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`
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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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