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

How do you differentiate $\mathrm{tan}2x=\mathrm{cos}3y$
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Using the chain rule, where $u=2x$ and $\frac{d}{dx}=\frac{d}{du}\cdot \frac{du}{dx}$
$\frac{d}{du}\left(\mathrm{tan}\left(u\right)\right)\cdot \frac{du}{dx}=\frac{d}{dx}\left(\mathrm{cos}\left(3y\right)\right)$
${\mathrm{sec}}^{2}\left(u\right)\cdot 2=\frac{d}{dx}\left(\mathrm{cos}\left(3y\right)\right)$
$2{\mathrm{sec}}^{2}\left(2x\right)=\frac{d}{dx}\left(\mathrm{cos}\left(3y\right)\right)$
Using the chain rule, where $v=3y$ and $\frac{d}{dx}=\frac{d}{dv}\cdot \frac{dv}{dy}\cdot \frac{dy}{dx}$
$2{\mathrm{sec}}^{2}\left(2x\right)=\frac{d}{dv}\left(\mathrm{cos}\left(v\right)\right)\cdot \frac{d}{dy}\left(3y\right)\cdot \frac{dy}{dx}$
$2{\mathrm{sec}}^{2}\left(2x\right)=-3\mathrm{sin}\left(v\right)\cdot \frac{dy}{dx}$
$-\frac{2{\mathrm{sec}}^{2}\left(2x\right)\cdot \mathrm{csc}\left(3y\right)}{3}=\frac{dy}{dx}$