Question

Multiple Choice Let \(f(x)=x-\frac{1}{x} .\) Find \(f^{\prime \prime}(x)\) \(\begin{array}{ll}{(\mathbf{A}) 1+\frac{1}{x^{2}}} & {(\mathbf{B}) 1-\frac{1}{x^{2}} \quad(\mathbf{C}) \frac{2}{x^{3}}} \\\ {(\mathbf{D})-\frac{2}{x^{3}}} & {\text { (E) does not exist }}\end{array}\)

Short answer

The second derivative of the function is \(f^{\prime \prime}(x)=\frac{2}{x^3}\), which matches option \((C)\).

Step-by-step solution

Find the first derivative

We find the first derivative \(f^{\prime}(x)\) using direct application of the derivative rules. The derivative of \(x\) is 1 while the derivative of \(-\frac{1}{x}\) or \(x^{-1}\) is \(-x^{-2}\) or \(-\frac{1}{x^2}\). Combining these, our first derivative is \(f^{\prime}(x)=1-\frac{1}{x^2}\).

Find the second derivative

To get \(f^{\prime \prime}(x)\), we need to differentiate \(f^{\prime}(x)=1-\frac{1}{x^2}\). The derivative of \(1\) is \(0\) and the derivative of \(-\frac{1}{x^2}\) or \(-x^{-2}\) is \(2x^{-3}\) or \(\frac{2}{x^3}\). Hence, \(f^{\prime \prime}(x)=0+\frac{2}{x^3}\) or simply \(f^{\prime \prime}(x)=\frac{2}{x^3}\).

Match with provided options

Looking at the answer options, \(f^{\prime \prime}(x)=\frac{2}{x^3}\) corresponds to choice \((\mathbf{C})\). Therefore, \((\mathbf{C})\) is the correct option.