Obviously, you’re going to take the derivative implicitly. This is easy with the right side of the equation. The left side will require, first, the Chain Rule and then the Product Rule.

d/dx (e^(xy)) = e^(xy) * d/dx(xy) (Chain Rule)

= e^(xy) (x dy/dx + y dx/dx) (Product Rule)

dx/dx is clearly 1, so you end up with

d/dx(e^(xy)) = (x dy/dx + y) e^(xy)

On the right, you get

d/dx (x – 2y – 1) = 1 – 2 dy/dx

So, after implicit differentiation, you get

(x dy/dx + y) e^(xy) = 1 – 2 dy/dx

Solve algebraically for dy/dx, plug in your values for x and y, and you’re done.

I’m having difficulty understanding what e is equal to? Because x-2y-1 is a line… and the slope of the tangent line of a line is the same as the slope of the line. And when you say e are you talking about the base of natural log. The definition of the tangent line is a line which intersects only one specific point on that function… when dealing with a line the tangent line is that same line… but to find the slope of the tangent line for non linear function at a specific point… you take the derivative of that function and sub in the point for the x and the y… and it should give you a specific value, and that value is the slope on that function at that specific point… i don’t know if that helped at all…