How do you use implicit differentiation to find dy/dx given #xsiny+x^2cosy=1#?

1 Answer
Oct 31, 2016

#(dy)/(dx)=-(siny+2xcosy)/((cosy)x-(siny)x^2)#

Explanation:

Differentiation of the expression is determined by using the product rule differentiation

#color(brown)(d/dx(u(x)+v(x))=d/dx(u(x))+d/dx(v(x)))#

#color(red)(d/dx(u(x)xxv(x))#
#color(red)(=(du(x))/dxxxv(x)+(dv(x))/dxxxu(x)#

Applying the differentiation of trigonometric functions
#color(blue)(d/dx(sinx)=cosx)#
#color(blue)(d/dx(cosx)=-sinx)#

#d/dx(xsiny+x^2cosy)=(d1)/dx#

#rArrcolor(brown)(d/dx(xsiny)+d/dx(x^2cosy)=0)#

#rArrcolor(red)(((dx)/dxxxsiny+(dsiny)/dxxxx)+((dx^2)/dxxxcosy+(dcosy)/dxxxx^2)=0#

#rArrsiny+(dy/dx)(cosy)x+2xcosy-(dy)/dx(siny)x^2=0#

#rArr(dy/dx)(cosy)x-(dy)/dx(siny)x^2=-siny-2xcosy#

#rArr(dy/dx)((cosy)x-(siny)x^2)=-siny-2xcosy#

#rArr(dy)/(dx)=-(siny+2xcosy)/((cosy)x-(siny)x^2)#