# How do you find dy/dx by implicit differentiation of xcosy=1 and evaluate at point (2, pi/3)?

Mar 1, 2017

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{2 , \frac{\pi}{3}} = \frac{1}{2 \sqrt{3}}$

#### Explanation:

Differentiate the equation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(x \cos y\right) = 0$

$\cos y - x \sin y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\cos y = x \sin y \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x t g y}$

In the point $\left(2 , \frac{\pi}{3}\right)$ the value is then:

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{2 , \frac{\pi}{3}} = \frac{1}{2 t g \left(\frac{\pi}{3}\right)} = \frac{1}{2 \sqrt{3}}$