# Given that xe^y = x+1, show that e^y + xe^ydy/dx = 1?

May 12, 2018

Given: $x {e}^{y} = x + 1$

Begin the implicit differentiation process:

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

The derivative of a constant is 0:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = \frac{\mathrm{dx}}{\mathrm{dx}}$

$\frac{\mathrm{dx}}{\mathrm{dx}}$ becomes 1:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = 1 \text{ }$

When attempting to compute a derivative of the form $\frac{d \left(u v\right)}{\mathrm{dx}}$, the product rule states that

$\frac{d \left(u v\right)}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}} v + u \frac{\mathrm{dv}}{\mathrm{dx}}$

The left side of equation  is in this form where $u = x$ and $v = {e}^{y}$

Substituting this into the product rule we obtain the equation:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = \frac{d \left(x\right)}{\mathrm{dx}} {e}^{y} + x \frac{d \left({e}^{y}\right)}{\mathrm{dx}}$

We know that $\frac{d \left(x\right)}{\mathrm{dx}} = 1$:

$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = {e}^{y} + x \frac{d \left({e}^{y}\right)}{\mathrm{dx}}$

${e}^{y}$ is a function of y, therefore, we must use the chain rule to compute $\frac{d \left({e}^{y}\right)}{\mathrm{dx}}$ as follows:

$\frac{d \left({e}^{y}\right)}{\mathrm{dx}} = \frac{d \left({e}^{y}\right)}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$

The exponential function is a special case where $\frac{d \left({e}^{y}\right)}{\mathrm{dy}} = {e}^{y}$:

$\frac{d \left({e}^{y}\right)}{\mathrm{dx}} = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

After using the product rule and the chain rule, we have$\frac{d \left(x {e}^{y}\right)}{\mathrm{dx}} = {e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$ and we substitute this into equation :

${e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

We shall stop here because the above is what we want.

May 12, 2018

We use implicit differentiation to see that

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - {e}^{y}}{x {e}^{y}}$

We now substitute into the given expression.

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

${e}^{y} + 1 - {e}^{y} = 1$

$1 = 1$

As required.

Hopefully this helps!

May 12, 2018

#### Explanation:

Another way of proving this is :

Let,

$f \left(x , y\right) = x {e}^{y} - x - 1$

And

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$

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

$\frac{\partial f}{\partial y} = x {e}^{y}$

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{y} - 1}{x {e}^{y}}$

Rearranging,

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

${e}^{y} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

May 12, 2018

Given expression:

$x {e}^{y} = x + 1$

Inspection reveals that differential of RHS with respect to $x$ gives us RHS of the expression required to be proved. Therefore, differentiating both sides with respect to variable $x$ we get

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

Using product rule and implicit differentiation on the LHS we get

$L H S = \frac{d}{\mathrm{dx}} \left(x {e}^{y}\right)$
$\implies L H S = x \frac{d}{\mathrm{dx}} {e}^{y} + {e}^{y} \frac{d}{\mathrm{dx}} x$
$\implies L H S = x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y}$

We see that it is same as $L H S$ of the expression which was to be proved.

Hence Proved.