# Taking a simple derivative

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?

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

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 $$x$$.

• I think your substitution of $L_{r}$ is not correct? Should be $1/C_{r}\omega_{r}^{2}$ instead of $1/\sqrt{C_{r}}\omega_{r}^{2}$ ? Jun 29 at 20:47
• 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 Jun 29 at 21:15
• @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. Jun 30 at 1:49