How do I differentiate this?

[KFS]

The problem asks to find dy/dx in y=sinh(x+y)/xy=1. What I do is: differentiate both sides using implicit differentiation which gives d/dx(sinh(x+y)/xy)=0. I differentiate this setting y' where I have to differentiate y with respect to x. Thus I get (xycosh(x+y)+xyy'cosh(x+y)-ysinh(x+y)-xy'sinh(x+y))/(xy)^2. Then I solve for y' and I get y'=(-xycosh(x+y)+ysinh(x+y))/(xycosh(x+y)-xsinh(x+y)). But the answer in the book says y'=(y-cosh(x+y))/(cosh(x+y)-x). What am I doing wrong? Is my procedure incorrect? Thank you.

[joigus]

You forgot to derive \( x+y \) using the chain rule in the first \( \cosh \).

Edit: No, wait a minute. You didn't.

I think the book "means",

\[y'=\frac{y\tanh\left(x+y\right)-xy}{xy-x\tanh\left(x+y\right)}\]

Edited November 3, 2020 by joigus

[studiot]

Why do you not get rid of the fraction first  ?

[joigus]

38 minutes ago, studiot said:

Why do you not get rid of the fraction first  ?

Good idea.

That's the ticket.

OK. I've been racking my brain for quite a while. And the reason is there is no hyperbolic-trigonometry relation that gives you equality between,

\[\frac{y\sinh\left(x+y\right)-xy\cosh\left(x+y\right)}{\cosh\left(x+y\right)xy-x\sinh\left(x+y\right)}\]

--for which you've applied correctly the chain rule--, and,

\[\frac{y-\cosh\left(x+y\right)}{\cosh\left(x+y\right)-x}\]

which is the simpler expression you will find if you follow @studiot's tip.

And the reason for that is that, if you want to find the simpler expression --Studiot's and your book's--, you must use the constraint given:

\[y'=\frac{y\sinh\left(x+y\right)-xy\cosh\left(x+y\right)}{\cosh\left(x+y\right)xy-x\sinh\left(x+y\right)}\]

\[xy=\sinh\left(x+y\right)\]

along your curve. So that,

\[y'=\frac{y\sinh\left(x+y\right)-\sinh\left(x+y\right)\cosh\left(x+y\right)}{\cosh\left(x+y\right)\sinh\left(x+y\right)-x\sinh\left(x+y\right)}=\]

\[=\frac{y-\cosh\left(x+y\right)}{\cosh\left(x+y\right)-x}\]

Consequently, your derivative was correct, but you must use the constraint once again if you want to get to the final expression that's in the book.

I hope that's clear.

It would remain for you to prove the simpler expression directly by using Studiot's tip. It's not difficult. I just wanted to dispel the "paradox."

Edited November 3, 2020 by joigus
emphasis added

[KFS]

How did you get xy=sinh(x+y)?

Edited November 3, 2020 by KFS

[joigus]

27 minutes ago, KFS said:

How did you get xy=sinh(x+y)?

Your constraint is F(x,y) = 1. Defines an implicit function y(x).

As F(x,y) = sinh(x+y)/xy, there you are.

[KFS]

Okay I see it now. Thank you, I was stuck in this.

[joigus]

20 hours ago, KFS said:

Okay I see it now. Thank you, I was stuck in this.

You're welcome.