The basic concepts are illustrated through a simple example. \\ & \hspace{2cm} \left. 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}}. Therefore, this value is finite. This equation implicitly defines $$\displaystyle y$$ as a function of $$\displaystyle x$$. \\ & \hspace{2cm} \left. 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*}. Then the following is true wherever the right side expression makes sense (see concept of equality conditional to existence of one side): . 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. All rights reserved. To reduce this to one variable, we use the fact that $$\displaystyle x(t)=e^{2t}$$ and $$\displaystyle y(t)=e^{−t}$$. Consider the ellipse defined by the equation $$\displaystyle x^2+3y^2+4y−4=0$$ as follows. 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. To find $$\displaystyle ∂z/∂v,$$ we use Equation \ref{chain2b}: \begin{align*} \dfrac{∂z}{∂v} =\dfrac{∂z}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}\dfrac{∂y}{∂v} \\[4pt] =2(6x−2y)+(−1)(−2x+2y) \\[4pt] =14x−6y. \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}}}. 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 \frac{\partial u}{\partial s}=(64)(2)+(16)(4)+(0)(0)=192. Let (x,y,z) lie on the ellipsoid \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, without solving for z, find \frac{\partial^2 z}{\partial x^2} and \frac{\partial ^2 z}{\partial x\partial y}., Exercise. $$\displaystyle \dfrac{∂w}{∂v}=\dfrac{15−33\sin 3v+6\cos 3v}{(3+2\cos 3v−\sin 3v)^2}$$, Example $$\PageIndex{4}$$: Drawing a Tree Diagram, \[ w=f(x,y,z),x=x(t,u,v),y=y(t,u,v),z=z(t,u,v) \nonumber. An alternative proof for the chain rule for multivariable functions Raymond Jensen Northern State University 2. \label{chian2b}\]. This diagram can be expanded for functions of more than one variable, as we shall see very shortly. The single-variable chain rule. Since $$\displaystyle f$$ is differentiable at $$\displaystyle P$$, we know that, $z(t)=f(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x−x_0)+f_y(x_0,y_0)(y−y_0)+E(x,y), \nonumber$, $\lim_{(x,y)→(x_0,y_0)}\dfrac{E(x,y)}{\sqrt{(x−x_0)^2+(y−y_0)^2}}=0. \nonumber$. \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 \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} 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}}. Download for free at http://cnx.org. 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 $$\displaystyle y$$. We will prove the Chain Rule, including the proof that the composition of two diﬁerentiable functions is diﬁerentiable. 1. Example. , Solution. \end{align*}\], $\dfrac{dy}{dx}=−\dfrac{∂f/∂x}{∂f/∂y}=−\dfrac{2x}{6y+4}=−\dfrac{x}{3y+2},$. How does the chain rule work when you have a composition involving multiple functions corresponding to multiple variables? Recall that the chain rule for the derivative of a composite of two functions can be written in the form, $\dfrac{d}{dx}(f(g(x)))=f′(g(x))g′(x).$. In this article, I cover the chain rule with several independent variables. Starting from the left, the function $$\displaystyle f$$ has three independent variables: $$\displaystyle x,y$$, and $$\displaystyle z$$. 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}}. $$\displaystyle z=f(x,y)=4x^2+3y^2,x=x(t)=\sin t,y=y(t)=\cos t$$, $$\displaystyle z=f(x,y)=\sqrt{x^2−y^2},x=x(t)=e^{2t},y=y(t)=e^{−t}$$, $$\displaystyle \dfrac{∂z}{∂x}=\dfrac{x}{\sqrt{x^2−y^2}}$$, $$\displaystyle \dfrac{∂z}{∂y}=\dfrac{−y}{\sqrt{x^2−y^2}}$$, $$\displaystyle \dfrac{dx}{dt}=−e^{−t}.$$. , Example. We will do it for compositions of functions of two variables. 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 $$\displaystyle t$$: \[\displaystyle z=f(x,y)=f(x(t),y(t))=4(x(t))^2+3(y(t))^2=4\sin^2 t+3\cos^2 t. \nonumber, $\displaystyle \dfrac{dz}{dt}=2(4\sin t)(\cos t)+2(3\cos t)(−\sin t)=8\sin t\cos t−6\sin t\cos t=2\sin t\cos t, \nonumber$. \end{align*}\]. Example. Let g:R→R2 and f:R2→R (confused?) \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}. Have questions or comments? We now practice applying the Multivariable Chain Rule. To use the chain rule, we again need four quantities—$$\displaystyle ∂z/∂x,∂z/dy,dx/dt,$$ and $$\displaystyle dy/dt:$$. 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. Let w=\ln(x+y), x=uv, y=\frac uv. What is \frac {\partial w}{\partial v}? Also related to the tangent approximation formula is the gradient of a function. 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. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. The upper branch corresponds to the variable $$\displaystyle x$$ and the lower branch corresponds to the variable $$\displaystyle y$$. Example 12.5.3 Using the Multivariable Chain Rule \nonumber. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Okay, so you know the chain rule from calculus 1, which takes the derivative of a composition of functions. 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. 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. If u=f(x,y), where x=e^s \cos t and y=e^s \sin t, show that $$\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=e^{-2s}\left[\frac{\partial ^2u}{\partial s^2}+\frac{\partial ^2u}{\partial t^2}\right].$$, Example. In the last couple videos, I talked about this multivariable chain rule, and I give some justification. Theorem. How would we calculate the derivative in these cases? David Smith (Dave) has a B.S. 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 $$\displaystyle f(x,y)=0$$. +\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. We’ll start with the chain rule that you already know from ordinary functions of one variable. If w=f\left(\frac{r-s}{s}\right), show that r\frac{\partial w}{\partial r}+s\frac{\partial w}{\partial s}=0., Exercise. ... Show proof Implicit function theorem. \\ & \hspace{2cm} \left. 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. Therefore, there are nine different partial derivatives that need to be calculated and substituted. As in single variable calculus, there is a multivariable chain rule. Find Textbook Solutions for Calculus 7th Ed. , Solution. 14.4) I Review: Chain rule for f : D ⊂ R → R. I Chain rule for change of coordinates in a line. If we treat these derivatives as fractions, then each product “simplifies” to something resembling $$\displaystyle ∂f/dt$$. What is the equation of the tangent line to the graph of this curve at point $$\displaystyle (2,1)$$? \end{align*}. EXPECTED SKILLS: and M.S. then 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}{∂u} =(6x−2y)e^u\sin v−2xe^u\cos v+8ze^u \\[4pt] =(6e^u\sin v−2eu\cos v)e^u\sin v−2(e^u\sin v)e^u\cos v+8e^{2u} \\[4pt] =6e^{2u}\sin^2 v−4e^{2u}\sin v\cos v+8e^{2u} \\[4pt] =2e^{2u}(3\sin^2 v−2\sin v\cos v+4). Calculate $$dz/dt$$ given the following functions. Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. \end{align*}. \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). 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*}. \\ & \hspace{2cm} \left. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Chain rule for functions of 2, 3 variables (Sect. Therefore, three branches must be emanating from the first node. 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. 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}. Proof of the chain rule: Just as before our argument starts with the tangent approximation at the point (x 0,y 0). o Δu ∂y o ∂w Finally, letting Δu → 0 gives the chain rule for . Let’s now return to the problem that we started before the previous theorem. 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. State the chain rules for one or two independent variables. In this diagram, the leftmost corner corresponds to $$\displaystyle z=f(x,y)$$. Receive free updates from Dave with the latest news! Two objects are traveling in elliptical paths given by the following parametric equations 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. Write out the chain rule for the function t=f(u,v) where u=u(x,y,z,w) and v=v(x,y,z,w)., Exercise. 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. f(g(x+h))−f(g(x)) h . Neural networks are one of the most popular and successful conceptual structures in machine learning. 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? Alternative Proof of General Form with Variable Limits, using the Chain Rule. To use the chain rule, we need four quantities—$$\displaystyle ∂z/∂x,∂z/∂y,dx/dt$$, and $$\displaystyle dy/dt$$: Now, we substitute each of these into Equation \ref{chain1}: \[\dfrac{dz}{dt}=\dfrac{\partial z}{\partial x} \cdot \dfrac{dx}{dt}+\dfrac{\partial z}{\partial y} \cdot \dfrac{dy}{dt}=(8x)(\cos t)+(6y)(−\sin t)=8x\cos t−6y\sin t. \nonumber, This answer has three variables in it. Section 7-2 : Proof of Various Derivative Properties. 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. Calculate $$\displaystyle dz/dt$$ for each of the following functions: a. \nonumber\]. \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}. +\frac{\partial u}{\partial y}\left(-e^s \sin t\right)+\left[\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\right]e^s \cos t\right] \\ & =e^{-2s}\left[\frac{\partial u}{\partial x}e^s \cos t+\frac{ \partial ^2 u}{\partial x^2}e^{2s} \cos ^2 t +\frac{\partial ^2 u}{\partial x \partial y}\left(e^{2s} \cos t \sin t\right) \right. However, we can get a better feel for it … Proof of the Chain Rule •If we define ε to be 0 when Δx = 0, the ε becomes a continuous function of Δx. Chain rule: let f be differentiable wrt. \[ \begin{align*} z =f(x,y)=x^2−3xy+2y^2 \\[4pt] x =x(t)=3\sin2t,y=y(t)=4\cos 2t \end{align*}. The first term in the equation is $$\displaystyle \dfrac{∂f}{∂x} \cdot \dfrac{dx}{dt}$$ and the second term is $$\displaystyle \dfrac{∂f}{∂y}⋅\dfrac{dy}{dt}$$. \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*}. \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: $$\displaystyle y=−\dfrac{11}{4}x+\dfrac{25}{4}$$, $$\displaystyle \dfrac{dz}{dt}=\dfrac{∂z}{∂x}⋅\dfrac{dx}{dt}+\dfrac{∂z}{∂y}⋅\dfrac{dy}{dt}$$, $$\displaystyle \dfrac{dz}{du}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂u}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂u}\dfrac{dz}{dv}=\dfrac{∂z}{∂x}⋅\dfrac{∂x}{∂v}+\dfrac{∂z}{∂y}⋅\dfrac{∂y}{∂v}$$, $$\displaystyle \dfrac{∂w}{∂t_j}=\dfrac{∂w}{∂x_1}\dfrac{∂x_1}{∂t_j}+\dfrac{∂w}{∂x_2}\dfrac{∂x_1}{∂t_j}+⋯+\dfrac{∂w}{∂x_m}\dfrac{∂x_m}{∂t_j}$$. \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}. 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}.. David is the founder and CEO of Dave4Math. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. (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 $$\frac{\partial w}{\partial t_1}=\sum_{k=1}^n \frac{\partial w}{\partial x_k}\frac{\partial x_k}{\partial t_1}, \quad … \quad , \frac{\partial w}{\partial t_m}=\sum_{k=1}^n \frac{\partial w}{\partial x_k}\frac{\partial x_k}{\partial t_m}.$$, Example. Chain Rule for Multivariable Functions. Exercise. b. b. Let w=f(t) be a differentiable function of t, where t =\left(x^2+y^2 +z^2\right)^{1/2}. Show that $$\left( \frac{d w}{d t} \right)^2=\left( \frac{\partial w}{\partial x} \right)^2+\left( \frac{\partial w}{\partial y} \right)^2+\left(\frac{\partial w}{\partial z} \right)^2.$$, Exercise. \end{align*} \]. Since $$\displaystyle f$$ has two independent variables, there are two lines coming from this corner. First the one you know. Solution A: We'll use theformula usingmatrices of partial derivatives:Dh(t)=Df(g(t))Dg(t). The general form of Leibniz's Integral Rule with variable limits can be derived as a consequence of the basic form of Leibniz's Integral Rule, the Multivariable Chain Rule, and the First Fundamental Theorem of Calculus. \\ & \hspace{2cm} \left. ∂u Ambiguous notation Using x=r \cos \theta and y=r \sin \theta we can state the chain rule to be used: \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 }. Theorem. 6. \begin{align*}\dfrac{∂w}{∂t} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂t}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂t} \\[4pt] \dfrac{∂w}{∂u} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂u}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂u} \\[4pt] \dfrac{∂w}{∂v} =\dfrac{∂w}{∂x}\dfrac{∂x}{∂v}+\dfrac{∂w}{∂y}\dfrac{∂y}{∂v} \end{align*}. Theorem 1. The proof of this chain rule is motivated by appealing to a previously proven chain rule with one independent variable. Dave4Math » Calculus 3 » Chain Rule for Multivariable Functions. Dave will teach you what you need to know. $$\displaystyle \dfrac{dz}{dt}=\dfrac{∂f}{∂x}\dfrac{dx}{dt}+\dfrac{∂f}{∂y}\dfrac{dy}{dt}$$, $$\displaystyle =(2x−3y)(6\cos2t)+(−3x+4y)(−8\sin2t)$$, $$\displaystyle =−92\sin 2t \cos 2t−72(\cos ^22t−\sin^22t)$$. In Note, the left-hand side of the formula for the derivative is not a partial derivative, but in Note it is. 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. +\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. (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. The chain rule for functions of more than one variable involves the partial derivatives with respect to all the independent variables. There is an important difference between these two chain rule theorems. The chain rule for the case when $n=4$ and $m=2$ yields the following the partial derivatives: \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} and \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}. Since $$\displaystyle x(t)$$ and $$\displaystyle y(t)$$ are both differentiable functions of $$\displaystyle t$$, both limits inside the last radical exist. The following theorem gives us the answer for the case of one independent variable. However, it is simpler to write in the case of functions of the form Theorem. Missed the LibreFest? 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 … Now suppose that $$\displaystyle f$$ is a function of two variables and $$\displaystyle g$$ is a function of one variable. And it might have been considered a little bit hand-wavy by some. Thus, for a differentiable function f, we can write Δy = f’(a) Δx + ε Δx, where ε 0 as x 0 (1) •and ε is a continuous function of Δx. 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. 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}. 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. 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). THE CHAIN RULE - Multivariable Differential Calculus - Beginning with a discussion of Euclidean space and linear mappings, Professor Edwards (University of Georgia) follows with a thorough and detailed exposition of multivariable differential and integral calculus. Calculate $$\displaystyle ∂z/∂x,\; ∂z/∂y,\; ∂x/∂u,\; ∂x/∂v,\; ∂y/∂u,$$ and $$\displaystyle ∂y/∂v$$, then use Equation \ref{chain2a} and Equation \ref{chain2b}. In this equation, both $$\displaystyle f(x)$$ and $$\displaystyle g(x)$$ are functions of one variable. Includes full solutions and score reporting., Solution. When $t=\pi ,$ the partial derivatives of $s$ are as follows. This gives us Equation. If $f$ is differentiable and $z=u+f\left(u^2v^2\right)$, show that u\frac{\partial z}{\partial u}-v\frac{\partial z}{\partial v}=u. Statements Statement of product rule for differentiation (that we want to prove) uppose and are functions of one variable. 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