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

Question

3568 views around the world

You can reuse this answer
Creative Commons License

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))#