How do you take the derivative of $tan (2 x)$ $tan(2x)$ ?

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Answer 1 · 2015-09-05T18:40:17

Using the chain rule $tan (2 x) = 2 {sec}^{2} (2 x)$ $tan(2x)=2sec^2(2x)$

The chain rule is as follows:

$\frac{d}{d x} f (g (x)) = [\frac{d}{d x} f (x)] ∣_{x = g (x)} \cdot \frac{d}{d x} g (x)$ $d/dx f(g(x)) = [d/dx f(x)]|_(x=g(x)) * d/dx g(x)$

...or, in words,
1) Get the derivative of the outer function, plug in the inner function...
2) ...multiplied by the derivative of the inner function.

In $tan (2 x)$ $tan(2x)$ , the outer function is $tan x$ $tan x$ and the inner function is $2 x$ $2x$ .

The derivative of $tan x$ $tan x$ is ${sec}^{2} x$ $sec^2 x$ .
Plug in $2 x$ $2x$ , and we have ${sec}^{2} (2 x)$ $sec^2 (2x)$ .

So, after our first step, we have:
$\frac{d}{d x} tan (2 x)$ $d/dx tan(2x)$
$= {sec}^{2} (2 x) \cdot \frac{d}{d x} (2 x)$ $=sec^2 (2x) * d/dx (2x)$

Then, we continue:
$\frac{d}{d x} tan (2 x)$ $d/dx tan(2x)$
$= {sec}^{2} (2 x) \cdot \frac{d}{d x} (2 x)$ $=sec^2 (2x) * d/dx (2x)$

$= {sec}^{2} (2 x) \cdot (2)$ $=sec^2 (2x) * (2)$

$= 2 {sec}^{2} (2 x)$ $=2sec^2(2x)$

How do you take the derivative of tan(2x) tan(2x)?