What is the implicit derivative of #1= xe^y-sin(xy) #?

1 Answer
Jun 28, 2017

#dy/dx = ( ycos(xy)-e^y)/(1+sin(xy)-xcos(xy))#

Explanation:

Write the equation as:

#xe^y = 1+sin(xy)#

and differentiate both sides with respect to #x#.

#d/dx (xe^y) = d/dx (1+sin(xy))#

#e^y+ xy'e^y = cos(xy) d/dx (xy)#

#e^y+ xy'e^y = cos(xy)(y+xy')#

#e^y+ xy'e^y = ycos(xy)+xy'cos(xy)#

# xy'e^y -xy'cos(xy) = ycos(xy)-e^y #

# xy'(e^y -cos(xy)) = ycos(xy)-e^y #

#y' = ( ycos(xy)-e^y)/(x(e^y -cos(xy)))#

and substituting #xe^y# from the original equation:

#y' = ( ycos(xy)-e^y)/(1+sin(xy)-xcos(xy))#