What is the derivative (dy/dx) of xcos(2x+3y) = ysinx ?

1 Answer
Mar 20, 2018

#dy/dx = (cos(2x+3y)-2xsin(2x+3y)-ycosx)/(3xsin(2x+3y)+sinx)#

Explanation:

Differentiate both sides of the equation with respect to #x#:

#cos(2x+3y)-xsin(2x+3y)(2+3dy/dx) = ycosx+sinxdy/dx#

#cos(2x+3y)-2xsin(2x+3y)-3xsin(2x+3y)dy/dx = ycosx+sinxdy/dx#

#cos(2x+3y)-2xsin(2x+3y)-ycosx = (3xsin(2x+3y)+sinx)dy/dx#

#dy/dx = (cos(2x+3y)-2xsin(2x+3y)-ycosx)/(3xsin(2x+3y)+sinx)#