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December 9th, 2009, 09:22 AM  #1 
Newbie Joined: Dec 2009 Posts: 12 Thanks: 0  Tangent equation of a tricky equation
The question is below: Find the tangent equation for siny=e^xy at (0,pie/2). Thanks. (I know the first step is to use Natural log for both sides, then I got confused.) 
December 9th, 2009, 09:35 AM  #2 
Senior Member Joined: Nov 2008 Posts: 265 Thanks: 0  Re: Tangent equation of a tricky equation
i don't think you have to use logarithmic differentiation just take the derivative of both sides: cos y (dy/dx) = [y + x (dy/dx)]e^xy now solve for dy/dx Edit: Actually you could take the natural log of both sides and get ln siny = xy ln e So: (1/sin y)(cos y (dy/dx)) = y + x (dy/dx) ***ln e is just 1 and again solve for dy/dx so you could go both ways 
December 9th, 2009, 09:49 AM  #3 
Newbie Joined: Dec 2009 Posts: 12 Thanks: 0  Re: Tangent equation of a tricky equation
But I still can't find dy/dx since dy/dx=ye^xy/(cosyxe^xy) at x=0 which goes to infinity. That's the case.

December 9th, 2009, 09:58 AM  #4 
Senior Member Joined: Nov 2008 Posts: 265 Thanks: 0  Re: Tangent equation of a tricky equation
i get the same

December 9th, 2009, 10:01 AM  #5 
Senior Member Joined: Nov 2008 Posts: 265 Thanks: 0  Re: Tangent equation of a tricky equation
maybe the tangent line doesn't pass through that point?

December 9th, 2009, 10:12 AM  #6 
Newbie Joined: Dec 2009 Posts: 12 Thanks: 0  Re: Tangent equation of a tricky equation
I don't know, as u said, maybe there is no tangent line for that point, just like y=tanx at x=pie/2. Thank you either.

December 9th, 2009, 10:17 AM  #7 
Senior Member Joined: Nov 2008 Posts: 265 Thanks: 0  Re: Tangent equation of a tricky equation
i'm sorry i couldn't help, but I know a lot of other people on the forums that can help when they log in

December 11th, 2009, 01:10 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,613 Thanks: 2071 
The derivative doesn't exist, but the tangent at (0, pi/2) is the line x = 0 (i.e., the yaxis). You can make the problem totally straightforward by using x as the dependent variable and y as the independent variable (where sin(y) > 0).

December 11th, 2009, 07:04 PM  #9 
Newbie Joined: Dec 2009 Posts: 12 Thanks: 0  Re: Tangent equation of a tricky equation
Oh yes skipjack, thank u. I have got the same answer from my brother. As u said, actually the derivative DNE which goes to infinity and the tangent line is x=0 . Thank you for u help. (dy/dx=y/cotyx, x=0, y=pi/2) >infinity 

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