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

##### 1 Answer
May 3, 2018

$m = \frac{1}{10}$

#### Explanation:

Given that the curve contains the point $\left(1 , 1\right)$, we find $C = 9$

The equation is equivalent to

${x}^{2} + 4 x y + 4 {y}^{2} = 9 {e}^{x - {y}^{2}}$.

Differentiate implicitly to get

$2 x + 4 y + 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 9 {e}^{x - {y}^{2}} \left(1 - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

At $\left(1 , 1\right)$ we have

$2 + 4 + 4 \frac{\mathrm{dy}}{\mathrm{dx}} + 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 9 - 18 \frac{\mathrm{dy}}{\mathrm{dx}}$

so, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{10}$