Find dy/dx by using implicit differentiation: sinx=x(1+tany)  TOPIC_SOLVED

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Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Postby Cafe au lait on Mon Jan 03, 2011 11:15 pm

Find dy/dx by using implicit differentiation: sinx=x(1+tany)

Here's what I have.

dy/dx[sinx]=dy/dx[x(1+tany)]
cosx=x[(sec^2y)dy/dx]+(1+tany)
cosx-tany-1=dy/dx[x(sec^2y)]
dy/dx=(cosx-tany-1)/(xsec^2y)

I haven't done this in a while so I am not confident in my answer. Did I do it right?
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Re: Find dy/dx by using implicit differentiation: sinx=x(1+t

Postby nona.m.nona on Mon Jan 03, 2011 11:55 pm

Looks good to me. Thank you for showing your work so nicely!
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Re: Find dy/dx by using implicit differentiation: sinx=x(1+t  TOPIC_SOLVED

Postby Martingale on Tue Jan 04, 2011 1:07 am

Cafe au lait wrote:dy/dx[sinx]=dy/dx[x(1+tany)]


your notation here is not correct

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Re: Find dy/dx by using implicit differentiation: sinx=x(1+t

Postby Cafe au lait on Tue Jan 04, 2011 2:19 am

Ooh, thanks for catching that Martingale! Fixed that on my paper.
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