My textbook states:

"$\zeta = x+\lambda_1y$ and $\eta = x+\lambda_2y$

From the chain rule we have:

$\frac{\partial}{\partial x} = \frac{\partial}{\partial \zeta}+\frac{\partial}{\partial \eta} $ and

$\frac{\partial}{\partial y} = \lambda_1\frac{\partial}{\partial \zeta}+\lambda_2\frac{\partial}{\partial \eta} $"

---End quote

How does the chain rule result in these equations?

My understanding of the chain rule was : $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial p}\frac{\partial p}{\partial x}$ where $p$ itself is a function of $x$.

I dont see how this relates to the above reasoning


1 Answer 1


You are confusing the multivariant and single-variant chain rules in your fomula. You don't get the multivariant version simply by replacing "$d$" with "$\partial$". $f$ is function of two variables, not just one, and $\zeta$ and $\eta$ are functions of both $x$ and $y$, not just one or the other. The chain rule for two variables is: $$\frac{\partial f}{\partial x} = \frac{\partial \zeta}{\partial x}\frac{\partial f}{\partial \zeta} + \frac{\partial \eta}{\partial x}\frac{\partial f}{\partial \eta}$$ $$\frac{\partial f}{\partial y} = \frac{\partial \zeta}{\partial y}\frac{\partial f}{\partial \zeta} + \frac{\partial \eta}{\partial y}\frac{\partial f}{\partial \eta}$$

In the version they gave, they are talking about the operators themselves, rather than the operators evaluated for a dummy function $f$, so they drop the $f$. And they've already calculated and substituted the various values for $\frac{\partial \zeta}{\partial x}, \frac{\partial \eta}{\partial x}, \frac{\partial \zeta}{\partial y}, \frac{\partial \eta}{\partial y}$.

  • $\begingroup$ Thanks a lot. I was thinking in terms of total derivatives... $\endgroup$
    – user32882
    Commented Nov 27, 2015 at 1:18
  • $\begingroup$ How about $\frac{\partial^2}{\partial x^2}$? $\endgroup$
    – user32882
    Commented Nov 27, 2015 at 1:35
  • $\begingroup$ Apply $\frac{\partial}{\partial x}$ to the expression above for $\frac{\partial f}{\partial x}$ and start cranking. It's a nasty mixture of partial derivatives, but you can do it. $\endgroup$ Commented Nov 27, 2015 at 3:15

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