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I think I know what is going on but please point out my mistakes and gaps in understanding!

1 the author decomposes the partial using CHAIN RULE

$$\frac{\partial \psi}{\partial x} = \frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}$$

2 the author uses the PRODUCT RULE somehow to get this?

$$\frac{\partial^2 \psi}{\partial x^2} = \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial \psi}{\partial r}\frac{\partial^2 r}{\partial x^2}$$

how does he use the product rule?: I am guessing he assigns:

$$u = \frac{\partial \psi}{\partial r}$$ $$v = \frac{\partial r}{\partial x}$$ $$\frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial x}\right) = \frac{\partial}{\partial x}\left( \frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right) = u \frac{\partial}{\partial x}\left(v\right)+v \frac{\partial}{\partial x}\left(u\right)$$

$$ \frac{\partial \psi}{\partial r}\times \frac{\partial}{\partial x}\left(\frac{\partial r}{\partial x}\right) + \frac{\partial r}{\partial x}\times \frac{\partial}{\partial x}\left(\frac{\partial \psi}{\partial r}\right) $$

I do not understand how the author turned $$ \frac{\partial \psi}{\partial r}\times \frac{\partial}{\partial x}\left(\frac{\partial r}{\partial x}\right)$$

into

$$ \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 $$

my understanding now from reading the answers is that

$$\frac{\partial}{\partial x} \frac{\partial \psi}{\partial r}= \frac{\partial^2}{\partial x \partial r } $$

$$\frac{\partial^2 \psi}{\partial x \partial r } = \frac{\partial}{\partial r}\frac{\partial \psi}{\partial x}$$

but $$\frac{\partial \psi}{\partial x} = \left(\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right)$$

so $$\frac{\partial}{\partial r} \left(\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}\right)$$

$$\frac{\partial^2 \psi}{\partial r^2}\frac{\partial r}{\partial x}$$

multiplying this by the $$\frac{\partial r}{\partial x}$$ that I left out

results in $$ \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 $$

could I start an infinite loop by factoring out another $$\frac{\partial \psi}{\partial x}$$ and replace it with $$\frac{\partial \psi}{\partial r}\frac{\partial r}{\partial x}$$

Community
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Conor
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    Indeed, that's how the product rule is being used here. How does your result differ from the author's result when you apply the product rule as you suggest? (By the way, the chain rule is rather unorthodox in this form. If you're transforming from the single variable $r$ to the single variable $x$, you should use total derivatives instead of partial derivatives. If there are further variables, they should appear in the chain rule, and the given form is correct only if the derivative of the other variables with respect to $x$ vanishes.) – joriki Aug 10 '16 at 13:38
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    I edited my answer to answer your last question. – Sanderr Aug 10 '16 at 14:03

2 Answers2

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$$ \frac{\partial^2 \psi}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial \psi}{\partial r} \frac{\partial r}{\partial x} \right) = \frac{\partial^2 \psi}{\partial x \partial r}\frac{\partial r}{\partial x} + \frac{\partial \psi}{\partial r} \frac{\partial^2 r}{\partial x^2} = \frac{\partial^2 \psi}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2 + \frac{\partial \psi}{\partial r} \frac{\partial^2 r}{\partial x^2} $$

The first equality is the chain rule, the second is the product rule, then the last one is the chain rule again. Do you see how the last equality follows from the chain rule?

Edit: $$\frac{\partial^2 \psi}{\partial x \partial r} = \frac{\partial}{\partial r}\frac{\partial \psi}{\partial x} = \frac{\partial}{\partial r} \left(\frac{\partial \psi }{\partial r}\frac{\partial r}{\partial x}\right) = \frac{\partial^2 \psi}{\partial r^2} \frac{\partial x}{\partial r} +\frac{\partial \psi}{\partial r}\frac{\partial}{\partial x} \frac{\partial r}{\partial r} $$ The last term is equal to zero.

Sanderr
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  • I thought at first that an infinite loop would occur involving factoring out $$\frac{\partial \psi}{\partial x}$$ I still am worried that this might happen – Conor Aug 10 '16 at 14:16
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$\frac{\partial }{\partial x}\frac{\partial \psi}{\partial r}=\frac{\partial^2 \psi}{\partial r^2}\frac{\partial r}{\partial x}$

mathshungry
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