Find the derivative of the function. \[ g(x)=\frac{3}{x^{6}}+\frac{2}{x^{2}}+9 \sqrt[3]{x} \] \( g^{\prime}(x)=\square \)

To find the derivative \( g'(x) \) of the function \( g(x)=\frac{3}{x^{6}}+\frac{2}{x^{2}}+9 \sqrt[3]{x} \), we can rewrite the function in a simpler form using negative exponents: \[ g(x) = 3x^{-6} + 2x^{-2} + 9x^{\frac{1}{3}}. \] Now, applying the power rule (where the derivative of \( x^n \) is \( n x^{n-1} \)), we differentiate each term: 1. For \( 3x^{-6} \): \[ \frac{d}{dx}(3x^{-6}) = 3 \cdot (-6)x^{-6-1} = -18x^{-7}. \] 2. For \( 2x^{-2} \): \[ \frac{d}{dx}(2x^{-2}) = 2 \cdot (-2)x^{-2-1} = -4x^{-3}. \] 3. For \( 9x^{\frac{1}{3}} \): \[ \frac{d}{dx}(9x^{\frac{1}{3}}) = 9 \cdot \frac{1}{3} x^{\frac{1}{3}-1} = 3x^{-\frac{2}{3}}. \] Now combine these results: \[ g'(x) = -18x^{-7} - 4x^{-3} + 3x^{-\frac{2}{3}}. \] Thus, the derivative is: \[ g'(x) = -\frac{18}{x^7} - \frac{4}{x^3} + \frac{3}{x^{\frac{2}{3}}} \] So finally, \( g^{\prime}(x)= -\frac{18}{x^{7}} - \frac{4}{x^{3}} + \frac{3}{x^{\frac{2}{3}}} \).