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 Question from Michael, a student: f(x)=sin^3(3x^2) find f ' (x)

Hi Michael,

I am going to change the problem slightly

g(x) = tan2(4/x) find g' (x)

The key to both of these problems is the chain rule, in fact for both problems you need to use the chain rule twice.

First I want to rewrite g(x) writing 4/x as 4x-1.

g(x) = tan2( 4x-1)

What I immediately see is that g(x) is some function squared.

g(x) = [tan( 4x-1)]2

Using the chain rule

g'(x) = 2[tan( 4x-1)]1 d/dx[tan( 4x-1)]

Now to find d/dx[tan( 4x-1)] I need to differentiate the tangent function. The derivative of the tangent function is the square of the secant function so, again using the chain rule

g'(x) = 2[tan( 4x-1)]1 d/dx[tan( 4x-1)]
= 2 tan( 4x-1) sec2(4x-1) d/dx(4x-1)

Finally I need to differentiate 4x-1.

g'(x) = 2[tan( 4x-1)]1 d/dx[tan( 4x-1)]
= 2 tan( 4x-1) sec2(4x-1) d/dx(4x-1)
= 2 tan( 4x-1) sec2(4x-1) (-4x-2)

My original question didn't have negative exponents so I would write my answer as

g'(x) = -8/x2 tan( 4/x) sec2(4/x)