How do you implicitly differentiate # sin x - cos y = e^(xy)sin x cos y#?

1 Answer
Dec 16, 2017

#dy/dx=[(sinxcosy)(ye^(xy))+(cosxcosy)(e^(xy))-cosx]/(siny+(sinxsiny)(e^(xy))-(xsinxcosy)(e^(xy)))#

Explanation:

use trig differentiation rules, e^x differentiation, chain rule, and product rule

#d/dx(sinx)-d/dx(cosy)=d/dx(e^(xy)sinxcosy)#

#cosx+siny(dy/dx)=(sinxcosy)(d/dx(e^(xy)))+(e^(xy))d/dx(sinxcosy)#

#cosx+siny(dy/dx)=(sinxcosy)(e^(xy)*d/dx(xy))+(e^(xy))(cosy*d/dx(sinx)+sinx*d/dx(cosy))#

#cosx+siny(dy/dx)=(sinxcosy)(e^(xy)*(y+x(dy/dx))+(e^(xy))(cosy*cosx+sinx*(-siny)*dy/dx)#

#cosx+siny(dy/dx)=(sinxcosy)(ye^(xy)+xe^(xy)(dy/dx))+cosxcosy*e^(xy)-sinxsiny*e^(xy)*dy/dx#

#siny(dy/dx)+sinxsiny*e^(xy)*dy/dx=(sinxcosy)(ye^(xy))+xsinxcosy*e^(xy)(dy/dx)+cosxcosy*e^(xy)-cosx#

#siny(dy/dx)+sinxsiny*e^(xy)*dy/dx-xsinxcosy*e^(xy)(dy/dx)=(sinxcosy)(ye^(xy))+cosxcosy*e^(xy)-cosx#

#(dy/dx)(siny+sinxsiny*e^(xy)-xsinxcosy*e^(xy))=(sinxcosy)(ye^(xy))+cosxcosy(e^(xy))-cosx#

#dy/dx=[(sinxcosy)(ye^(xy))+(cosxcosy)(e^(xy))-cosx]/(siny+(sinxsiny)(e^(xy))-(xsinxcosy)(e^(xy)))#