Question

Finding $f^{\prime }\left(x\right)$ for $f\left(x\right)=x+\frac{1}{x}$

• A. $f^{\prime }\left(x\right)=1-\frac{1}{x^2}$
  
• B. $f^{\prime }\left(x\right)=\frac{x^2-1}{x^2}$
• C. $f^{\prime }\left(x\right)=\frac{x^2+1}{x^2}$
• D. $\text{Both A and B}$

Answer 1

The correct option is D $\text{Both A and B}$

Given,

$f\left(x\right)=x+\frac{1}{x}$

Now differentiating both sides with respect to $x$ we get,

$f^{\prime }\left(x\right)=1-\frac{1}{x^2}$

or, $f^{\prime }\left(x\right)=\frac{x^2-1}{x^2}$.