Question

If $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\left(1-\mathrm{x}\right)}$, then the derivative of the composite function $\mathrm{f}\left[\mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}\right]$ is

• A. 0
• B. $\frac{1}{2}$
• C. 1
• D. 2

Answer 1

The correct option is C

1

Explanation for the correct option:

Step 1: Finding the value of $f\left[f\right(x\left)\right]$

Given function: $\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\left(1-\mathrm{x}\right)}$

Clearly$f\left(x\right)$is not defined at $x=1$

For $\mathrm{n}\in \mathrm{R}-\left\{1\right\}$

Put, $x=\frac{1}{x-1}$in the above function

$\begin{array}{c}\therefore f\left[f\left(x\right)\right]=\frac{1}{1-\frac{1}{1-x}}\\ =\frac{1-x}{1-x-1}\text{}\\ =\frac{1-x}{-x}\text{}\\ =\frac{x-1}{x}\end{array}$

Step 2: Find $\mathrm{f}\left[\mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}\right]$

Put, $x=\frac{x-1}{x}$in the function $f\left[f\right\{f\left(x\right)\}]$

$\begin{array}{rcl}\therefore \mathrm{f}\left[\mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}\right]& =& \mathrm{f}\left[\frac{\mathrm{x}-1}{\mathrm{x}}\right]\\ & =& \frac{1}{1-{\displaystyle \frac{\mathrm{x}-1}{\mathrm{x}}}}\\ & =& \frac{\mathrm{x}}{\mathrm{x}-\mathrm{x}+1}\\ & =& \mathrm{x}\end{array}$

Differentiate the function $\mathrm{f}\left[\mathrm{f}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}\right]$ with respect to x,

$\begin{array}{c}\therefore \frac{d}{dx}\left[f\{f\left(f\left(x\right)\right)\right]=x\left[\because \frac{d}{dx}\left(x^n\right)=n{x}^{n-1}\right]\\ =1\mathrm{for}\mathrm{all}\mathrm{n}\in \mathrm{R}-\left\{1\right\}\end{array}$

Hence, the correct option is option (c).