If,

`y = cos root(3)(x)`

to take the derivative of y, apply the formula `(cos u) ' = -sin u * u'` .

`y ' = (cos root (3)(x) )'`

`y ' = -sin root(3)(x) * (root(3)(x)) '`

To determine `(root(3)(x))'` , express the cube root as exponent of x....

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If,

`y = cos root(3)(x)`

to take the derivative of y, apply the formula `(cos u) ' = -sin u * u'` .

`y ' = (cos root (3)(x) )'`

`y ' = -sin root(3)(x) * (root(3)(x)) '`

To determine `(root(3)(x))'` , express the cube root as exponent of x. Then, apply power formula of derivative which is `(u^n)' = n * u^(n-1)* u'` .

`y' =-sinroot(3)(x) *(x^(1/3))`

`y' =-sinroot(3)(x) *1/3 * x^(-2/3) * x'`

Note that `x' = 1` .

`y ' = -sin root(3)(x) * 1/3 * x^(-2/3) * 1`

`y' = -(sin root(3)(x)*x^(-2/3))/3`

Then, apply the negative exponent rule `( a^(-m) = 1/a^m)` .

`y' = -(sin root(3)(x))/(3x^(2/3))`

Then, express the `x^(2/3)` as radicals.

`y' = -(sinroot(3)(x))/(3 root(3)(x^2))`

**Hence, `y'= -(sinroot(3)(x))/(3 root(3)(x^2)) ` .**