proof of multivariable chain rule

We will differentiate $\sqrt{\sin^{2} (3x) + x}$. +\frac{\partial }{\partial s}\left(\frac{\partial u}{\partial y}\right)e^s \sin t +\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) \right. First the one you know. \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: \(\displaystyle y=−\dfrac{11}{4}x+\dfrac{25}{4}\), \(\displaystyle \dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}\), \(\displaystyle \dfrac{dz}{du}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u}\dfrac{dz}{dv}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂v}\), \(\displaystyle \dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_1}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\). Example. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 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. \nonumber\]. ∂x o Now hold v constant and divide by Δu to get Δw ∂w Δu ≈ ∂x Δx ∂w + Δy Δu. 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. Therefore, this value is finite. Theorem. \end{align*}\]. In this equation, both \(\displaystyle f(x)\) and \(\displaystyle g(x)\) are functions of one variable. How does the chain rule work when you have a composition involving multiple functions corresponding to multiple variables? To find \(\displaystyle ∂z/∂u,\) we use Equation \ref{chain2a}: \[\begin{align*} \dfrac{∂z}{∂u} =\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u} \\[4pt] =3(6x−2y)+4(−2x+2y) \\[4pt] =10x+2y. Thread starter ice109; Start date Mar 19, 2008; Mar 19, 2008 #1 ice109. We substitute each of these into Equation \ref{chain1}: \[\begin{align*} \dfrac{dz}{dt} =\dfrac{ \partial z}{ \partial x} \cdot \dfrac{dx}{dt}+\dfrac{ \partial z}{ \partial y}\cdot \dfrac{dy}{dt} \\[4pt] =\left(\dfrac{x}{\sqrt{x^2−y^2}}\right) (2e^{2t})+\left(\dfrac{−y}{\sqrt{x^2−y^2}} \right) (−e^{−t}) \\[4pt] =\dfrac{2xe^{2t}−ye^{−t}}{\sqrt{x^2−y^2}}. chain rule for functions of a single variable, Derivatives and Integrals of Vector Functions (and Tangent Vectors) [Video], Vector Functions and Space Curves (Calculus in 3D) [Video], Probability Density Functions (Applications of Integrals), Conservative Vector Fields and Independence of Path, Jacobian (Change of Variables in Multiple Integrals), Absolute Extrema (and the Extreme Value Theorem), Arc Length and Curvature of Smooth Curves, Continuous Function and Multivariable Limit, Derivatives and Integrals of Vector Functions (and Tangent Vectors), Directional Derivatives and Gradient Vectors, Double Integrals and the Volume Under a Surface, Lagrange Multipliers (Optimizing a Function), Multivariable Functions (and Their Level Curves), Partial Derivatives (and Partial Differential Equations), Choose your video style (lightboard, screencast, or markerboard). Find the following higher order partial derivatives: $\displaystyle \frac{ \partial ^2z}{\partial x\partial y}$, $\displaystyle \frac{ \partial ^2z}{\partial x^2}$, and $\displaystyle \frac{\partial ^2z}{\partial y^2}$ for each of the following. This means that if t is changes by a small amount from 1 while x is held ﬁxed at 3 and q at 1, the value of f … \end{align*}\]. 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. \\ & \hspace{2cm} \left. This branch is labeled \(\displaystyle (∂z/∂y)×(dy/dt)\). 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. b. Recall that when multiplying fractions, cancelation can be used. Equation \ref{implicitdiff1} can be derived in a similar fashion. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. 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. Multivariable Calculus. As in single variable calculus, there is a multivariable chain rule. We have \(\displaystyle f(x,y,z)=x^2e^y−yze^x.\) Therefore, \[\begin{align*} \dfrac{∂f}{∂x} =2xe^y−yze^x \\[4pt] \dfrac{∂f}{∂y} =x^2e^y−ze^x \\[4pt] \dfrac{∂f}{∂z} =−ye^x\end{align*}\], \[\begin{align*} \dfrac{∂z}{∂x} =−\dfrac{∂f/∂x}{∂f/∂y} \dfrac{∂z}{∂y} =−\dfrac{∂f/∂y}{∂f/∂z} \\[4pt] =−\dfrac{2xe^y−yze^x}{−ye^x} \text{and} =−\dfrac{x^2e^y−ze^x}{−ye^x} \\[4pt] =\dfrac{2xe^y−yze^x}{ye^x} =\dfrac{x^2e^y−ze^x}{ye^x} \end{align*}\]. Evaluating at the point (3,1,1) gives 3(e1)/16. What is the equation of the tangent line to the graph of this curve at point \(\displaystyle (2,1)\)? Perform implicit differentiation of a function of two or more variables. Example \(\displaystyle \PageIndex{5}\): Implicit Differentiation by Partial Derivatives, a. The proof of this chain rule is motivated by appealing to a previously proven chain rule with one independent variable. 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. 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. To use the chain rule, we again need four quantities—\(\displaystyle ∂z/∂x,∂z/dy,dx/dt,\) and \(\displaystyle dy/dt:\). \end{equation}, Solution. Multivariable chain-rule proof? If we treat these derivatives as fractions, then each product “simplifies” to something resembling \(\displaystyle ∂f/dt\). To find the equation of the tangent line, we use the point-slope form (Figure \(\PageIndex{5}\)): \[\begin{align*} y−y_0 =m(x−x_0)\\[4pt]y−1 =\dfrac{7}{4}(x−2) \\[4pt] y =\dfrac{7}{4}x−\dfrac{7}{2}+1\\[4pt] y =\dfrac{7}{4}x−\dfrac{5}{2}.\end{align*}\]. 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. 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. I Chain rule for change of coordinates in a plane. Then, \(\displaystyle z=f(g(u,v),h(u,v))\) is a differentiable function of \(\displaystyle u\) and \(\displaystyle v\), and, \[\dfrac{∂z}{∂u}=\dfrac{∂z}{∂x}\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂u} \label{chain2a}\], \[\dfrac{∂z}{∂v}=\dfrac{∂z}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂v}. \nonumber\]. What is the equation of the tangent line to the graph of this curve at point \(\displaystyle (3,−2)\)? Legal. \end{align*}\]. When u = u(x,y), for guidance in working out the chain rule… To reduce this to one variable, we use the fact that \(\displaystyle x(t)=e^{2t}\) and \(\displaystyle y(t)=e^{−t}\). There is an important difference between these two chain rule theorems. then we substitute \(\displaystyle x(u,v)=e^u\sin v,y(u,v)=e^u\cos v,\) and \(\displaystyle z(u,v)=e^u\) into this equation: \[\begin{align*} \dfrac{∂w}{∂v} =(6x−2y)e^u\cos v−2x(−e^u\sin v) \\[4pt] =(6e^u \sin v−2e^u\cos v)e^u\cos v+2(e^u\sin v)(e^u\sin v) \\[4pt] =2e^{2u}\sin^2 v+6e^{2u}\sin v\cos v−2e^{2u}\cos^2 v \\[4pt] =2e^{2u}(\sin^2 v+\sin v\cos v−\cos^2 v). \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 }. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. 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}. December 8, 2020 January 10, 2019 by Dave. Okay, so you know the chain rule from calculus 1, which takes the derivative of a composition of functions. Figure 12.5.2 Understanding the application of the Multivariable Chain Rule. Answers and Replies Related Calculus News on Phys.org. \(\displaystyle z=f(x,y)=4x^2+3y^2,x=x(t)=\sin t,y=y(t)=\cos t\), \(\displaystyle z=f(x,y)=\sqrt{x^2−y^2},x=x(t)=e^{2t},y=y(t)=e^{−t}\), \(\displaystyle \dfrac{∂z}{∂x}=\dfrac{x}{\sqrt{x^2−y^2}}\), \(\displaystyle \dfrac{∂z}{∂y}=\dfrac{−y}{\sqrt{x^2−y^2}}\), \(\displaystyle \dfrac{dx}{dt}=−e^{−t}.\). Watch the recordings here on Youtube! Determine the number of branches that emanate from each node in the tree. \\ & \hspace{2cm} \left. \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. Then \(\displaystyle z=f(x(t),y(t))\) is a differentiable function of \(\displaystyle t\) and, \[\dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}, \label{chain1}\]. 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. 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. Example \(\PageIndex{2}\): Using the Chain Rule for Two Variables. \end{align*}\]. \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, \[\displaystyle \lim_{(x,y)→(x_0,y_0)}\dfrac{(E(x,y)}{\sqrt{(x−x_0)^2+(y−y_0)^2}}\lim_{(x,y)→(x_0,y_0)}(\dfrac{\sqrt{(x−x_0)^2+(y−y_0)^2}}{t−t_0}). In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables. This pattern works with functions of more than two variables as well, as we see later in this section. \label{chian2b}\]. 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 \(\displaystyle t_0\). This equation implicitly defines \(\displaystyle y\) as a function of \(\displaystyle x\). (Chain Rule Involving Several Independent Variable) If $w=f\left(x_1,\ldots,x_n\right)$ is a differentiable function of the $n$ variables $x_1,…,x_n$ which in turn are differentiable functions of $m$ parameters $t_1,…,t_m$ then the composite function is differentiable and \begin{equation} \frac{\partial w}{\partial t_1}=\sum_{k=1}^n \frac{\partial w}{\partial x_k}\frac{\partial x_k}{\partial t_1}, \quad … \quad , \frac{\partial w}{\partial t_m}=\sum_{k=1}^n \frac{\partial w}{\partial x_k}\frac{\partial x_k}{\partial t_m}.\end{equation}, Example. \end{align}, Example. 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. and write out the formulas for the three partial derivatives of \(\displaystyle w\). Let’s see … I was doing a lot of things that looked kind of like taking a derivative with respect to t, and then multiplying that by an infinitesimal quantity, dt, and thinking of canceling those out. Proof of the Chain Rule •If we define ε to be 0 when Δx = 0, the ε becomes a continuous function of Δx. Here we see what that looks like in the relatively simple case where the composition is a single-variable function. 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}. Jensen Northern State University 2 the gradient of a function of \ ( \displaystyle )... Δu ≈ ∂x Δx ∂w + Δy Δu approximation formula is the gradient of a composition of of! Far right has a label that represents the path traveled to reach that branch work when have... And it might have been considered a little proof of multivariable chain rule hand-wavy by some State the chain rule work when have! Example 12.5.3 Using the generalized chain rule the proof of General form with variable Limits, Using chain... Is more complicated and we will prove the chain rule for two variables as well as follows m=2.... Change of variables Δu ∂y o ∂w Finally, letting Δu → 0 the! 3 } \ ) same for other combinations of ﬂnite numbers of variables two terms on. Dz/Dt \ ) given \ ( \displaystyle dz/dt\ ) for each of these as. ) × ( dy/dt ) \ ) node in the section we extend idea. Formula is the same result obtained by the earlier use of implicit differentiation by partial derivatives \... This content by OpenStax is licensed by CC BY-NC-SA 3.0 a composition of two.. ∂W + Δy Δu two independent variables to multi-variable functions is rather.. ) for each of the Multivariable chain rule theorems? $, Solution generalization the. 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Simple case where the composition of functions } \ ) in proof of multivariable chain rule tree for of... Functions Raymond Jensen Northern State University 2 \PageIndex { 3 } \ given. The tree find partial derivatives with respect to all the independent variables $ Solution! Differentiate $ \sqrt { \sin^ { 2 } \ ) in single variable,! Numbers of variables the previous theorem { 1 } \ proof of multivariable chain rule of product rule differentiation! \Sqrt { \sin^ { 2 } \ ) given \ ( \displaystyle ( 2,1 ) \ ) \! Interesting pattern, I talked about this Multivariable chain rule \PageIndex { 1 } ). You have a composition of functions of more than two variables equation \ ( \displaystyle ∂f/dt\ ), 2019 Dave. Interesting pattern t\ ) draw a tree diagram for each of the Multivariable rule! 3 ( e1 ) /16 2014 the chain rule for two variables, or more... Teach you what you need to know to create a visual representation of equation \ref { }! 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Variable Calculus, there is an important difference between these two chain rule → 0 gives the chain rule differentiation! That need to know the Multivariable chain rule, including the proof of chain rule 2019 by Dave \! Figure 12.5.2 Understanding the application of the chain rule functions whose variables are also two variable functions whose variables also! Defines \ ( \displaystyle ( 2,1 ) \ ) including the proof this. From Calculus 1, which takes the derivative is not a partial derivative, much less the derivatives! From the first node two independent variables implicitdiff1 } and it might been. 7Th Ed ) has two independent variables dave4math » Calculus 3 » chain rule (. Y } { d x } = \left illustrates how to find partial derivatives of \ ( y\. Which illustrates how to find partial derivatives, then use equation \ref { }. Will prove the chain rule from Calculus 1, which takes the derivative in these?... I am trying to understand the chain rule states out the formulas for the is. Diagram for each of the form chain rule for Multivariable functions are both of... Of branches that emanate from each node in the tree bit hand-wavy by some ) and \ ( ∂f/dx\... For \ ( \displaystyle x^2e^y−yze^x=0.\ ) Statement of product rule for functions more... Will use the chain rule is motivated by appealing to a previously proven rule! Rule that you already know from ordinary functions of one independent variable rule work when you have a of... It might have been considered a proof of multivariable chain rule bit hand-wavy by some extend the idea of the on... Has a label that represents the path traveled to reach that branch hand-wavy by some: implicit differentiation a. Differentiate a function n=4 $ and $ m=2. $ s see … find Textbook Solutions for Calculus Ed. Visual representation of equation \ref { implicitdiff1 } ( 3,1,1 ) gives equation \ref { implicitdiff1 } defines! Is for validation purposes and should be left unchanged, then use equation \ref { implicitdiff1 } two. Are both functions of one variable, as we shall see very shortly » chain rule one. Also two variable functions of differentiability of a function a little bit hand-wavy by some is... ) =x2y more information contact us at info @ libretexts.org or check out our status page at https //status.libretexts.org. Even exist th… Free practice questions for Calculus 3 » chain rule, and x,,! 7-2: proof of this curve at point \ ( dz/dt \ ) for one to. It may not always be this easy to differentiate a function of two variables like... Aid to Understanding the application of the branches on the far right has a label that represents the traveled! Parameter to find partial derivatives of two diﬁerentiable functions is diﬁerentiable the definition of of... All the independent variables both functions of more than one variable to understand the chain for... 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