Question

Let $f\left(x\right)={\tan }^{-1}x,$ then $f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)$ is equal to $0$, where $x$is equal to

• A. $0$
• B. $1$
• C. $i$
• D. $-i$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Calculating the value of $x$:

Step 1: Differentiating $f\left(x\right)$ with respect to $x$

Given,

$f\left(x\right)={\tan }^{-1}x$,

$f’\left(x\right)+f’’\left(x\right)=0...\left(1\right)$

Differentiating $f\left(x\right)$ we get,

$f\text{'}\left(x\right)=\frac{1}{1+x^2}$

Again differentiating $f\text{'}\left(x\right)$ with respect to $x$

$f\text{'}\text{'}\left(x\right)=-\frac{1}{{\left(1+x^2\right)}^2}\times \left(2x\right)$

Step 2: Calculating $f’\left(x\right)+f’’\left(x\right)$

$$\begin{gathered} f\text{'}\left(x\right)+f\text{'}\text{'}\left(x\right)=\frac{1}{\left(1+x^2\right)}-\frac{2x}{{\left(1+x^2\right)}^2} \\ =\frac{1+x^2-2x}{{\left(1+x^2\right)}^2} \\ =\frac{{\left(1-x\right)}^2}{{\left(1+x^2\right)}^2}.......\left(2\right) \end{gathered}$$

Step 3: Equating equations $\left(1\right)\&\left(2\right)$

$\begin{array}{rcl}\frac{{\left(1-x\right)}^2}{{\left(1+x^2\right)}^2}& =& 0\\ 1-x& =& 0\\ x& =& 1\end{array}$

Hence, option (B) is the correct answer