I'm using variable substitution to solve a problem. Finding the relationships between the first derivatives of both sets is straightforward using the chain rule, but I'm uncertain if the way I'm determining the second derivative relationships is correct.(adsbygoogle = window.adsbygoogle || []).push({});

Given a description of a problem expressed in the x and y variables, I make the substitution as follows

[itex]r = 3x + y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = x + \frac{1}{5}y[/itex]

The chain rule gives

[itex]\frac{\partial}{\partial x} = 3 \frac{\partial}{\partial r} + \frac{\partial}{\partial s} \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\partial}{\partial y} = \frac{\partial}{\partial r} + \frac{1}{5} \frac{\partial}{\partial s}[/itex]

Solving for [itex]\frac{\partial}{\partial r}[/itex] and [itex]\frac{\partial}{\partial s}[/itex] I get

[itex]\frac{\partial}{\partial r} = - \frac{1}{2} \frac{\partial}{\partial x} + \frac{5}{2} \frac{\partial}{\partial y} \ \ \ \ \ \ \ \ \frac{\partial}{\partial s} = \frac{5}{2} \frac{\partial}{\partial x} - \frac{15}{2} \frac{\partial}{\partial y}[/itex]

Are the second derivative relationships from the chain rule then ?

[itex]\frac{\partial^2}{\partial x^2} = - \frac{1}{2} \frac{\partial^2}{\partial r^2} + \frac{5}{2} \frac{\partial^2}{\partial s^2} \ \ \ \ \ \ \ \ \frac{\partial^2}{\partial y^2} = \frac{5}{2} \frac{\partial^2}{\partial r^2} - \frac{15}{2} \frac{\partial^2}{\partial s^2}[/itex]

Maybe this a strange example as the [itex]\frac{5}{2} \frac{\partial}{\partial y}[/itex] and [itex]\frac{5}{2} \frac{\partial}{\partial x}[/itex] in the separate equations seem to make the relationships symmetrical.

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# Second derivative relationships of substituted varaibles

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