# What is the derivative of \tan(2x) ?

What is the derivative of $\mathrm{tan}\left(2x\right)$ ?
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Orlando Paz
Assuming that you know the derivative rule: $\frac{d}{dx}\left(\mathrm{tan}x\right)={\mathrm{sec}}^{2}\left(x\right)$
$\frac{d}{dx}\left(\mathrm{tan}\left(2x\right)\right)$ will simply be ${\mathrm{sec}}^{2}\left(2x\right)\cdot \frac{d}{dx}\left(2x\right)$ according to the chain rule.
Then $\frac{d}{dx}\left(\mathrm{tan}\left(2x\right)\right)=2{\mathrm{sec}}^{2}\left(2x\right)$
If you want to easily understand chain rule, just remember my tips: take the normal derivative of the outside (ignoring whatever is inside the parenthesis) and then multiply it by the derivative of the inside (stuff inside the parenthesis)

Jack Maxson
The first thing to realize is that we're dealing with a composite function $f\left(g\left(x\right)\right)$, where
$f\left(x\right)=\mathrm{tan}x$ and $g\left(x\right)=2x$
When we differentiate a composite function, we use the Chain Rule
${f}^{\prime }\left(g\left(x\right)\right)\cdot {g}^{\prime }\left(x\right)$
From the definition of tangent and an application of the Quotient Rule, we know that ${f}^{\prime }\left(x\right)={\mathrm{sec}}^{2}x$
We also know that ${g}^{\prime }\left(x\right)=2$. Now, we have everything we need to plug into the Chain Rule:
${\mathrm{sec}}^{2}\left(2x\right)\cdot 2$, which can be rewritten as
$2{\mathrm{sec}}^{2}\left(2x\right)$
Hope this helps!