Given H(x) = (x+x^-1)^3

Differentiate with respect to x on both sides

d/dx (H(x)) = d/dx (x+x^-1)^3

H'(x) = 3(x+x^-1)^2 * [1-(x^-2)]

= 3[(x+1/x)^2] [1-(1/x^2)]

There are two ways to differentiate this function. One way is to multiply it out and apply a Power Rule. The other way is to use a Chain Rule.

`H(x)=(x+x^(-1))^(3) `

So,

`H'(x)=3(x+x^(-1))^(2)*(1-(1)/(x^(2))) `

Simplifying, you get:

`H'(x)=3(1-(1)/(x^(2)))*(x+x^(-1))^(2) `

The `H(x)` function can be rewritten as:``

which means; `H(x) = x^3 + 3*x + 3*x^-1 + x^-3`

Differentianting the function `H(x)` :

`H'(x) = 3*x^2 + 3 + (-1)*3*x^-2 + (-3)*x^(-4)`

Adjusting the format:

`H'(x) = (3x^2 - 3/x^2) - (3/x^4 +3)`

`H'(x) = 3(1-x^-4)(x^2+1)`

`H(x) = (x+x^(-1))^3`

= `x^3 + x^(-3) + 3x + 3/x`

so,

`H'(x)= 3(x^2) + (-3)x^(-4)+ 3 - 3/(x^2)`