How do you find $(dy)/(dx)$ given $x^2sinx+y^2cosy=1$ ?

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Answer 1 · 2016-11-22T09:12:39

$(dy)/(dx)=-(x(2sinx+xcosx))/(y(2cosy-ysiny))$

differentiate both sides with respect to $""x""$ . For the terms on the $" "LHS$ the product rule will have to be employed twice.

$d/(dx)(x^2sinx+y^2cosy)=d/(dx)(1)$

$d/(dx)(x^2sinx)+d/(dx)(y^2cosy)=d/(dx)(1)$

$2xsinx+x^2cosx+2y(dy)/(dx)cosy+y^2(-siny)(dy)/(dx)=0$

now rearrange for $""(dy)/(dx)$

$(dy)/(dx)(2ycosy-y^2siny)=-(2xsinx+x^2cosx)$

$(dy)/(dx)=-(x(2sinx+xcosx))/(y(2cosy-ysiny))$

How do you find (dy)/(dx)(dy)/(dx) given x^2sinx+y^2cosy=1x^2sinx+y^2cosy=1?