What is the derivative of $y = sin(tan(3x))$ ?

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Answer 1 · 2015-08-11T21:46:59

$y^' = 3 cos(tan(3x)) * sec^2(3x)$

You can differentiate this function by using the chain rule three times.

First, start by writing your function as $y = sinu$ , where $u = tan(3x)$ . Its derivative will take the form

$d/dx(sinu) = d/(du)(sinu) * d/dx(u)$

$d/dx(sinu) = cosu * d/dx(tan(3x))$

Now focus on $d/dx(tan(3x))$ , which can be written as $tan(u_1)$ , with $u_1 = 3x$ .

$d/dx(tanu_1) = d/(du_1)(tanu_1) * d/dx(u_1)$

$d/dx(tanu_1) = sec^2u_1 * d/dx(3x)$

$d/dx(tan(3x)) = sec^2(3x) * 3$

Plug this into your target derivative to get

$d/dx(sin(tan(3x))) = cos(tan(3x)) * 3 sec^2(3x)$

Therefore,

$y^' = d/dx(sin(tan(3x))) = color(green)(3 cos(tan(3x)) * sec^2(3x))$

What is the derivative of y = sin(tan(3x))y = sin(tan(3x))?