Where do I begin to derive this equation?

Question

anonymous · Answer

attachment

anonymous · Answer

Separate the natural log?

dumbcow · Answer

what does c_lim refer to?

anonymous · Answer

@dumbcow is just a limiting concentration of a solute

dumbcow · Answer

oh ok, can we treat it like a constant then

anonymous · Answer

The solution should be this, but I have no idea where it came from.

dumbcow · Answer

got it, alright are you familiar with product rule for derivatives

anonymous · Answer

Yeah

anonymous · Answer

I'm not sure how I will derive the fraction though.

dumbcow · Answer

ok lets apply that here

$$\rightarrow (c_A)' \ln \frac{c_{\lim}}{c_A} + c_A ( \ln \frac{c_{\lim}}{c_A})' = 0$$
derivative of c is 1

derivative of ln(u) is 1/u * du

derivative of c_lim/c = -c_lim/c^2

dumbcow · Answer

for fraction
$$(\frac{1}{x})' = (x^{-1})' = -x^{-2} = -\frac{1}{x^2}$$

anonymous · Answer

I see. I'll try it out now. Thanks!

dumbcow · Answer

yw, you get a lot of things to cancel out

at the end they split up the log(clim/cA) --> logclim - logcA