Question #b7ba3

1 Answer
Feb 6, 2018

#dy/dx=(y-siny)/(xcosy-x)#

Explanation:

#"differentiate "color(blue)"implicitly with respect to x"#

#"differentiate both sides using the "color(blue)"product rule"#

#"given "y=f(x)g(x)" then"#

#dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"#

#rArrxcosydy/dx+siny=xdy/dx+y#

#rArrxcosydy/dx-xdy/dx=y-siny#

#rArrdy/dx(xcosy-x)=y-siny#

#rArrdy/dx=(y-siny)/(xcosy-x)#