# If xe^y-11x+5y=0, what is dy/dx?

Dec 3, 2016

The derivative of $y$ with respect to $x$ is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{11 - {e}^{y}}{x {e}^{y} + 5} .$

#### Explanation:

If $y$ and $x$ are implicitly related ("depend on each other") as in the equation $x {e}^{y} - 11 x + 5 y = 0$, it makes sense that the rate at which $y$ changes (with respect to $x$) would depend on both $x$ and $y$, and not just on $x$.

To find this rate of change $\frac{\mathrm{dy}}{\mathrm{dx}}$, what we need to do is take the derivative of both sides with respect to $x$, treating $y$ as a function of $x$.

$\text{ "xe^y" "-" "11x+" "5y=" } 0$

$\implies \textcolor{b r o w n}{\frac{d}{\mathrm{dx}} \left(x {e}^{y}\right)} - \textcolor{b l u e}{\frac{d}{\mathrm{dx}} 11 x} + \textcolor{g r e e n}{\frac{d}{\mathrm{dx}} 5 y} = \textcolor{m a \ge n t a}{\frac{d}{\mathrm{dx}} 0}$

$\implies \textcolor{b r o w n}{\left(1\right) {e}^{y} + x {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)} - \textcolor{b l u e}{11} + \textcolor{g r e e n}{5 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)} = \textcolor{m a \ge n t a}{0}$

Notice: since $y$ is considered a function of $x$, the derivative of $y$ (with respect to $x$) remains its own term in the equation. To make reading the equation easier, we will use $y '$ in place of $\frac{\mathrm{dy}}{\mathrm{dx}}$. Continuing, we have:

${e}^{y} + x {e}^{y} y ' - 11 + 5 y ' = 0$

$\implies \text{ "xe^yy'+5y'=11-e^y" }$ (isolate the $y '$ terms)

$\implies \text{ "y'(xe^y+5)=11-e^y" }$ (factor out $y '$)

$\implies \text{ } y ' = \frac{11 - {e}^{y}}{x {e}^{y} + 5}$

Therefore, the derivative of $y$ with respect to $x$ is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{11 - {e}^{y}}{x {e}^{y} + 5} .$