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

2 Answers
Jul 20, 2017

$1$

Explanation:

Making $z = x - y$ we have

$z {e}^{z} = C$ or $z = W \left(C\right)$ where $W \left(\cdot\right)$ is the Lambert Function.

then $z = {z}_{0} = W \left(C\right)$

Now ${x}_{0} = x - y$ and $\mathrm{dx} = \mathrm{dy}$ or $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

and finally

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Jul 20, 2017

The slope of the tangent line is $m = 1$

Explanation:

Differentiate the equation implicitly:

$\frac{d}{\mathrm{dx}} \left(\left(x - y\right) {e}^{x - y}\right) = 0$

$\left(x - y\right) \frac{d}{\mathrm{dx}} \left({e}^{x - y}\right) + \left(1 - y '\right) {e}^{x - y} = 0$

$\left(x - y\right) \left(1 - y '\right) {e}^{x - y} + \left(1 - y '\right) {e}^{x - y} = 0$

$\left(x - y + 1\right) \left(1 - y '\right) {e}^{x - y} = 0$

As ${e}^{t} \ne 0$ this implies:

$\left(x - y + 1\right) \left(1 - y '\right) = 0$

We have then two possible solutions:

$\left(1\right) \text{ } \left(1 - y '\right) = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$y = x + a$

or:

$\left(2\right) \text{ } \left(x - y + 1\right) = 0$

$y = x + 1$

which is just the particular case of $\left(1\right)$ for $a = 1$.

Thus:

$y = x + a$

is the general solution, which is always a straight line of slope $m = 1$ for any $C$ coincident with ts tangent.