KFS Posted November 3, 2020 Share Posted November 3, 2020 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. Link to comment Share on other sites More sharing options...
joigus Posted November 3, 2020 Share Posted November 3, 2020 (edited) 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 Link to comment Share on other sites More sharing options...
studiot Posted November 3, 2020 Share Posted November 3, 2020 Why do you not get rid of the fraction first ? 1 Link to comment Share on other sites More sharing options...
joigus Posted November 3, 2020 Share Posted November 3, 2020 (edited) 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 Link to comment Share on other sites More sharing options...
KFS Posted November 3, 2020 Author Share Posted November 3, 2020 (edited) How did you get xy=sinh(x+y)? Edited November 3, 2020 by KFS Link to comment Share on other sites More sharing options...
joigus Posted November 3, 2020 Share Posted November 3, 2020 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. Link to comment Share on other sites More sharing options...
KFS Posted November 3, 2020 Author Share Posted November 3, 2020 Okay I see it now. Thank you, I was stuck in this. Link to comment Share on other sites More sharing options...
joigus Posted November 4, 2020 Share Posted November 4, 2020 20 hours ago, KFS said: Okay I see it now. Thank you, I was stuck in this. You're welcome. Link to comment Share on other sites More sharing options...
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