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I get the idea behind partial differentiation but this one is really tricky!

If $z = xe^{-y}$, and $x = \cosh t$, and $y = \cos s$, then what is the partial of $z$ with respect to $s$, and partial of $z$ with respect to $t$?

Thanks

2 Answers 2

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There are two ways to do this:

1.) just plug in the expressions for $x$ and $y$ in terms of $s$ and $t$ and differentiate.

2.) use the multivariate chain rules:

$ \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} $ $ \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} $

Make sense?

By request, $ dz = d(xe^{-y}) =e^{-y}dx+-xe^{-y}dy$ however, since $x = \cosh(t)$ and $y = \cos(s)$ we find $ dx = \sinh(t)dt \qquad \text{and} \qquad dy = -\sin(s)ds $ plug those into the $dz$ formula to obtain: $ dz = e^{-y}\sinh(t)dt -xe^{-y}(-\sin(s)ds) $ We can read from the above the coefficient of $dt$ is $\frac{\partial z}{\partial t}$ and the coefficient of $ds$ is $\frac{\partial z}{\partial s}$. Of course I use $x$ and $y$ here as abbreviations for the $t,s$ formulas. Intuitively, the partial derivative w.r.t. $t$ is when $s$ is held constant so $ds=0$ and this is why this approach works. In advanced calculus we can give better answers in terms of the implicit or inverse function theorems. As a general principle, you can use differentials and proceed formally, this approach goes a long way.

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    @MykeArya none taken.2013-04-13
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$\frac{\partial{z(x,y)}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{s}}=\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{s}}+0=\{-xe^{-y}\}\{-\sin(s)\}$

$\frac{\partial{z(x,y)}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+\frac{\partial{z}}{\partial{y}}\cdot\frac{\partial{y}}{\partial{t}}=\frac{\partial{z}}{\partial{x}}\cdot\frac{\partial{x}}{\partial{t}}+0=\{e^{-y}\}\{-\sinh(s)\}$

It's just applying chain rule and noticing that $\frac{\partial{x}}{\partial{s}}=0$ because $x$ does not depend on $s$. Therfore is constant regarding this variable, and the derivative of a constant vanishes. And similarty $\frac{\partial{y}}{\partial{t}}=0$.

The trick when doing this kind of things, is realizing that partial differenciation means that the other variables are constants regarding this operation. Otherwise is just being careful and not to miss anything in the steps.