Posted by Ifi on Saturday, October 27, 2012 at 9:56pm.
so we are differentiating
cosx siny + x^5 = 11 implicitly ?
cosx(cosy) dy/dx + siny(-sinx) + 5x^4 = 0
dy/dx = (sinx siny - 5x^3)/(cosx cosy)
which agrees with my Wolfram result
http://www.wolframalpha.com/input/?i=find+dy%2Fdx+for+%28cos%28x%29%29%28sin%28y%29%29+%2B+x%5E5+%3D+11
just noticed you wanted dx/dy (strange)
cosx siny + x^5 = 11
cosx(cosy) + siny(-sinx) dx/dy = 5x^4 dx/dy = 0
dx/dy (5x4 - sinxsiny) = - cosxcosy
dx/dy = -cosxcosy/(5x^4 - sinxsiny)
the same as yours.
At reiny .... Thanks
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