# How do you differentiate -2=yln(e^(x-y^3))+xe^(x-y)?

Jul 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x {e}^{x - y} + {e}^{x - y} + y}{x {e}^{x - y} + 4 {y}^{3} - x} . \mathmr{and} ,$

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

#### Explanation:

Knowing that, $\ln \left({e}^{x - {y}^{3}}\right) = \left(x - {y}^{3}\right) \ln e = x - {y}^{3} ,$ we have,

$- 2 = y \left(x - {y}^{3}\right) + x {e}^{x - y} , \mathmr{and} , x {e}^{x - y} = y \left({y}^{3} - x\right) - 2 , i . e . ,$

$x {e}^{x - y} = {y}^{4} - x y - 2.$

$\therefore \frac{d}{\mathrm{dx}} \left\{x {e}^{x - y}\right\} = \frac{d}{\mathrm{dx}} \left\{{y}^{4} - x y - 2\right\} .$

Using the Product and Sum/Difference Rule, we get,

$x \frac{d}{\mathrm{dx}} \left\{{e}^{x - y}\right\} + {e}^{x - y} \frac{d}{\mathrm{dx}} \left\{x\right\} = \frac{d}{\mathrm{dx}} \left({y}^{4}\right) - \frac{d}{\mathrm{dx}} \left(x y\right) - \frac{d}{\mathrm{dx}} \left(2\right) ,$ or,

$x \frac{d}{\mathrm{dx}} \left\{{e}^{x - y}\right\} + {e}^{x - y} = \frac{d}{\mathrm{dx}} \left({y}^{4}\right) - \left\{x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left(x\right)\right\} \ldots . \left(1\right)$

Here, by the Chain Rule, $\frac{d}{\mathrm{dx}} \left\{{e}^{x - y}\right\} = {e}^{x - y} \frac{d}{\mathrm{dx}} \left(x - y\right) ,$

$= {e}^{x - y} \left\{\frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(y\right)\right\} = {e}^{x - y} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) \ldots \ldots \ldots \ldots \ldots . . \left(2\right) .$

Also, $\frac{d}{\mathrm{dx}} \left({y}^{4}\right) = \frac{d}{\mathrm{dy}} \left({y}^{4}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 4 {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right) .$

Utilising $\left(2\right) \mathmr{and} \left(3\right)$ in $\left(1\right) ,$ we get,

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

$\therefore x {e}^{x - y} - x {e}^{x - y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x - y} = \left(4 {y}^{3} - x\right) \frac{\mathrm{dy}}{\mathrm{dx}} - y ,$

$\therefore x {e}^{x - y} + {e}^{x - y} + y = x {e}^{x - y} \frac{\mathrm{dy}}{\mathrm{dx}} + \left(4 {y}^{3} - x\right) \frac{\mathrm{dy}}{\mathrm{dx}} ,$

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

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

This Answer is quite fair, but it can be put as :

Using, xe^(x-y)=y^4-xy-2, &, e^(x-y)=(y^4-xy-2)/x,

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

$= \frac{x \left({y}^{4} - x y - 2\right) + \left({y}^{4} - x y - 2\right) + x y}{x \left({y}^{4} - x y - 2 + 4 {y}^{3} - x\right)} ,$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left({y}^{4} - x y - 2\right) + {y}^{4} - 2}{x \left({y}^{4} - x y - 2 + 4 {y}^{3} - x\right)} .$

Enjoy Maths.!