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I have a function $\phi(\mu, \sigma)$. $\mu$ and $\sigma$ are voltages (in mV in my case), so $\phi$ is a function of two voltages. $\phi$ itself, however, is in units of time (ms in my case). (This is because the voltage units are canceled by a division within the function, and the result is multiplied by a constant in units of time.)

If I take the partial derivative of $\phi$ with respect to $\mu$ (or $\sigma$), what will the units of the partial derivative, $\frac{\partial \phi}{\partial \mu}$, be?

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The derivative is the slope of the tangent line to the function at each point.

For a function $y(x)$, the slope is in units of $y / x$.

Hence, the units of the derivative of the function in your question are ms / mV.

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According as you say, I can assume the function $\phi$=k $\frac{\mu}{\sigma}$ where k has the units of ms.

Hence by partial differentiation, we get $\frac{\partial \phi}{\partial \mu}$ = $\frac{k}{\sigma}$ with units $ms/mV$.

and for $\frac{\partial \phi}{\partial \sigma}$ = -k$\frac{\mu}{\sigma^2}$ with units $ms*mV/(mV)^2$ = $ms/mV$

Hence the units of the partial derivative will be $ms/mV$.

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  • $\begingroup$ the function is nowhere near as simple as you're assuming. the division, for example, is in the limits of an integral inside the function. but you've captured the essence of the effects on the units here. $\endgroup$ – dbliss Sep 7 '15 at 16:52

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