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

Mar 16, 2016

Slope of the tangent line equals $0$ at point $\left(0 , 0\right)$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{x y - x - y}\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{x y - x - y}\right) + \frac{d}{\mathrm{dx}} \left({x}^{2}\right) \cdot {e}^{x y - x - y}$ (Product Rule)

$= {x}^{2} {e}^{x y - x - y} \cdot \left(\frac{\mathrm{dy}}{\mathrm{dx}} \cdot 1 + y - 1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) + 2 x \cdot {e}^{x y - x - y}$
(Chain Rule, Product Rule and Power Rule)

$= {x}^{2} {e}^{x y - x - y} \cdot \left(\cancel{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot 1} + y - 1 - \cancel{\frac{\mathrm{dy}}{\mathrm{dx}}}\right) + 2 x \cdot {e}^{x y - x - y}$

$= {x}^{2} {e}^{x y - x - y} \cdot \left(y - 1\right) + 2 x {e}^{x y - x - y}$

=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 $\left(x , y\right)$

Slope at point $\left(0 , 0\right)$ = ${e}^{0} \cdot 0 = 1 \cdot 0 = 0$

Mar 16, 2016

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

#### Explanation:

If the point $\left(0 , 0\right)$ is on the graph of ${x}^{2} {e}^{x y - x - y} = C$,

then $C = {\left(0\right)}^{2} {e}^{0 - 0 - 0} = 0$.

But now we have ${x}^{2} {e}^{x y - x - y} = 0$ which implies that ${x}^{2} = 0$ (because for all $u$, ${e}^{u} \ne 0$).

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.