Given that y=ln1/3(1-e^-2x), sohw that dy/dx= 2/3e^-y - 2

[Smithereens]

Question: Given that y=ln1/3(1-e^-2x), sohw that dy/dx= 2/3e^-y - 2.
Thank you very much!!

 

[stapel]

Question: Given that y=ln1/3(1-e^-2x),...

As posted, the function can be read as follows:

. . . . .\(\displaystyle y\, =\, \dfrac{\ln(1)}{3\left(1\, -\, e^{-2}x\right)}\)

...or as:

. . . . .\(\displaystyle y\, =\, \ln\left(\dfrac{1}{3}\right)\,\left(1\, -\, e^{-2}x\right)\)

...or in other ways. Is either of these correct?

...sohw that dy/dx= 2/3e^-y - 2.

As posted, the derivative can be read as follows:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{2}{3e^{-y}}\, -\, 2\)

...or as:

. . . . .\(\displaystyle \dfrac{dy}{dx}\, =\, \dfrac{2}{3}\, e^{-y}\, -\, 2\)

...or in other ways. Is either of these correct?

When you reply, please include a clear listing of your thoughts and steps in trying to find the derivative. Thank you!

 

[HallsofIvy]

Or should we presume that you mean e^(-2x) rather than e^(-2) x? And (1/3)(1- e^(-2x)) rather than 1/(3(1- e^(-2x))? Please use parentheses to clarify what you mean.
If you mean y= ln((1/3)(1- e^(-2x))) then, since the result you want is in terms of "y" rather than "x", I would write that as e^y= (1/3)(1- e^(-2x)) and differentiate both sides with respect to x. What is the derivative of e^y with respect to y? What is the derivative of e^y with respect to x?