Question

If $y=\frac{1-x}{x^2}$, then $\frac{dy}{dx}$is

• A. $\frac{2}{x}+\frac{2}{x^3}$
• B. $-\frac{2}{x^3}+\frac{1}{x^2}$
• C. $-\frac{2}{x^3}+\frac{2}{x^2}$
• D. None of these

Answer 1

The correct option is B

$-\frac{2}{x^3}+\frac{1}{x^2}$

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$.

Separate the fraction in the right hand side of the equation $y=\frac{1-x}{x^2}$ into two.

$\begin{array}{rcl}y& =& \frac{1-x}{x^2}\\ & \Rightarrow & y=\frac{1}{x^2}-\frac{x}{x^2}\\ & \Rightarrow & y=\frac{1}{x^2}-\frac{1}{x}\end{array}$

Now differentiate $\begin{array}{rcl}y& =& \frac{1}{x^2}-\frac{1}{x}\end{array}$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\frac{1}{x^2}-\frac{1}{x}\right)\\ & =& \frac{d}{dx}\left(\frac{1}{x^2}\right)-\frac{d}{dx}\left(\frac{1}{x}\right)\\ & =& \frac{-2}{x^3}-\frac{-1}{x^2}\\ & =& -\frac{2}{x^3}+\frac{1}{x^2}\end{array}$

Hence, the correct option is B.