# What is the implicit derivative of 1= x/y^3-e^(x-y)?

Mar 7, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{-} 3 - {e}^{x - y}}{3 x {y}^{-} 4 - {e}^{x - y}}$

#### Explanation:

We can use product rule and chain rule to fully expand this out:
$1 = \frac{x}{y} ^ 3 - {e}^{x - y}$
Differentiating both sides,
$0 = d \left(\frac{x}{y} ^ 3\right) - d \left({e}^{x - y}\right)$
$0 = \frac{1}{y} ^ 3 \setminus \mathrm{dx} + x \setminus d \left(\frac{1}{y} ^ 3\right) - {e}^{x} {\mathrm{de}}^{- y} - {e}^{- y} {\mathrm{de}}^{x}$
$0 = {y}^{- 3} \mathrm{dx} - 3 x {y}^{- 4} \mathrm{dy} + {e}^{x} {e}^{- y} \mathrm{dy} - {e}^{- y} {e}^{x} \mathrm{dx}$
$0 = \left({y}^{-} 3 - {e}^{x - y}\right) \mathrm{dx} - \left(3 x {y}^{-} 4 - {e}^{x - y}\right) \mathrm{dy}$
$\left(3 x {y}^{-} 4 - {e}^{x - y}\right) \mathrm{dy} = \left({y}^{-} 3 - {e}^{x - y}\right) \mathrm{dx}$
$\mathrm{dy} = \frac{\left({y}^{-} 3 - {e}^{x - y}\right)}{\left(3 x {y}^{-} 4 - {e}^{x - y}\right)} \mathrm{dx}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{-} 3 - {e}^{x - y}}{3 x {y}^{-} 4 - {e}^{x - y}}$

If we want to make this a little prettier, we can multiply both sides by ${y}^{4} {e}^{y}$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y {e}^{y} - {y}^{4} {e}^{x}}{3 x {e}^{y} - {y}^{4} {e}^{x}}$

but that's just my preference.