# How do you implicitly differentiate 11=(x)/(ye^x)?

Oct 12, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 11 y {e}^{x}}{11 {e}^{x}}$

#### Explanation:

$11 = \frac{x}{y {e}^{x}}$
can be rearranged as;
$11 y {e}^{x} = x$

We can now differentiate wrt x;

$\frac{d}{\mathrm{dx}} \left\{\left(11 y\right) \left({e}^{x}\right)\right\} = \frac{d}{\mathrm{dx}} \left(x\right)$

We can use the product rule to differentiate the LHS and we know already know how to differentiate the RHS:

So, $\frac{d}{\mathrm{dx}} \left(11 y {e}^{x}\right) = \frac{d}{\mathrm{dx}} \left(x\right)$
$\therefore \left(11 y\right) \left(\frac{d}{\mathrm{dx}} {e}^{x}\right) + \left(\frac{d}{\mathrm{dx}} 11 y\right) \left({e}^{x}\right) = \frac{d}{\mathrm{dx}} \left(x\right)$
$\therefore \left(11 y\right) {e}^{x} + \left(\frac{d}{\mathrm{dx}} 11 y\right) \left({e}^{x}\right) = 1$

We can then use the chain rule as $\frac{\mathrm{dA}}{\mathrm{dx}} = \frac{\mathrm{dA}}{\mathrm{dy}} . \frac{\mathrm{dy}}{\mathrm{dx}}$; so
$11 y {e}^{x} + \left(\frac{d}{\mathrm{dy}} 11 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} \left({e}^{x}\right) = 1$
$11 y {e}^{x} + 11 \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} = 1$
$11 {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - 11 y {e}^{x}$
Hence, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 11 y {e}^{x}}{11 {e}^{x}}$