# Find dy/dx by implicit differentiation. 7 cos x sin y = 1 y'=?

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### 1 Answer

Find `(dy)/(dx)` if `7cosxsiny=1` :

Differentiate both sides with respect to `x` :

`d/(dx)7cosxsiny=7d/(dx)[cosxsiny]`

Use the product rule: `d/(dx)[u*v]=u'v+uv'` :

`=7[d/(dx)cosx*siny+cosx*d/(dx)siny]`

`=7[-sinxsiny+cosxcosy(dy)/(dx)]` using the chain rule on `d/(dx)siny`

So the derivative of the left hand side is `-7sinxsiny+7cosxcosy(dy)/(dx)` and teh derivative of the right hand side is 0.

`-7sinxsiny+7cosxcosyy'=0`

`cosxcosyy'=sinxsiny`

`y'=(sinxsiny)/(cosxcosy)`

`y'=tanxtany`

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**Given `7cosxsiny=1` then `(dy)/(dx)=tanxtany` **

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