How do you differentiate #-3=(x-y)e^(x-y)#?

To differentiate the equation -3=(x-y)e^(x-y), you can use the product rule and the chain rule. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by (u'v + uv'). The chain rule states that if you have a function g(x) composed with another function f(x), then the derivative of this composition is given by g'(f(x)) * f'(x).

Applying the product rule and the chain rule, you get:

d/dx [-3] = d/dx [(x-y)e^(x-y)] 0 = (1 - dy/dx)e^(x-y) + (x-y)e^(x-y)(1 - dy/dx)

Now, you can solve for dy/dx. First, factor out e^(x-y) from both terms:

0 = e^(x-y)[1 - dy/dx + x - y - dy/dx(x - y)]

Now, divide both sides by e^(x-y):

0 = 1 - dy/dx + x - y - dy/dx(x - y)

Now, rearrange terms and solve for dy/dx:

0 = 1 + x - y - dy/dx(1 + x - y)

dy/dx(1 + x - y) = 1 + x - y

Finally, divide both sides by (1 + x - y):

dy/dx = (1 + x - y) / (1 + x - y)

So, dy/dx = 1.