How do you differentiate #sinx+siny=1#?

Question

4788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-29T07:35:31

The answer is #=-cosx/(-sin^2x+2sinx)#

#sinx+siny=1#

#siny=1-sinx#

#cosy=sqrt(1-(1-sinx)^2)=-sin^2x+2sinx#

Differentiating wrt #x#

#cosx+dy/dxcosy=0#

#=>#, #dy/dxcosy=-cosx#

#=>#, #dy/dx=-cosx/cosy#

#=>#, #dy/dx=-cosx/(-sin^2x+2sinx)#