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

##### 1 Answer
Jun 21, 2016

$y ' = - \setminus \frac{2 {\sec}^{2} \left(2 x\right)}{3 \sqrt{1 - {\tan}^{2} \left(2 x\right)}}$

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \setminus \tan \left(2 x\right) = 2 {\sec}^{2} \left(2 x\right)$

if you are uncomfortable with this, make the sub $u = 2 x$ and use the chain rule so $\setminus \frac{d}{\mathrm{dx}} \setminus \tan \left(2 x\right) = \setminus \frac{d}{\mathrm{dx}} \setminus \tan \left(u \left(x\right)\right) = \setminus \frac{d}{\mathrm{du}} \setminus \tan \left(u\right) \setminus \frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} u \left(2\right) = 2 {\sec}^{2} \left(2 x\right)$

similarly: $\setminus \frac{d}{\mathrm{dx}} \setminus \cos \left(3 y\right) = - 3 \sin \left(3 y\right) \setminus y '$

$\setminus \implies y ' = - \setminus \frac{2 {\sec}^{2} \left(2 x\right)}{3 \sin \left(3 y\right)}$

using the identity ${\cos}^{2} + {\sin}^{2} = 1$ we can say that $3 \sin \left(3 y\right) = 3 \sqrt{1 - {\cos}^{2} \left(3 y\right)} = 3 \sqrt{1 - {\tan}^{2} \left(2 x\right)}$

so $y ' = - \setminus \frac{2 {\sec}^{2} \left(2 x\right)}{3 \sqrt{1 - {\tan}^{2} \left(2 x\right)}}$