the \(\displaystyle t\) branch. \end{equation}, Example. If we compose a differentiable function with a differentiable function , we get a function whose derivative is. 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}. 8.2 Chain Rule For functions of one variable, the chain rule allows you to di erentiate with respect to still another variable: ya function of xand a function of tallows dy dt = dy dx dx dt (8:3) You can derive this simply from the de nition of a derivative. \end{align*}\]. More formal treatment of multivariable chain rule. $\begingroup$ @guest There are a lot of ways to word the chain rule, and I know a lot of ways, but the ones that solved the issue in the question also used notation that the students didn't know. So I was looking for a way to say a fact to a particular level of students, using the notation they understand. \end{equation} as desired. The chain rule consists of partial derivatives . $(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. The notation df /dt tells you that t is the variables Calculate \(dz/dt \) given the following functions. \begin{align} \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt} + \frac{df}{dy}\frac{dy}{dt} \end{align} I found multiple derivation of this results online using differentials and mean value theorem, but they don't look like rigorous to me. The chain rule consists of partial derivatives .

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