Since z is a function of the two variables x and y, the derivatives in the Chain Rule for z with respect to x and y are partial derivatives. \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. Or perhaps they are both functions of two variables, or even more. To use the chain rule, we again need four quantities—\(\displaystyle ∂z/∂x,∂z/dy,dx/dt,\) and \(\displaystyle dy/dt:\). In this diagram, the leftmost corner corresponds to \(\displaystyle z=f(x,y)\). Missed the LibreFest? where the ordinary derivatives are evaluated at \(\displaystyle t\) and the partial derivatives are evaluated at \(\displaystyle (x,y)\). 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. The Multivariable Chain Rule is used to differentiate functions with inputs of multiple variables. In calculus-online you will find lots of 100% free exercises and solutions on the subject Multivariable Chain Rule that are designed to help you succeed! 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. > Example: Consider a parameterized curve (u,v)=g(t), and a parameterized. \end{align}. In this section we extend the Chain Rule to functions of more than one variable. \end{align*}\], The left-hand side of this equation is equal to \(\displaystyle dz/dt\), which leads to, \[\dfrac{dz}{dt}=f_x(x_0,y_0)\dfrac{dx}{dt}+f_y(x_0,y_0)\dfrac{dy}{dt}+\lim_{t→t_0}\dfrac{E(x(t),y(t))}{t−t_0}. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. This is the simplest case of taking the derivative of a composition involving multivariable functions. (i) As a rule, e.g., “double and add 1” (ii) As an equation, e.g., f(x)=2x+1 (iii) As a table of values, e.g., x 012 5 20 … Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. surface (x,y,z)=f(u,v). b. Exercise. That is 2xy, and maybe I should point out that this is w sub x, times dx over dt plus -- Well, w sub y is x squared times dy over dt plus w sub z, which is going to be just one, dz over dt. Most of us last saw calculus in school, but derivatives are a critical part of machine learning, particularly deep neural networks, which are trained by optimizing a loss function. 0. If we compose a differentiable function with a differentiable function , we get a function whose derivative is. If we treat these derivatives as fractions, then each product “simplifies” to something resembling \(\displaystyle ∂f/dt\). 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 \(\displaystyle t\): \[\displaystyle z=f(x,y)=f(x(t),y(t))=4(x(t))^2+3(y(t))^2=4\sin^2 t+3\cos^2 t. \nonumber\], \[\displaystyle \dfrac{dz}{dt}=2(4\sin t)(\cos t)+2(3\cos t)(−\sin t)=8\sin t\cos t−6\sin t\cos t=2\sin t\cos t, \nonumber\]. 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. \nonumber\], The slope of the tangent line at point \(\displaystyle (2,1)\) is given by, \[\displaystyle \dfrac{dy}{dx}∣_{(x,y)=(2,1)}=\dfrac{3(2)−1+2}{2−1+3}=\dfrac{7}{4} \nonumber\]. \end{align}, Example. 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. EXPECTED SKILLS: Be able to compute partial derivatives with the various versions of the multivariate chain rule. First the one you know. Calculate \(dz/dt \) given the following functions. \end{align*}\], The formulas for \(\displaystyle ∂w/∂u\) and \(\displaystyle ∂w/∂v\) are, \[\begin{align*} \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}. Chapter 5 … Then the derivative of y with respect to t is the derivative of y with respect to x multiplied by the derivative of x with respect to t … Notation, domain, and range; Parameterized surfaces; Quadratic surfaces; Surfaces; Surfaces in other coordinate systems; Traces, contours, and level sets; Differentiation of multivariable functions. To derive the formula for \(\displaystyle ∂z/∂u\), start from the left side of the diagram, then follow only the branches that end with \(\displaystyle u\) and add the terms that appear at the end of those branches. 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]. 2 $\begingroup$ I am trying to understand the chain rule under a change of variables. \end{align*}\], Then we substitute \(\displaystyle x(u,v)=3u+2v\) and \(\displaystyle y(u,v)=4u−v:\), \[\begin{align*} \dfrac{∂z}{∂v} =14x−6y \\[4pt] =14(3u+2v)−6(4u−v) \\[4pt] =18u+34v \end{align*}\]. 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. which is the same solution. \end{align*}\]. \end{align*}\]. Calculate \(\displaystyle dy/dx\) if y is defined implicitly as a function of \(\displaystyle x\) via the equation \(\displaystyle 3x^2−2xy+y^2+4x−6y−11=0\). 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. Well, the chain rule tells us that dw/dt is, we start with partial w over partial x, well, what is that? Vector form of the multivariable chain rule. This connection between parts (a) and (c) provides a multivariable version of the Chain Rule. Dave will help you with what you need to know, Calculus (Start Here) – Enter the World of Calculus, Mathematical Proofs (Using Various Methods), Chinese Remainder Theorem (The Definitive Guide), Math Solutions: Step-by-Step Solutions to Your Problems, Math Videos: Custom Made Videos For Your Problems, LaTeX Typesetting: Trusted, Fast, and Accurate, LaTeX Graphics: Custom Graphics Using TikZ and PGFPlots. Example 12.5.3 Using the Multivariable Chain Rule. \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 }. \end{align*}\]. Calculate \(\displaystyle ∂f/dx\) and \(\displaystyle ∂f/dy\), then use Equation \ref{implicitdiff1}. 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. For the function f(x,y) where x and y are functions of variable t , we first differentiate the function partially with respect to one variable and then that variable is differentiated with respect to t . \[\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}. … David is the founder and CEO of Dave4Math. \\ & \hspace{2cm} \left. Jump to:navigation, search. However, since x = x(t) and y = y(t) are functions of the single variable t, their derivatives are the standard derivatives of functions of one variable. Set \(\displaystyle f(x,y)=3x^2−2xy+y^2+4x−6y−11=0,\) then calculate \(\displaystyle f_x\) and \(\displaystyle f_y: f_x=6x−2y+4\) \(\displaystyle f_y=−2x+2y−6.\), \[\displaystyle \dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y}=\dfrac{6x−2y+4}{−2x+2y−6}=\dfrac{3x−y+2}{x−y+3}. When you compute df /dt for f(t)=Cekt, you get Ckekt because C and k are constants. The Multivariable Chain Rule Nikhil Srivastava February 11, 2015 The chain rule is a simple consequence of the fact that di erentiation produces the linear approximation to a function at a point, and that the derivative is the coe cient appearing in this linear approximation. \frac { 2 x + y + 7 } { 2 y - x + 3 } \right| _ { ( 3 , - 2 ) } = \dfrac { 2 ( 3 ) + ( - 2 ) + 7 } { 2 ( - 2 ) - ( 3 ) + 3 } = - \dfrac { 11 } { 4 } \nonumber\], Equation of the tangent line: \(\displaystyle y=−\dfrac{11}{4}x+\dfrac{25}{4}\), \(\displaystyle \dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}\), \(\displaystyle \dfrac{dz}{du}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u}\dfrac{dz}{dv}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂v}\), \(\displaystyle \dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_1}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}\). Last, each of the branches on the far right has a label that represents the path traveled to reach that branch. The simplest version of the Multivariable Chain Rule is as follows: d d t f ( x ( t ) , y ( t ) ) = ∂ f ∂ x ∗ \end{equation}. Let $w=f(t)$ be a differentiable function of $t$, where $t =\left(x^2+y^2 +z^2\right)^{1/2}.$ Show that \begin{equation} \left( \frac{d w}{d t} \right)^2=\left( \frac{\partial w}{\partial x} \right)^2+\left( \frac{\partial w}{\partial y} \right)^2+\left(\frac{\partial w}{\partial z} \right)^2.\end{equation}, Exercise. 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. If $f$ is differentiable and $z=u+f\left(u^2v^2\right)$, show that \begin{equation} u\frac{\partial z}{\partial u}-v\frac{\partial z}{\partial v}=u. Multivariable chain rule intuition. Then, \(\displaystyle z=f(g(u,v),h(u,v))\) is a differentiable function of \(\displaystyle u\) and \(\displaystyle v\), and, \[\dfrac{∂z}{∂u}=\dfrac{∂z}{∂x}\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂u} \label{chain2a}\], \[\dfrac{∂z}{∂v}=\dfrac{∂z}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂v}. If we are given the function y = f(x), where x is a function of time: x = g(t). \end{align*} \]. Hot Network Questions Advice for first electric guitar How do I use an advice to … Then \(\displaystyle z=f(x(t),y(t))\) is a differentiable function of \(\displaystyle t\) and, \[\dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}, \label{chain1}\]. Again, 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 \(\displaystyle t\): \[\begin{align*} z =f(x,y) \\[4pt] =f(x(t),y(t)) \\[4pt] =\sqrt{(x(t))^2−(y(t))^2} \\[4pt] =\sqrt{e^{4t}−e^{−2t}} \\[4pt] =(e^{4t}−e^{−2t})^{1/2}. +\frac{\partial u}{\partial y}\left(-e^s \sin t\right) +\frac{\partial ^2 u}{\partial x \partial y}\left(-e^{2s} \cos t \sin t\right) +\frac{\partial ^2 u}{\partial y^2}e^{2s} \cos ^2 t\right] \\ & =e^{-2s}\left[\frac{ \partial ^2 u}{\partial x^2}e^{2s} \cos ^2 t +\frac{ \partial ^2 u}{\partial y^2}e^{2s} \sin ^2 t +\frac{ \partial ^2 u}{\partial x^2}\left(e^{2s} \sin ^2 t\right) +\frac{ \partial ^2 u}{\partial y^2}e^{2s} \cos ^2 t\right] \\ & =\frac{ \partial ^2u}{\partial x^2}+\frac{ \partial ^2u}{\partial y^2}. Note, that the sizes of the matrices are automatically of the right. Okay, so you know the chain rule from calculus 1, which takes the derivative of a composition of functions. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. Dave4Math » Calculus 3 » Chain Rule for Multivariable Functions. Recall that the chain rule for the derivative of a composite of two functions can be written in the form, \[\dfrac{d}{dx}(f(g(x)))=f′(g(x))g′(x).\]. Starting from the left, the function \(\displaystyle f\) has three independent variables: \(\displaystyle x,y\), and \(\displaystyle z\). \end{align*}\]. We have \(\displaystyle f(x,y,z)=x^2e^y−yze^x.\) Therefore, \[\begin{align*} \dfrac{∂f}{∂x} =2xe^y−yze^x \\[4pt] \dfrac{∂f}{∂y} =x^2e^y−ze^x \\[4pt] \dfrac{∂f}{∂z} =−ye^x\end{align*}\], \[\begin{align*} \dfrac{∂z}{∂x} =−\dfrac{∂f/∂x}{∂f/∂y} \dfrac{∂z}{∂y} =−\dfrac{∂f/∂y}{∂f/∂z} \\[4pt] =−\dfrac{2xe^y−yze^x}{−ye^x} \text{and} =−\dfrac{x^2e^y−ze^x}{−ye^x} \\[4pt] =\dfrac{2xe^y−yze^x}{ye^x} =\dfrac{x^2e^y−ze^x}{ye^x} \end{align*}\]. 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 \(\displaystyle t_0\). If $z=x^2y+3x y^4,$ where $x=e^t$ and $y=\sin t$, find $\frac{d z}{d t}.$. 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. \end{equation} At what rate is the distance between the two objects changing when $t=\pi ?$, Solution. What is the equation of the tangent line to the graph of this curve at point \(\displaystyle (3,−2)\)? This gives us Equation. Recall from implicit differentiation provides a method for finding \(\displaystyle dy/dx\) when \(\displaystyle y\) is defined implicitly as a function of \(\displaystyle x\). 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). All we need to do is use the formula for multivariable chain rule. So I was looking for a way to say a fact to a particular level of students, using the notation they understand. In this case the chain rule says: DF (t) = (Df)(g(t)) . 12.5: The Multivariable Chain Rule. Because $z=f(x,y)$ is differentiable, we can write the increment $\Delta z$ in the following form: \begin{equation} \Delta z=\frac{\partial z}{\partial x}\Delta x+\frac{\partial z}{\partial y}\Delta y+\epsilon_1\Delta x+\epsilon_2\Delta y \end{equation} where $\epsilon_1\to 0$ and $\epsilon_2\to 0$ as both $\Delta x\to 0$ and $\Delta y\to 0.$ Dividing by $\Delta t\neq 0,$ we obtain \begin{equation} \frac{\Delta z}{\Delta t}=\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}. Partial Derivatives-II ; Differentiability-I; Differentiability-II; Chain rule-I; Chain rule-II; Unit 3. I think you're mixing up the chain rule for single- and multivariable functions. Essentially the same procedures work for the multi-variate version of the Chain Rule. able chain rule helps with change of variable in partial diﬀerential equations, a multivariable analogue of the max/min test helps with optimization, and the multivariable derivative of a scalar-valued function helps to ﬁnd tangent planes and trajectories. In this multivariable calculus video lesson we will explore the Chain Rule for functions of several variables. which is the same result obtained by the earlier use of implicit differentiation. It is often useful to create a visual representation of Equation for the chain rule. 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 }. Computing the derivatives shows df dt = (2x) (2t) + (2y) (4t3). Each of these three branches also has three branches, for each of the variables \(\displaystyle t,u,\) and \(\displaystyle v\). The generalization of the chain rule to multi-variable functions is rather technical. To get the formula for \(\displaystyle dz/dt,\) add all the terms that appear on the rightmost side of the diagram. Have questions or comments? \end{align*}\]. 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). Curvature. \(\displaystyle \dfrac{dz}{dt}=\dfrac{∂f}{∂x}\dfrac{dx}{dt}+\dfrac{∂f}{∂y}\dfrac{dy}{dt}\), \(\displaystyle =(2x−3y)(6\cos2t)+(−3x+4y)(−8\sin2t)\), \(\displaystyle =−92\sin 2t \cos 2t−72(\cos ^22t−\sin^22t)\). 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. And it's not just any old scalar calculus that pops up---you need differential matrix calculus, the shotgun wedding of linea… In the multivariate chain rule one variable is dependent on two or more variables. And this is known as the chain rule. The Chain Rule with One Independent Variable, The Chain Rule with Two Independent Variables, The Chain Rule with Several Independent Variables, Choose your video style (lightboard, screencast, or markerboard), 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), Triple Integrals in Cylindrical and Spherical Coordinates. Multivariable chain rule and directional derivatives. \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 proof of this theorem uses the definition of differentiability of a function of two variables. Concepts for multivariable functions. 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. Somehow … Let us remind ourselves of how the chain rule works with two dimensional functionals. +\frac{\partial u}{\partial y}e^s \sin t +\left[\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\right]e^s \sin t \right. $(1) \quad f(x,y)=\left(1+x^2+y^2\right)^{1/2}$ 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. One way of describing the chain rule is to say that derivatives of compositions of differentiable functions may be obtained by linearizing. 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. Partial Derivatives-II ; Differentiability-I; Differentiability-II; Chain rule-I; Chain rule-II; Unit 3. 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. (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. \[\dfrac { d y } { d x } = \left. 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}. 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. The online Chain rule derivatives calculator computes a derivative of a given function with respect to a variable x using analytical differentiation. In the real world, it is very difficult to explain behavior as a function of only one variable, and economics is no different. We substitute each of these into Equation \ref{chain1}: \[\begin{align*} \dfrac{dz}{dt} =\dfrac{ \partial z}{ \partial x} \cdot \dfrac{dx}{dt}+\dfrac{ \partial z}{ \partial y}\cdot \dfrac{dy}{dt} \\[4pt] =\left(\dfrac{x}{\sqrt{x^2−y^2}}\right) (2e^{2t})+\left(\dfrac{−y}{\sqrt{x^2−y^2}} \right) (−e^{−t}) \\[4pt] =\dfrac{2xe^{2t}−ye^{−t}}{\sqrt{x^2−y^2}}. As Preview Activity 10.3.1 suggests, the following version of the Chain Rule holds in general. \end{equation}. The ones that used notation the students knew were just plain wrong. \end{equation}. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. However, it may not always be this easy to differentiate in this form. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. \nonumber\]. We want to describe behavior where a variable is dependent on two or more variables. Multivariable Chain Rule. 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? Let \(z=x^2y+x\text{,}\) where \(x=\sin(t)\) and \(y=e^{5t}\text{. Find $dy/ dx $, assuming each of the following the equations defines $y$ as a differentiable function of $x.$$(1) \quad \left(x^2-y\right)^{3/2}+x^2y=2$ $(2) \quad \tan ^{-1}\left(\frac{x}{y}\right)=\tan ^{-1}\left(\frac{y}{x}\right)$, Exercise. \\ & \hspace{2cm} \left. (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 … Then z = f(x(t), y(t)) is differentiable at t and dz dt = ∂z ∂xdx dt + ∂z ∂y dy dt. IMOmath: Training materials on chain rule in multivariable calculus. \end{align*}\]. 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. \\ & \hspace{2cm} \left. \label{chian2b}\]. For the single variable case, [tex] (f \circ g)'(x) = f'(g(x))g'(x) Solution A: We'll use theformula usingmatrices of partial derivatives:Dh(t)=Df(g(t))Dg(… In this instance, the multivariable chain rule says that df dt = @f @x dx dt + @f @y dy dt. We’ll start with the chain rule that you already know from ordinary functions of one variable. To implement the chain rule for two variables, we need six partial derivatives—\(\displaystyle ∂z/∂x,\; ∂z/∂y,\; ∂x/∂u,\; ∂x/∂v,\; ∂y/∂u,\) and \(\displaystyle ∂y/∂v\): \[\begin{align*} \dfrac{∂z}{∂x} =6x−2y \dfrac{∂z}{∂y}=−2x+2y \\[4pt] \displaystyle \dfrac{∂x}{∂u} =3 \dfrac{∂x}{∂v}=2 \\[4pt] \dfrac{∂y}{∂u} =4 \dfrac{∂y}{∂v}=−1. Partial Derivative / Multivariable Chain Rule Notation. 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. Solution. 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}. Example. The proof of this chain rule is motivated by appealing to a previously proven chain rule with one independent variable. The Multivariable Chain Rule allows us to compute implicit derivatives easily by just computing two derivatives. 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. The Multivariable Chain Rule is used to differentiate functions with inputs of multiple variables. This branch is labeled \(\displaystyle (∂z/∂y)×(dy/dt)\). More specific economic interpretations will be discussed in the next section, but for now, we'll just concentrate on developing the techniques we'll be using. Viewed 136 times 5. Therefore, there are nine different partial derivatives that need to be calculated and substituted. Subsection 10.5.1 The Chain Rule. Limits for multivariable functions-I; Limits for multivariable functions-II; Continuity of multivariable functions; Partial Derivatives-I; Unit 2. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 14.5: The Chain Rule for Multivariable Functions, [ "article:topic", "generalized chain rule", "intermediate variable", "license:ccbyncsa", "showtoc:no", "authorname:openstaxstrang" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FCalculus%2FBook%253A_Calculus_(OpenStax)%2F14%253A_Differentiation_of_Functions_of_Several_Variables%2F14.5%253A_The_Chain_Rule_for_Multivariable_Functions, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 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. 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