# What is the slope of the tangent line of  (x-2)(y-3)-e^y= C , where C is an arbitrary constant, at (-2,1)?

Jan 16, 2017

$y ' {|}_{\left(- 2 , 1\right)} = - \frac{2}{e + 4}$

#### Explanation:

We can use implicit differentiation (and the product rule for the first term):

$\left(y - 3\right) + \left(x - 2\right) y ' - y ' {e}^{y} = 0$

$\left(y - 3\right) = y ' \left({e}^{y} - x + 2\right)$

$y ' = \frac{y - 3}{{e}^{y} - x + 2}$

$y ' {|}_{\left(- 2 , 1\right)} = \frac{1 - 3}{{e}^{1} - \left(- 2\right) + 2} = - \frac{2}{e + 4}$

we can also check this by evaluating the normal vector $m a t h b f n$ as

$m a t h b f n = \nabla \left(\left(x - 2\right) \left(y - 3\right) - {e}^{y}\right) = \left\langle y - 3 , x - 2 - {e}^{y}\right\rangle$

$m a t h b f {n}_{\left(- 2 , 1\right)} = \left\langle - 2 , - 4 - e\right\rangle$

and for tangent vector $m a t h b f t$, we say:

$m a t h b f n \cdot m a t h b f t = 0 \implies \setminus m a t h b f t = \left\langle 4 + e , - 2 ,\right\rangle$, same answer