Question

How do I derive this equation?

Answer 1

Thats a nice equation

Answer 2

\[d\frac{ (Ca*\ln \frac{ cL }{ Ca }) }{ dCa}\]

Answer 3

I know it's the product rule. but the natural log has me confused.

Answer 4

d/dx (ln y) = 1/y dy/dx

Answer 5

@phi So is the answer Ca*1/(cL/Ca)+ln(cL/Ca)

Answer 6

or should I just simply the ln fraction

Answer 7

is cL a variable or a constant ?

Answer 8

Constant

Answer 9

\[ d\frac{ (Ca\ln \frac{ cL }{ Ca }) }{ dCa} = Ca \frac{d \ln \frac{ cL }{ Ca }}{d Ca} + \ln \frac{ cL }{ Ca }\]

and
\[ \frac{d \ln \frac{ cL }{ Ca }}{d Ca} = \frac{1}{\frac{cL}{Ca}} \frac{d}{d Ca} \frac{ cL }{ Ca }\\
= \frac{cL}{Ca} cL \frac{d (Ca)^{-1}}{d Ca} \\
= \frac{(cL)^2}{Ca} (-(Ca)^{-2}) \\
= - \frac{(cL)^2}{(Ca)^3}
\]

Answer 10

oops, I did not flip the fraction

Answer 11

I think I have made a mistake as well...

Answer 12

\[ \frac{d \ln \frac{ cL }{ Ca }}{d Ca} = \frac{1}{\frac{cL}{Ca}} \frac{d}{d Ca} \frac{ cL }{ Ca }\\
= \frac{Ca}{cL} cL \frac{d (Ca)^{-1}}{d Ca} \\
= Ca (-(Ca)^{-2}) \\
= - \frac{1}{Ca}
\]

Answer 13

The initial equation I gave you should equal to zero

Answer 14

From using the Product rule. I should get d/Ca (Ca*lncL-Ca*lnCa)=0

Answer 15

so (with any luck) we get
\[ d\frac{ (Ca\ln \frac{ cL }{ Ca }) }{ dCa} = Ca \frac{d \ln \frac{ cL }{ Ca }}{d Ca} + \ln \frac{ cL }{ Ca } = Ca \cdot - \frac{1}{Ca} +\ln \frac{ cL }{ Ca } \\
= \ln \frac{ cL }{ Ca }-1
\]

Answer 16

yes, starting with
\[ C_a \ln cL-C_a \ln C_a \]
makes it easier
\[ \frac{d}{C_a}\left(C_a \ln cL-C_a \ln C_a \right) \\
= \ln cL - \left( C_a \cdot \frac{1}{C_a} + \ln C_a\right) \\
= \ln cL - 1 - \ln C_a \\
=\ln\left( \frac{cL}{C_a}\right) -1
\]

Answer 17

and if that equals zero,
\[ \ln\left( \frac{cL}{C_a}\right) -1 =0 \\ \ln\left( \frac{cL}{C_a}\right)=1\\
\frac{cL}{C_a}= e \\ cL= C_a\cdot e
\]