# How do you differentiate tan 2x = cos 3y?

Oct 14, 2015

$- \frac{2 {\sec}^{2} \left(2 x\right) \cdot \csc \left(3 y\right)}{3} = \frac{\mathrm{dy}}{\mathrm{dx}}$

#### Explanation:

Using the chain rule, where $u = 2 x$ and $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{du}} \left(\tan \left(u\right)\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\cos \left(3 y\right)\right)$

${\sec}^{2} \left(u\right) \cdot 2 = \frac{d}{\mathrm{dx}} \left(\cos \left(3 y\right)\right)$

$2 {\sec}^{2} \left(2 x\right) = \frac{d}{\mathrm{dx}} \left(\cos \left(3 y\right)\right)$

Using the chain rule, where $v = 3 y$ and $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dv}} \cdot \frac{\mathrm{dv}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$2 {\sec}^{2} \left(2 x\right) = \frac{d}{\mathrm{dv}} \left(\cos \left(v\right)\right) \cdot \frac{d}{\mathrm{dy}} \left(3 y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$2 {\sec}^{2} \left(2 x\right) = - 3 \sin \left(v\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$- \frac{2 {\sec}^{2} \left(2 x\right) \cdot \csc \left(3 y\right)}{3} = \frac{\mathrm{dy}}{\mathrm{dx}}$