How do you implicitly differentiate #y= (x-y)^2- e^(x y) #?

Question

How do you implicitly differentiate #y= (x-y)^2-   e^(x y)   #?

Charles Arocha · Accepted Answer

The answer #y'=-(e^(xy)*y+2y-2x)/(x*e^(xy)-2y+2x+1# #d/dx(y)=d/dx((x-y)^2-e^(xy)))# # y'=2(x-y)*d/dx(x-y)-e^(xy)*d/dx(xy)# #y'=2(x-y).(1-y')-(x.y'+y).e^(xy)# solve the equation for y' #y'=-(e^(xy)*y+2y-2x)/(x*e^(xy)-2y+2x+1#

Isaiah Allen · Answer

undefined To implicitly differentiate $ y = (x - y)^2 - e^{xy} $ with respect to $ x $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the chain rule and product rule where necessary.
3. Solve for $ \frac{{dy}}{{dx}} $.

Here's the breakdown:

$ y = (x - y)^2 - e^{xy} $

1. Differentiate both sides:
   $ \frac{{d}}{{dx}}[y] = \frac{{d}}{{dx}}[(x - y)^2] - \frac{{d}}{{dx}}[e^{xy}] $

2. Apply the chain rule and product rule where necessary:
   $ \frac{{dy}}{{dx}} = 2(x - y) \frac{{d}}{{dx}}[x - y] - e^{xy} \left( y \frac{{d}}{{dx}}[x] + x \frac{{d}}{{dx}}[y] \right) $

3. Simplify and solve for $ \frac{{dy}}{{dx}} $:
   $ \frac{{dy}}{{dx}} = 2(x - y) - e^{xy}(y + x \frac{{dy}}{{dx}}) $
   $ \frac{{dy}}{{dx}} + e^{xy}x \frac{{dy}}{{dx}} = 2(x - y) - e^{xy}y $
   $ \frac{{dy}}{{dx}}(1 + x e^{xy}) = 2(x - y) - e^{xy}y $
   $ \frac{{dy}}{{dx}} = \frac{{2(x - y) - e^{xy}y}}{{1 + x e^{xy}}} $

That's the implicit differentiation of the given equation with respect to $ x $.