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I have the following relation: $ c = 2c_1-1 $

I also know that $c_1 = c_1(z,t)$. I want a way to simplify: $$ \frac{\partial^2}{\partial z^2}\left(\frac{\partial g(c_1)}{\partial c}\right) $$ which I know is going to involve the chain rule and then probably a product rule for the second one. My attempt is as follows: First, I notice that $c_1 = \frac{c-1}{2} $ so I have $\frac{\partial g}{\partial c}=2\frac{\partial g}{\partial c_1} $. Then we get $$ 2\frac{\partial}{\partial z}\frac{\partial}{\partial z}\left(\frac{\partial g}{\partial c_1}\right)=2\frac{\partial}{\partial z}\left[\frac{\partial^2 g}{\partial c_1^2}\cdot\frac{\partial c_1}{\partial z}\right] $$ Applying the second partial with respect to $z$, we then have to use the product rule: $$ = 2\left[\frac{\partial^3g}{\partial c_1^3}\cdot\left(\frac{\partial c_1}{\partial z}\right)^2+\frac{\partial^2 g}{\partial c_1^2}\cdot\frac{\partial^2 c_1}{\partial z^2}\right] $$ Is this correct? Is there a simpler way to do this?

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Firstly: $\dfrac{\mathrm d c_1}{\mathrm d c}=\dfrac{\mathrm d ((c+1)/2)}{\mathrm d c}=\dfrac 12$


There's not really a faster way to derive it than step-by-step, save perhaps using a shorthand notation to ... keep it short.

Let's say $b=c_1$ and use the subscript notations that $f_z=\tfrac{\partial f}{\partial z}$ and such, then:

$$\begin{align}(g_c)_{zz}&=(g_b~b_c)_{zz}&&\text{chain rule}\\&=\tfrac 12 (g_{b})_{zz}&&b_c=1/2\\&=\tfrac 12(g_{bb}~b_z)_z&&\text{chain rule}\\&=\tfrac 12(g_{bbz}~b_{z}+g_{bb}~b_{zz})&&\text{product rule}\\&=\tfrac 12(g_{bbb}~(b_z)^2+g_{bb}~b_{zz})&&\text{chain rule}\\[3ex]\therefore\qquad\dfrac{\partial^2~~}{\partial z~^2}\dfrac{\mathrm d g}{\mathrm d c}&=\dfrac 12\left(\dfrac{\mathrm d^3 g}{\mathrm d {c_1}^3}\left(\dfrac{\partial c_1}{\partial z}\right)^2+\dfrac{\mathrm d^2 g}{\mathrm d{c_1}^2}\dfrac{\partial^2 c_1}{{\partial z}^2}\right)\end{align}$$

As you had, except for the scalar factor.


PS: Long hand

$$\begin{align}\dfrac{\partial^2~~}{\partial z~^2}\dfrac{\mathrm d g}{\mathrm d c}&=\dfrac{\partial^2~~}{\partial z~^2}\left(\dfrac{\mathrm d g}{\mathrm d b}\cdot\dfrac{\mathrm d b}{\mathrm d c}\right)&&\text{chain rule}\\&=\dfrac 12 \dfrac{\partial^2~~}{\partial z~^2}\dfrac{\mathrm d g}{\mathrm d b}\\&=\dfrac 12\dfrac{\partial~~}{\partial z}\left(\dfrac{\mathrm d^2 g}{\mathrm d b^2}\cdot\dfrac{\partial b}{\partial z}\right)&&\text{chain rule}\\&=\dfrac 12\left(\dfrac{\partial~~}{\partial z}\dfrac{\mathrm d^2 g}{\mathrm d b^2}\cdot\dfrac{\partial b}{\partial z}+\dfrac{\mathrm d^2 g}{\mathrm d b^2}\cdot\dfrac{\partial^2 b}{\partial z^2}\right)&&\text{product rule}\\&=\dfrac 12\left(\dfrac{\mathrm d^3 g}{\mathrm d {b}^3}\left(\dfrac{\partial b}{\partial z}\right)^2+\dfrac{\mathrm d^2 g}{\mathrm d{b}^2}\dfrac{\partial^2 b}{{\partial z}^2}\right)&&\text{chain rule}\end{align}$$

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  • $\begingroup$ Thanks for sharing this clear answer! Can't believe the one mistake I had was in the scalar part $\endgroup$
    – Mjoseph
    Oct 14, 2021 at 13:32

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