# What is the slope of the tangent line of x^2e^(xy)= C , where C is an arbitrary constant, at (-1,1)?

Dec 11, 2017

The slope of the tangent is $m = - 1$.

#### Explanation:

The slope of the tangent line to the curve is the value of the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ in the point $P \left(- 1 , 1\right)$.

We can calculate the derivative using implicit differentiation:

${x}^{2} {e}^{x y} = C$

Differentiate both sides with respect to $x$:

$2 x {e}^{x y} + {x}^{2} {e}^{x y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$x {e}^{x y} \left(2 + x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

Now, ${e}^{x y} \ne 0$ because the exponential is never null, and $x \ne 0$ otherwise we would have $C = 0$, and $y$ indeterminate, which means for $C = 0$ the curve defined by the equation is the line $x = 0$ that does not pass through $P$. So:

$2 + x y + {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 + x y}{x} ^ 2$

In the point $P \left(- 1 , 1\right)$ then the value of the derivative is:

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{- 1 , 1} = - \frac{2 + \left(- 1\right) \cdot 1}{- 1} ^ 2 = - 1$