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))#