Question

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

Answer 1

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

Explanation:

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))`