Find \( f^{\prime}(x) \) for \( f(x)=-\frac{3}{4 x}-7 x^{-4}-\sqrt[3]{x} \)

To find \( f^{\prime}(x) \) for the function \( f(x) = -\frac{3}{4 x} - 7 x^{-4} - \sqrt[3]{x} \), we will differentiate each term separately. 1. The derivative of \( -\frac{3}{4 x} \) can be rewritten as \( -\frac{3}{4} x^{-1} \). Differentiating gives: \[ \frac{d}{dx} \left( -\frac{3}{4} x^{-1} \right) = -\frac{3}{4} \cdot (-1) x^{-2} = \frac{3}{4 x^2} \] 2. For the term \( -7 x^{-4} \), the derivative is: \[ \frac{d}{dx} \left( -7 x^{-4} \right) = -7 \cdot (-4) x^{-5} = 28 x^{-5} \] 3. The derivative of \( -\sqrt[3]{x} \) (which is \( -x^{1/3} \)) is: \[ \frac{d}{dx} \left( -x^{1/3} \right) = -\frac{1}{3} x^{-2/3} \] Now we can combine these results to find \( f^{\prime}(x) \): \[ f^{\prime}(x) = \frac{3}{4 x^2} + 28 x^{-5} - \frac{1}{3} x^{-2/3} \] Thus, the final result is: \[ f^{\prime}(x) = \frac{3}{4 x^2} + \frac{28}{x^5} - \frac{1}{3} x^{-2/3} \]