e^(xy) = x – 2y -1

Differentiating with respect to x:

e^(xy)[ x dy/dx + y ) = 1 – 2y dy/dx

[x e^(xy) + 2y] (dy/dx) = 1 – y e^(xy)

dy / dx = [ 1 – y e^(xy) ] / [x e^(xy) + 2y].

When x = 2 and y = 0:

dy / dx = (1 – 0) / (2 + 0)

= 1 / 2.

i’m having concern expertise what e is comparable to? because of the fact x-2y-one million is a line… and the slope of the tangent line of a line is the comparable because of the fact the slope of the line. And once you’re saying e are you speaking on the subject of the backside of organic log. The definition of the tangent line is a line which intersects in common terms one particular component on that function… while coping with a line the tangent line is that comparable line… yet to locate the slope of the tangent line for non linear function at a particular component… you’re taking the spinoff of that function and sub interior the component for the x and the y… and that’s going to grant you a particular value, and that value is the slope on that function at that particular component… i do no longer comprehend if that helped in any respect…