I have the following function that I wish to take the derivative of $$ z(\omega)=\frac{1-C_{c}L_{r}\omega^{2}-\frac{\omega^{2}}{\omega_{r}^{2}}}{i\omega C_{c}\left(1-\frac{\omega^{2}}{\omega_{r}^{2}}\right)} $$ where $i$ is the imaginary unit. I wish to take the derivative $\frac{dz(\omega)}{d\omega_{r}}$. However, $\omega_{r}$ is also dependent on $L_{r}$ such that $$ \omega_{r} = \frac{1}{\sqrt{L_{r}C_{r}}} $$ I understand that there should be some chain-rule that happens when I take the derivative of $z$ with respect to $\omega_{r}$. But since $L_{r}$ is also dependent on $\omega_{r}$, how should I take the derivative properly?

  • $\begingroup$ is $C_c$ really supposed to be $C_r$ ? $\endgroup$
    – WW1
    Jun 29 at 20:31
  • $\begingroup$ No. $C_{c}$ is this case is really $C_{c}$ it is not dependent on $C_{r}$. $\endgroup$
    – kowalski
    Jun 29 at 20:37
  • $\begingroup$ If $C_r$ is constant, you can just replace $C_{c}L_{r}\omega^{2}$ with $\frac{C_{c}}{C_r} \left (\frac{\omega}{\omega_r} \right ) ^{2}$ $\endgroup$
    – WW1
    Jun 29 at 20:51
  • $\begingroup$ $C_{r}$ unfortunately also depends on $\omega_{r}$ and subsequently $L_{r}$. I think this is a vicious circle :( $\endgroup$
    – kowalski
    Jun 29 at 21:16

1 Answer 1


First, I'd rearrange the expression by plugging $\omega_r$ in terms of $L_r$, that is:

$$z(\omega)=\dfrac{1-C_c\frac{1}{\sqrt{C_r}\omega_r^2}-\frac{\omega^2}{\omega_r^2}}{i\omega C_c\Big(1-\frac{\omega^2}{\omega^2_r}\Big)}$$

You can further simplify by separating the terms in the numerator as follows:

$$z(\omega)=\dfrac{1}{i\omega C_c\Big(1-\frac{\omega^2}{\omega^2_r}\Big)}-\dfrac{\frac{1}{\omega_r^2}\Big(\frac{1}{\sqrt{C_r}}+\omega^2\Big)}{i\omega C_c \Big(1-\frac{\omega^2}{\omega^2_r}\Big)}=\dfrac{1}{i\omega C_c\Big(1-\frac{\omega^2}{\omega^2_r}\Big)}-\dfrac{\frac{1}{\sqrt{C_r}}+\omega^2}{i\omega C_c(\omega_r^2+\omega^2)}=x-y$$

where $x$ and $y$ are defined as above.

You can rewrite $y$, as a function of $\omega_r$ as $y(\omega_r)=f(g(\omega_r))$, where $g(\omega_r)=\omega_r^2+\omega^2$, and $f(g)=\dfrac{\frac{1}{\sqrt{C_r}}+\omega^2}{i\omega C_cg}$. You can verify that $y(\omega_r)=f(g(\omega_r))$. Then to calculate the derivative with respect to $\omega_r$ we can use the chain rule as follows: $\dfrac{d y(\omega_r)}{d\omega_r}=\dfrac{df}{dg}\dfrac{dg}{d\omega_r}=f'(g(\omega_r))g'(\omega_r)$. Note that the $f'$ is evaluated at $g(\omega_r)$, not $\omega_r$.

In your case, that would be $g'(\omega_r)=2\omega_r$ and $f'(g(\omega_r))=-\dfrac{\frac{1}{\sqrt{C_r}}+\omega^2}{i\omega C_c(g(\omega_r))^2}=-\dfrac{\frac{1}{\sqrt{C_r}}+\omega^2}{i\omega C_c(\omega_r^2+\omega^2)^2}$. Then you put together those expressions and simplify if necessary. I'll let you do $x$.

  • $\begingroup$ I think your substitution of $L_{r}$ is not correct? Should be $1/C_{r}\omega_{r}^{2}$ instead of $1/\sqrt{C_{r}}\omega_{r}^{2}$ ? $\endgroup$
    – kowalski
    Jun 29 at 20:47
  • $\begingroup$ Barring the typos, I understand what you are trying to do. But $C_{r}$ is also dependent on $\omega_{r}$, so this method doesn't work because you're treating $C_{r}$ as a constant $\endgroup$
    – kowalski
    Jun 29 at 21:15
  • $\begingroup$ @kowalski yes, my substitution is incorrect. It wasn't specified that $C_r$ also depends on $\omega_r$. However, even for that case the procedure would be very similar. In your case, you would have a $C_r'$ term. If you want an expression only in terms of $\omega_r$, you need two equations, one for $L_r$, another for $C_r$; with one equation you end up going in circles, as you mentioned the comments linked to the question. $\endgroup$
    – Schach21
    Jun 30 at 1:49

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