Question

Where do I begin to derive this equation?

Answer 1

Separate the natural log?

Answer 2

what does c_lim refer to?

Answer 3

@dumbcow is just a limiting concentration of a solute

Answer 4

oh ok, can we treat it like a constant then

Answer 5

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

Answer 6

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

Answer 7

Yeah

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

yw, you get a lot of things to cancel out

at the end they split up the log(clim/cA) --> logclim - logcA