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In the book Mathematical Methods for Physics and Engineering by RIley Hobson and Bence, an example is given in chapter 5 of using partial differentiation. The example problem

Now the first partial derivatives are evaluated, which appears perfectly understandable to me. The issue arises when these are then used to calculate the value of the two senond derivatives as below:

Calculation of the second derivative wrt x

Plane polar coordinates, ρ and φ, and Cartesian coordinates, x and y, are related by the expressions $ x = ρ cos φ$, $y = ρ sin φ$. An arbitrary function $f(x, y)$ can be re-expressed as a function $g(ρ, φ)$. Transform the expression $$\frac{∂^{2}f}{ ∂x^{2}} + \frac{∂^{2}f }{∂y^{2}}$$ into one in ρ and φ.

So after differentiation I get the same as the book for the first derivative wrt x: $$\frac{∂}{∂x} = cos φ \frac{∂}{∂ρ} − sinρ\frac{φ}{∂φ}$$

Then it says:

Now it's only a matter of writing: $$\frac{∂^{2}f}{∂x^{2}} = \frac{∂}{∂x}(\frac{∂f}{∂x}) = \frac{∂}{∂x}(\frac{∂}{∂x})f$$ $$=(cos φ \frac{∂}{∂ρ} − sinρ\frac{φ}{∂φ})(cos φ \frac{∂}{∂ρ} − sinρ\frac{φ}{∂φ})g$$ $$=cos^{2} φ \frac{∂^{2}g}{∂ρ^{2}} − \frac{2 cos φ sin φ}{ρ^{2}}\frac{∂g}{∂φ} - \frac{2 cos φ sin φ}{ρ}\frac{∂^{2}g}{∂ρ∂φ} + \frac{sin^{2}φ}{ρ}\frac{∂g}{∂ρ}+\frac{sin^{2}φ}{ρ^{2}}\frac{∂^{2}g}{∂φ^{2}}$$

I don't understand why we can just square the first derivate to get the second. I understand that there are specific examples where this is true, but it seems completely incorrect as general rule.

It would be great if someone could explain the intuition behind this and under what circumstances this rule applies.

ekke
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He is not squaring the first derivative to get the second derivative, he is squaring the "derivative with respect to $x$ operator", which is just notation for composing it with itself. Let $T$ be an operator which eats a function to give you a function, you are confusing $(T(f))^2 $ with $T^2(f) =T(T(f))$.

Suppose $T= d/dx $, the "first derivative operator". Then $T(f)= \frac{df}{dx}=f'$ and $$T^2(f)=\left(\frac{d}{dx}\right)^2f=\frac{d}{dx}\left(\frac{df}{dx}\right)=\frac{d^2f}{dx^2}=f^{"},$$ while $$(T(f))^2 =\left( \frac{df}{dx}\right)^2=\left(f^{\prime}\right)^2.$$

GFR
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  • Can you explain how the act of multiplying the operators is equivalent to composition? In the example they multiply the derivative operators together to get the second derivative operator. I don't understand how this results in composition. – ekke Mar 28 '19 at 18:49
  • It is little more than notation. When you "multiply" two differential operators you should think of applying one after the other, innermost first. It is useful to think of what they would do to a function $f$. So for example the bit $(- \sin\rho \partial_\phi)(\cos \phi\partial_\rho) f$ gives you $-\sin\rho ( \partial_\phi(\cos\phi)\partial_\rho f +\cos\phi \partial_\phi\partial_\rho f) $. Composition equal to multiplication is then essentially the fact that by convention $\partial_x ^2 \equiv \partial_x\circ \partial_x$. – GFR Mar 29 '19 at 10:09