Differentiation of a function with respect to itself

1. Aug 20, 2009

bitrex

In one of my electronics textbooks I have the following equation related to feedback in amplifiers:

$$K_f = \frac{K}{1-K\beta}$$

$$\frac{dK_f}{K_f} = \frac{1}{1-K\beta}\frac{dK}{K}$$

I'm not sure how this was derived - how was Kf differentiated with respect to itself?

2. Aug 20, 2009

jgens

$K_f$ wasn't differentiated with respect to itself, it was differentiated with repect to $K$. Here's what they did . . .

$$K_f = \frac{K}{1 - \beta K}$$

$$\frac{\mathrm{d}K_f}{\mathrm{d}K} = \frac{(1 - \beta K) + (\beta K)}{(1 - \beta K)^2}$$

$$\frac{\mathrm{d}K_f}{\mathrm{d}K} = \frac{1}{(1 - \beta K)^2}$$

$$\frac{\mathrm{d}K_f}{K_f} = \frac{\mathrm{d}K}{K(1 - \beta K)}$$

Basically, it's just an application of the quotient rule for differentiation.

3. Aug 20, 2009

bitrex

Ah, I see now. They took the derivative of Kf with respect to K, and then expressed that derivative as a ratio to get dKf/Kf. Thank you!