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

To find the derivative \( f'(x) \) of the function \[ f(x) = -\frac{3}{4x} - 7x^{-4} - \sqrt[3]{x}, \] we'll differentiate each term separately. 1. **First Term: \( -\frac{3}{4x} \)** Rewrite the term using a negative exponent: \[ -\frac{3}{4x} = -\frac{3}{4} x^{-1}. \] Differentiate: \[ \frac{d}{dx} \left( -\frac{3}{4} x^{-1} \right) = -\frac{3}{4} \cdot (-1) x^{-2} = \frac{3}{4} x^{-2}. \] 2. **Second Term: \( -7x^{-4} \)** Differentiate: \[ \frac{d}{dx} \left( -7x^{-4} \right) = -7 \cdot (-4) x^{-5} = 28 x^{-5}. \] 3. **Third Term: \( -\sqrt[3]{x} \)** Rewrite the term using a fractional exponent: \[ -\sqrt[3]{x} = -x^{1/3}. \] Differentiate: \[ \frac{d}{dx} \left( -x^{1/3} \right) = -\frac{1}{3} x^{-2/3}. \] Combining all the derivatives, we get: \[ f'(x) = \frac{3}{4} x^{-2} + 28 x^{-5} - \frac{1}{3} x^{-2/3}. \] Alternatively, this can be written using positive exponents in the denominators: \[ f'(x) = \frac{3}{4x^2} + \frac{28}{x^5} - \frac{1}{3x^{2/3}}. \] **Final Answer:** \[ f\,'(x) = \frac{3}{4\,x^{2}} + \frac{28}{x^{5}} - \frac{1}{3\,x^{\,2/3}} \]