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December 9th, 2009, 09:22 AM   #1
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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.)
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December 9th, 2009, 09:35 AM   #2
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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
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December 9th, 2009, 09:49 AM   #3
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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.
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December 9th, 2009, 09:58 AM   #4
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Re: Tangent equation of a tricky equation

i get the same
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December 9th, 2009, 10:01 AM   #5
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Re: Tangent equation of a tricky equation

maybe the tangent line doesn't pass through that point?
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December 9th, 2009, 10:12 AM   #6
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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.
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December 9th, 2009, 10:17 AM   #7
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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
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December 11th, 2009, 01:10 PM   #8
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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).
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December 11th, 2009, 07:04 PM   #9
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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
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