Question

What is the slope of the tangent line of `x^2e^(xy-x-y)= C`, where C is an arbitrary constant, at `(0,0)`?

Answer 1

Slope of the tangent line equals `0` at point `(0,0)`

Explanation:

By implicit differentiation:

`d/dx (x^2 e^(xy-x-y)) = x^2 d/dx(e^(xy-x-y)) + d/dx(x^2) * e^(xy-x-y)` (Product Rule)

`=x^2e^(xy-x-y) * (dy/dx*1 +y -1 -dy/dx) + 2x * e^(xy-x-y)`
(Chain Rule, Product Rule and Power Rule)

`=x^2e^(xy-x-y) * (cancel(dy/dx*1) +y -1 -cancel(dy/dx)) + 2x * e^(xy-x-y)`

`=x^2e^(xy-x-y) * (y - 1) +2xe^(xy-x-y)`

`=e^(xy-x-y) (x^2(y-1) + 2x))`

Therefore `e^(xy-x-y) (x^2(y-1) + 2x))` defines the slope of the tangent to the function at point `(x,y)`

Slope at point `(0,0)` = `e^0 * 0 = 1 * 0 = 0`

Answer 2

The tangent line is vertical. The slope is not defined.

Explanation:

If the point `(0,0)` is on the graph of `x^2e^(xy-x-y)=C`,

then `C = (0)^2 e^(0-0-0) = 0`.

But now we have `x^2e^(xy-x-y)=0` which implies that `x^2 = 0` (because for all `u`, `e^u != 0`).

So the equation is equivalent to `x=0` which is the `y`-axis. The tangent line to a line is the line. So, the tangent line to the vertical axis is the vertical axis, the slope of which is not defined.