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Can somebody give me an intuitive explanation for the below equations. I'm not sure how they come about and how they can be perceived logically.

$$\frac{\partial z}{\partial s} =\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial s} \ \ \text{and} \ \ \frac{\partial z}{\partial t} =\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} $$

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    $\begingroup$ Presumably $z = f$. $\endgroup$ – Yuval Filmus Jan 28 '15 at 23:38
  • $\begingroup$ @YuvalFilmus Yes, my apologies. $\endgroup$ – Mark Jan 28 '15 at 23:39
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    $\begingroup$ @YuvalFilmus I don't think it's correct to say that $z = f$ ; rather, $z(s,t) = f(x(s,t),y(s,t))$. $f$ and $z$ aren't the same function. People often aren't careful to distinguish $z$ and $f$ in this context but I find that to be confusing. $\endgroup$ – littleO Jan 29 '15 at 0:18
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Suppose for simplicity that $z = f(x,y)$ and $x,y$ are functions of $s$. The partial derivatives of $z$ given a first-order approximation for $z$: $$ f(x+\Delta x,y+\Delta y) \approx f(x,y) + \frac{\partial f}{\partial x}(x,y) \Delta x + \frac{\partial f}{\partial y}(x,y) \Delta y. $$ The error in this approximation should be "small", say $o(\Delta x+\Delta y)$ (if you don't know what this means, it's not important). Similarly, $$ x(s+\Delta s) \approx x(s) + \frac{\partial x}{\partial s}(s) \Delta s, \quad y(s+\Delta s) \approx y(s) + \frac{\partial y}{\partial s}(s) \Delta s. $$ Finally, $\frac{\partial f}{\partial s}$ satisfies $$ f(x(s+\Delta s),y(s+\Delta s)) \approx f(x(s),y(s)) + \frac{\partial f(x,y)}{\partial s}(s) \Delta s. $$ We can now prove the formula: $$ \begin{align*} f(x(s+\Delta s),y(s+\Delta s)) &\approx f(x(s) + \frac{\partial x}{\partial s}(s) \Delta s, y(s) + \frac{\partial y}{\partial s}(s) \Delta s) \\ &\approx f(x(s),y(s)) + \frac{\partial f}{\partial x}(x,y) \frac{\partial x}{\partial s}(s) \Delta s + \frac{\partial f}{\partial y}(x,y) \frac{\partial y}{\partial s}(s) \Delta s. \end{align*} $$

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According to the chain rule $$(g \circ f)'(x_0) = g'(f(x_0))f'(x_0)$$

The matrix of the composition of two linear funtions is the product of their respective matrices. Hence the matrix $(g \circ f)'(x_0)$ of $d_{x_0}(g \circ f)$ is

$$\begin{pmatrix}\frac{\partial z}{\partial s} & \frac{\partial z}{\partial t}\end{pmatrix} = \begin{pmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{pmatrix} \begin{pmatrix}\frac{\partial x}{\partial s} \\\frac{\partial y}{\partial t}\end{pmatrix} $$

To illustrate suppose $f$ and $g$ are given by

$$w = g(x,y,z), \ \ x = f_1 (s,t), \ \ y = f_2(s,t), \ \ z = f_3(s,t) $$

Then by the chain rule

$$\begin{pmatrix}\frac{\partial w}{\partial s} & \frac{\partial w}{\partial t}\end{pmatrix} = \begin{pmatrix}\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}& \frac{\partial g}{\partial z}\end{pmatrix} \begin{pmatrix}\frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t}\\\frac{\partial z}{\partial s} & \frac{\partial z}{\partial t}\end{pmatrix} $$

Which yields

$$\begin{align}\frac{\partial w}{\partial s} &= \frac{\partial g}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial g}{\partial y}\frac{\partial y}{\partial s}+ \frac{\partial g}{\partial z}\frac{\partial z}{\partial s}\\\frac{\partial w}{\partial t} &=\frac{\partial g}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial g}{\partial y}\frac{\partial y}{\partial t}+ \frac{\partial g}{\partial z}\frac{\partial z}{\partial t} \end{align}$$

*Note: Take a close look to transformation matrices and Jacobian Matrices and this may also help.

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Just be aware that the $\partial f$ in $\frac{\partial f}{\partial x}$ is different from (and independent of) the $\partial f$ in $\frac{\partial f}{\partial y}$. Think of the former $\partial f$ as a change in $f$ due to a change in $x$ and the latter as a change in $f$ due to a change in $y$. Because a change in $s$ causes both a change in $y$ and a change in $x$, we need to add the change in $f$ due to a change in $x$ (which in turn was caused by a change in $s$) to the change in $f$ due to a change in $y$ (which in turn was caused by a change in $s$). If you need more clarification, re-read my first sentence.

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\begin{align} z(s + \Delta s, t) &= f(x(s + \Delta s, t),y(s + \Delta s, t)) \\& \approx f \left(x(s,t) + \frac{\partial x(s,t)}{\partial s} \Delta s,y(s,t) + \frac{\partial y(s,t)}{\partial s} \Delta s \right) \\ \tag{$\spadesuit$}&\approx f(x(s,t),y(s,t)) + \frac{\partial f(x(s,t),y(s,t))}{\partial x} \frac{\partial x(s,t)}{\partial s} \Delta s \\& \qquad \qquad \qquad \quad+ \frac{\partial f(x(s,t),y(s,t))}{\partial y}\frac{\partial y(s,t)}{\partial s} \Delta s. \end{align}

Comparing this with \begin{equation} z(s + \Delta s, t) \approx z(s,t) + \frac{\partial z(s,t)}{\partial s} \Delta s \end{equation} we discover that \begin{equation} \frac{\partial z(s,t)}{\partial s} = \frac{\partial f(x(s,t),y(s,t))}{\partial x} \frac{\partial x(s,t)}{\partial s} + \frac{\partial f(x(s,t),y(s,t))}{\partial y}\frac{\partial y(s,t)}{\partial s} . \end{equation}

The key step is in line ($\spadesuit$), where we use the approximation \begin{equation} f(x + \Delta x, y + \Delta y) \approx f(x,y) + \frac{\partial f(x,y)}{\partial x} \Delta x + \frac{\partial f(x,y)}{\partial y}{\Delta y}. \end{equation}

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  • $\begingroup$ Why does the approximation $\spadesuit$ hold? $\endgroup$ – fishiwhj Jul 24 '15 at 2:44

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