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find dy/dx by implicit differentiation 3cosxsiny=1
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product rule is needed for the cos(x)*sin(y) part chain rule is needed for the sin(y) part
\[3 \cdot [ \sin(y) \frac{d}{dx}(\cos(x))+\cos(x) \frac{d}{dx}( \sin(y))]=\frac{d}{dx}(1)\]
so easy question what is the derivative of (cos(x)) w.r.t x?
-sinx
ok so now we have \[3[\sin(y)(-\sin(x))+\cos(x)\frac{d}{dx}(\sin(y))]=0 \]
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last part for differentiating what is the derivative of sin(y) w.r.t x
cosy(y')
yep
\[3[-\sin(y)\sin(x)+\cos(x)\cos(y)y']=0 \]
now you can solve for y' i suggest distributing the 3 first though
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thanks so much these problems seem so scary at first but they are getting easier
cool stuff they aren't too scary :)
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