Question

differentiate:
\frac{1}{x}\frac{1}{x^{3}}=

Explanation:

Step1: Rewrite the function

Rewrite \(\frac{1}{x}-\frac{1}{x^{3}}\) as \(x^{- 1}-x^{-3}\).

Step2: Apply the power rule

The power rule for differentiation is \(\frac{d}{dx}(x^{n})=nx^{n - 1}\).
For \(y = x^{-1}\), \(\frac{dy}{dx}=-1\times x^{-1 - 1}=-x^{-2}\).
For \(y=-x^{-3}\), \(\frac{dy}{dx}=-(-3)x^{-3 - 1}=3x^{-4}\).

Step3: Combine the results

The derivative of \(y=x^{-1}-x^{-3}\) is \(\frac{dy}{dx}=-x^{-2}+3x^{-4}\).
Rewrite it as \(\frac{-1}{x^{2}}+\frac{3}{x^{4}}\).

Answer:

\(\frac{-1}{x^{2}}+\frac{3}{x^{4}}\)