Find dy/dx by implicit differentiation: e^x/y = 9x − y

For this we should take derivative at both sides, and solve using chain rule.

U might me used to solving the problem like $\frac{de^x}{dx}$ = $e^x$

For $\frac{de^ \frac xy}{dx}$, we need to multiply both numerator and denomenator by $d (\frac xy)$. This is chain rule! To match the variable (x/y) we chained d(x/y).

$\frac{de^ \frac xy}{dx}$ = $\frac{de^ \frac xy}{d (\frac xy)} \times \frac{d (\frac xy)}{dx}$

= $e^ \frac xy \times \frac{d (\frac xy)}{dx}$

Now, to solve $\frac{d (\frac xy)}{dx}$, let's use quotient rule,

= $e^ \frac xy \times (\frac{y - x\frac{dy}{dx}}{y^2})$

Going back to original question,

$e^ \frac xy$ = 9x − y

Taking derivative on both sides,

$\frac{de^ \frac xy}{dx}$ = $9\frac{dx}{dx} \, - \, \frac{dy}{dx}$

We already solved $\frac{de^ \frac xy}{dx}$ above. so, let's use it!

or, $e^ \frac xy \times (\frac{y - x\frac{dy}{dx}}{y^2})$ = 9 - $\frac{dy}{dx}$

or, $e^ \frac xy (y - x\frac{dy}{dx})$ = $9y^2 \, - \, y^2 \frac{dy}{dx}$

or, $\frac{dy}{dx}$ = $\frac{y.e^ \frac xy \, - \, 9y^2}{x.e^ \frac xy \, - \, y^2}$

OPTIONAL step. We've given $e^ \frac xy$ = 9x − y in the question. So let's substitute it!

or, $\frac{dy}{dx}$ = $\frac{y(9x \, - \, y) \, - \, 9y^2}{x(9x \, - \, y) \, - \, y^2}$

so, $\frac{dy}{dx}$ = $\frac{9xy \, - \, 10y^2}{9x^2 \, - \, xy \, - \, y^2}$