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

Apr 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {y}^{4} {e}^{x - y}}{3 {y}^{2} - {y}^{4} {e}^{x - y} + 4 {e}^{x - y} {y}^{3}}$

#### Explanation:

Here,

$1 = \frac{x}{y} ^ 3 - y {e}^{x - y}$

$\implies {y}^{3} = x - {y}^{4} {e}^{x - y}$

Diff.w.r.t. $x$, $\text{using "color(blue) "product and chain rule}$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \left[{y}^{4} \frac{d}{\mathrm{dx}} \left({e}^{x - y}\right) + {e}^{x - y} \frac{d}{\mathrm{dx}} \left({y}^{4}\right)\right]$

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \left[{y}^{4} {e}^{x - y} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) + {e}^{x - y} 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

3y^2(dy)/(dx)=1-y^4e^(x-y)+y^4e^(x-y)(dy)/(dx)-4e^(x- y)y^3(dy)/(dx)

3y^2(dy)/(dx)-y^4e^(x-y)(dy)/(dx)+4e^(x-y)y^3(dy)/(dx)=1- y^4e^(x-y)

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[3 {y}^{2} - {y}^{4} {e}^{x - y} + 4 {e}^{x - y} {y}^{3}\right] = 1 - {y}^{4} {e}^{x - y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {y}^{4} {e}^{x - y}}{3 {y}^{2} - {y}^{4} {e}^{x - y} + 4 {e}^{x - y} {y}^{3}}$