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

Question

Charles Arocha · Accepted Answer

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

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)))#

Emilia Aguilar · Answer

undefined To implicitly differentiate the equation sin(x) - cos(y) = e^(xy)sin(x)cos(y), follow these steps:

1. Differentiate both sides of the equation with respect to x.
2. Apply the chain rule whenever necessary.
3. Solve for dy/dx.

Differentiating sin(x) with respect to x gives cos(x), and differentiating e^(xy)sin(x)cos(y) with respect to x using the product rule gives e^(xy)(ysin(x) + cos(x)) + e^(xy)cos(x)sin(y)(-y).

So, the implicit derivative is:

cos(x) + e^(xy)(ysin(x) + cos(x)) + e^(xy)cos(x)sin(y)(-y) = -sin(y)dy/dx

Now, solve for dy/dx:

dy/dx = [cos(x) + e^(xy)(ysin(x) + cos(x)) + e^(xy)cos(x)sin(y)(-y)] / -sin(y)