# How do you find (dy)/(dx) given xy=tanxy?

Oct 6, 2016

$- \frac{y}{x}$

#### Explanation:

$x y = \tan x y \to y = y \left(x\right)$ so

$f \left(y \left(x\right) , x\right) = y \left(x\right) x - \tan \left(y \left(x\right) x\right) = 0$

now

d/(dx)f(y(x),x) = y(x) + x y(x) - Sec(x y(x))^2 (y(x) + x y'(x)) = 0

Solving for $y ' \left(x\right)$ we obtain

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

We can also obtain the same result straightforward

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}} = - \frac{y - y S e c {\left(x y\right)}^{2}}{x - x S e c {\left(x y\right)}^{2}} = - \frac{y}{x}$