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

Oct 18, 2016

$y = 2 x + 5$

Explanation:

${x}^{2} / \left({e}^{x - y}\right) = C$
$\therefore {x}^{2} = C {e}^{x - y}$
$\therefore {x}^{2} = C {e}^{x} {e}^{-} y$

Differentiating implicitly involves differentiating everything wrt $x$ and using the chain rule and product rule:

So, $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(C {e}^{x} {e}^{-} y\right)$
$\therefore 2 x = C \frac{d}{\mathrm{dx}} \left({e}^{x} {e}^{-} y\right)$
$\therefore 2 x = C \left({e}^{x} \frac{d}{\mathrm{dx}} {e}^{-} y + {e}^{-} y \frac{d}{\mathrm{dx}} {e}^{x}\right)$ using the product rule
$\therefore 2 x = C \left({e}^{x} \frac{d}{\mathrm{dy}} {e}^{-} y \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} y {e}^{x}\right)$ using the chain rule
$\therefore 2 x = C \left(- {e}^{x} {e}^{-} y \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} y {e}^{x}\right)$
$\therefore 2 x = {x}^{2} / \left({e}^{x - y}\right) \left(- {e}^{x} {e}^{-} y \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} y {e}^{x}\right)$

We don't need an explicit expression for $\frac{\mathrm{dy}}{\mathrm{dx}}$ so don't attempt to simplify further. We can just substitute $x = - 2$ and $y = 1$, to find $\frac{\mathrm{dy}}{\mathrm{dx}}$ at that value.

So, at $\left(- 2 , 1\right)$ we have:
$\therefore 2 x = {x}^{2} / \left({e}^{x - y}\right) \left(- {e}^{x} {e}^{-} y \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} y {e}^{x}\right)$
$\therefore 2 \left(- 2\right) = {\left(- 2\right)}^{2} / \left({e}^{- 2 - 1}\right) \left(- {e}^{-} 2 {e}^{-} 1 \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} 1 {e}^{-} 2\right)$
$\therefore - 4 = \frac{4}{{e}^{-} 3} \left(- {e}^{-} 3 \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{-} 3\right)$
$\therefore - 4 = 4 \left(- \frac{\mathrm{dy}}{\mathrm{dx}} + 1\right)$
$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} + 1 = - 1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 2$

So the tangent passes through $\left(- 2 , 1\right)$ and has slope $m = 2$ so we use the formula $y - {y}_{1} = m \left(x - {x}_{1}\right)$ to get the equation:

$y - 1 = 2 \left(x - - 2\right)$
$\therefore y - 1 = 2 \left(x + 2\right)$
$\therefore y - 1 = 2 x + 4$
$\therefore y = 2 x + 5$