To find the implicit derivative, we'll use the chain rule and product rule. It's important to remember that when differentiating a yy term, a dy/dxdydx term will be spit out thanks to the chain rule.
d/dx[-3=xye^(x-y)]ddx[−3=xyex−y]
0=ye^(x-y)d/dx[x]+xe^(x-y)d/dx[y]+xyd/dx[e^(x-y)]0=yex−yddx[x]+xex−yddx[y]+xyddx[ex−y]
Find each derivative.
d/dx[x]=1ddx[x]=1
d/dx[y]=dy/dxddx[y]=dydx
d/dx[e^(x-y)]=e^(x-y)d/dx[x-y]=e^(x-y)(1-dy/dx)=e^(x-y)-e^(x-y)dy/dxddx[ex−y]=ex−yddx[x−y]=ex−y(1−dydx)=ex−y−ex−ydydx
Plug them back in and solve for dy/dxdydx.
0=ye^(x-y)+xe^(x-y)dy/dx+xye^(x-y)-xye^(x-y)dy/dx0=yex−y+xex−ydydx+xyex−y−xyex−ydydx
-xye^(x-y)-ye^(x-y)=dy/dx(xe^(x-y)-xye^(x-y))−xyex−y−yex−y=dydx(xex−y−xyex−y)
(-xye^(x-y)-ye^(x-y))/(xe^(x-y)-xye^(x-y))=dy/dx−xyex−y−yex−yxex−y−xyex−y=dydx
dy/dx=(e^(x-y)(-xy-y))/(e^(x-y)(x-xy))dydx=ex−y(−xy−y)ex−y(x−xy)
dy/dx=-(y(x+1))/(x(1-y))dydx=−y(x+1)x(1−y)
