cosy=xcos(a+y) Find dy/dx
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cosy=xcos(a+y)
Differentiating both sides wrt 'x'
=>d/dx(cos y) =cos(a+y) dx/dx+ xd/dx [cos(a+y) ] (RHS differentiated on the basis of product rule)
=> -siny*dy/dx= cos (a+y)- xsin(a+y)*dy/dx
=> dy/dx [x sin(a+y)-siny] = cos(a+y)
Now x= (cosy)/cos(a+y)
=> dy/dx {[ sin(a+y).cosy-siny.cos(a+y )] / cos(a+y)}=cos(a+y)
=> dx/dy [ sin(a+y-y )]= cos2(a+y) { using sine formula sin(x-y)= sinx.cosy-siny.cosx}
=>dy/d x= cos2(a+y)/ sin(a)
@skb
Differentiating both sides wrt 'x'
=>d/dx(cos y) =cos(a+y) dx/dx+ xd/dx [cos(a+y) ] (RHS differentiated on the basis of product rule)
=> -siny*dy/dx= cos (a+y)- xsin(a+y)*dy/dx
=> dy/dx [x sin(a+y)-siny] = cos(a+y)
Now x= (cosy)/cos(a+y)
=> dy/dx {[ sin(a+y).cosy-siny.cos(a+y )] / cos(a+y)}=cos(a+y)
=> dx/dy [ sin(a+y-y )]= cos2(a+y) { using sine formula sin(x-y)= sinx.cosy-siny.cosx}
=>dy/d x= cos2(a+y)/ sin(a)
@skb
RabbitPanda:
Thnx 4 brainlist
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