Find derivation (dB/decade) for given amplitude characteristic of low pass filter [Hz, -] I am trying to find derivation (differential attenuation) for frequency's 600 and 2000 Hz for given amplitude characteristic of low pass filter, which look like this: 

I assume, that I should transform my approximation (graph) to dB form, but how should I do this (if I am right).
 A: You need to find the slope of the tangent lines to the curve at the frequencies $600$ and $2000$ Hz represented in a Bode plot with two logarithmic scales instead of two linear ones as in your plot. The horizontal axis of the low pass filter should measure the frequency in decades and the vertical axis should measure the amplitude in decibels (dB). The frequencies whose magnitudes differ by a factor of ten such as  $1, 10, 100, 1000, 10000$ Hz are equaly spaced and in general two frequencies $f_1,f_2$ are 
$$\log_{10}\left(\frac{f_1}{f_2}\right)$$
decades apart. So from $100$ and $600$ Hz are $\log_{10}\left(\frac{600}{100}\right)\approx 0.778$ decades and from $2000$ to $1000$ Hz are $\log_{10}\left(\frac{2000}{1000}\right)\approx 0.301$ decades. To transform the amplitude $A_1$ in a dB change with respect to the amplitude $A_0$ use the conversion formula (if the amplitude in your plot is a voltage and not a power)
$$\text{Gain}_\mathrm{ dB} =20 \log_{10} \left (\frac{A_1}{A_0} \right ), $$ 
because  a change in voltage by a factor of $10$ is a $20$ dB change and means a change in power by a factor of $100=10^2$, assuming that the power is proportional to the square of the voltage. (See Wikipedia). The attenuation is the opposite of the gain.
