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Let: $\log(s)=z$

I understand that

$$\frac{\partial}{\partial s}=\frac{\partial}{\partial z} \frac{\partial z}{\partial s} = e^{-z}\frac{\partial}{\partial z}$$

What is the second derivative, ie $\frac{\partial^2}{\partial s^2}$

Applying the Product rule I reach the following:

$$\frac{\partial^2}{\partial s^2} = \frac{\partial^2}{\partial z^2} \frac{\partial z}{\partial s} + \frac{\partial^2}{\partial z^2} \frac{\partial z}{\partial s} = e^{-z}\left(\frac{\partial^2}{\partial z^2} - \frac{\partial}{\partial z} \right ) $$

Is this correct or should we reach to:

$$\frac{\partial^2}{\partial s^2} =e^{-2z}\left(\frac{\partial^2}{\partial z^2} - \frac{\partial}{\partial z} \right )$$

PPP
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Anastasiosyal
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  • You're trying to partial differentiante what ? The symbol $;\frac\partial{\partial s};$ seems to be lacking the target function... – DonAntonio May 24 '14 at 00:10
  • This is a fragment of the step from transforming the black scholes PDE to log space and constant coefficients in order to get to the heat equation.. I just want to express d/ds as a function of d/dz given that s=e^z – Anastasiosyal May 24 '14 at 00:16

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