How do you differentiate #e^y sin(x) = x + xy#?

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Answer 1 · 2015-09-03T15:40:27

#e^ysinx = x+xy#

Differentiating with respect to x, differentiate terms in y with respect to y, and multiply by #dy/dx#, and differentiate terms in x as per normal.

LHS: #e^y(dy/dx)sinx# + #e^ycosx# (using product rule)
RHS: #1+y+x(dy/dx)# (using product rule as well)

#e^y(dy/dx)sinx# + #e^ycosx# = #1+y+x(dy/dx)#

Make #dy/dx# the subject.

#e^y(dy/dx)sinx - x(dy/dx)# = #1+y-e^ycosx#
#dy/dx(e^ysinx - x)# = #1+y-e^ycosx#
#dy/dx# = #(1+y-e^ycosx)/(e^ysinx - x)#

I hope that helped!