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

May 21, 2016

$y = \left(12 {e}^{\frac{1}{3}} + 3 {e}^{\frac{1}{3}} x\right)$

#### Explanation:

Given $p = \left(- 2 , 1\right)$ if this point pertains to the graphic of $f \left(x , y\right) = {\left(x - y\right)}^{3} {e}^{y} - C$ then $f \left(- 2 , 1\right) = 0 \to {\left(- 2 - 1\right)}^{3} {e}^{1} - C = 0$ so $C = - 27 e$.
Also from ${\left(x - y\right)}^{3} {e}^{y} = C$ we obtain $x = {\left(C {e}^{- y}\right)}^{\frac{1}{3}} + y$.
A generic point ${p}_{0}$ pertaining to $f \left(x , y\right) = 0$ has the form
${p}_{0} = \left({\left(C {e}^{- {y}_{0}}\right)}^{\frac{1}{3}} + {y}_{0} , {y}_{0}\right)$.

Calculating $\frac{\mathrm{dy}}{\mathrm{dx}}$ we proceed as:
$\mathrm{df} \left(x , y\right) = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0 \to \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$ then
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{3 - x + y} = \frac{3}{3 - {\left(C {e}^{- y}\right)}^{\frac{1}{3}}}$
The tangent straight to the point ${p}_{0}$ gives
$\left(y - {y}_{0}\right) = \frac{3}{3 - {\left(C {e}^{- {y}_{0}}\right)}^{\frac{1}{3}}} \left(x - {x}_{0}\right)$ but
${x}_{0} = {\left(C {e}^{- {y}_{0}}\right)}^{\frac{1}{3}} + {y}_{0}$ so the tangent at ${p}_{0}$ reads
$y = {y}_{0} + \frac{3}{3 - {\left(C {e}^{- {y}_{0}}\right)}^{\frac{1}{3}}} \left(x - {\left(C {e}^{- {y}_{0}}\right)}^{\frac{1}{3}} + {y}_{0}\right)$
And for ${p}_{0} = \left(- 2 , 1\right)$ reads
$y = \left(12 {e}^{\frac{1}{3}} + 3 {e}^{\frac{1}{3}} x\right)$
/(6 e ^(1/3))#