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I am trying to write the term below, $$ \frac{\partial^2(a^n c)}{\partial (a^p x)^2} $$ in terms of $$ \frac{\partial^2c}{\partial x^2}$$ only.

How do I move $a^p$ and $a^n$ out of the derivatives? My understanding is that I can first write it as

$$ \frac{a^n \partial^2(c)}{\partial (a^p x)^2} $$ Then $$ \frac{a^n \partial^2(c)}{a^{2p} \partial (x)^2} $$ so, $$ \frac{\partial^2(a^n c)}{\partial (a^p x)^2} = a^{n-2p}\frac{\partial^2c}{\partial x^2} $$

Is this correct? I am a bit confused on why the expoential are treated differently.

  • Yes. You can tell just by the notation that the bottom is $(dx)^2$, so it’s two times while the top is $d^2 y$ which has one $y$. As for the actual reason, each time you take a derivative, you add an additional factor of $x$ on the bottom in the definition of the derivative. – Eric Apr 24 '21 at 02:01

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Suppose $f$ is a function of $x$, and $u = kx$, so $x = \frac uk$ and $\frac {dx}{du} = \frac 1k$. By the chain rule, $$\frac {df}{du} = \frac{df}{dx}\frac{dx}{du} = \frac 1k\frac{df}{dx}$$ This applies for any function $f$, including $\frac{df}{dx}$. So. $$\frac {d^2f}{du^2} = \frac{d}{du}\left(\frac{df}{du}\right) = \frac 1k\frac{d}{dx}\left(\frac{df}{du}\right) = \frac 1k\frac{d}{dx}\left(\frac1k\frac{df}{dx}\right) = \frac1{k^2}\frac {d^2f}{dx^2}$$ But on the other hand, by linearity of the derivative $$\frac {d^2(mf)}{du^2} = \frac{d}{du}\left(\frac{d(mf)}{du}\right) = \frac{d}{du}\left(m\frac{df}{du}\right) = m\frac{d}{du}\left(\frac{df}{du}\right) = m\frac {d^2f}{du^2}$$

As Eric says, this is an aspect of why we put $d^2f$ on top, but $(dx)^2$ on bottom. They behave differently.

Paul Sinclair
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