dervative help

[mark]

im having trouble with this problem and i dont know if my work is right

obtain dy/dx

e^y+e^x = sinhy

heres my work

e^y + e^x = sinhy

ln (e^y) + ln (e^x) = ln( sinhy)

y + x = ln(sinhy)

dy/dx + 1 = cothy dy/dx

from here i do not know what to do

 

[pka]

Why not do this?
\(\displaystyle \L
\begin{array}{l}
e^y y' + e^x = \cosh (y)y' \\
y' = \frac{{ - e^x }}{{e^y - \cosh (y)}} \\
\end{array}\)

 

[soroban]

Hello, mark!

im having trouble with this problem $\displaystyle \;$ . . . I'm not surprised!

Obtain $\displaystyle \frac{dy}{dx}:\;\;e^y\,+\,e^x \:=\:\sinh y$

heres my work

$\displaystyle e^y\,+\,e^x\;=\;\sinh y$

$\displaystyle \ln(e^y)\,+\,\ln(e^x)\;=\;\ln(\sinh y)\;\;$ . . . what?

$\displaystyle y\,+\,x\;=\;\ln(\sinh y)\;\;$ . . . WHAT?

$\displaystyle \frac{dy}{dx}\,+\,1\;=\;\coth y\left(\frac{dy}{dx}\right)\;\;$

from here i do not know what to do $\displaystyle \;\;$Yes, you do . . . solve for dy/dx . . .

We have: \(\displaystyle \L\,e^y\,+\,e^x\;=\;\sin y\)

Differentiate implicitly: \(\displaystyle \L\,e^y\left(\frac{dy}{dx}\right)\,+\,e^x\;=\;\cosh y\left(\frac{dy}{dx}\right)\)


Okay, one more time . . .

How do we solve for $\displaystyle \frac{dy}{dx}$ ?

We get all the $\displaystyle \frac{dy}{dx}$ terms on one side . . .

\(\displaystyle \L\,e^y\left(\frac{dy}{dx}\right)\,-\,\cosh y\left(\frac{dy}{dx}\right)\;=\;-e^x\)

Factor: \(\displaystyle \L\:\left(e^y\,-\,\cosh y\right)\left(\frac{dy}{dx}\right)\;=\;-e^x\)

Therefore: \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{-e^x}{e^y\,-\,\cosh y}\;\;\) . . . or: \(\displaystyle \L\,\frac{e^x}{\cosh y\,-\,e^y}\)

 

[CHAMI]

WHY E ^ X IS ITSELF AND NOT DY/DX IN FRONT OF IT? IS THAT BECAUSE YOU FIND Y INSTEAD OF X AND THEREFORE; YOU DO NOT GET DY/DX FOR E^X? What happen if you have two xy terms with like x^2y something like that with one with a power of 2 and one without?