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)

Now try your problem.

Harley