find the derivative of the function.\ny = \\frac{5}{x^{3}} - \\frac{4}{x}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Rewrite the function

Rewrite $y = \frac{5}{x^{3}}-\frac{4}{x}$ as $y = 5x^{- 3}-4x^{-1}$.

Step2: Apply power - rule for derivatives

The power - rule states that if $y = ax^{n}$, then $\frac{dy}{dx}=anx^{n - 1}$. For the first term $y_1 = 5x^{-3}$, $\frac{dy_1}{dx}=5\times(-3)x^{-3 - 1}=-15x^{-4}$. For the second term $y_2=-4x^{-1}$, $\frac{dy_2}{dx}=-4\times(-1)x^{-1 - 1}=4x^{-2}$.

Step3: Find the derivative of the whole function

$\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}=-15x^{-4}+4x^{-2}$.

Answer:

$- \frac{15}{x^{4}}+\frac{4}{x^{2}}$