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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/(cosy-xe^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: 20,937 Thanks: 2210 The derivative doesn't exist, but the tangent at (0, pi/2) is the line x = 0 (i.e., the y-axis). 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/coty-x, x=0, y=pi/2) ---->infinity Tags equation, tangent, tricky Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hampos Differential Equations 1 May 22nd, 2012 06:42 AM danneman91 Trigonometry 9 September 18th, 2011 05:25 PM julian21 Algebra 1 March 23rd, 2010 09:02 PM interchaz Applied Math 3 May 14th, 2009 09:49 PM interchaz Computer Science 1 December 31st, 1969 04:00 PM

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