Question

If $y={\tan }^{-1}\left(\frac{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+a^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{1-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{a}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)$, then $\frac{dy}{dx}=$

• A. $\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(1+x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}$
• B. $\frac{a}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(1+x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}$
• C. $-\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(1+x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}$
• D. $-\frac{a}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(1+x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}$

Answer 1

The correct option is A

$\frac{1}{3x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(1+x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}$

Explanation for the correct option.

Step 1. Simplify the given equation.

Using the identity ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$ the equation $y={\tan }^{-1}\left(\frac{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+a^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{1-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{a}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)$ can be simplified as:

$$\begin{gathered} y={\tan }^{-1}\left(\frac{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+a^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{1-x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{a}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right) \\ \Rightarrow y={\tan }^{-1}\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)+{\tan }^{-1}\left(a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right) \end{gathered}$$

Step 2. Find the value of $\frac{dy}{dx}$.

Differentiate both sides of the equation $y={\tan }^{-1}\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)+{\tan }^{-1}\left(a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left({\tan }^{-1}\left(x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)+{\tan }^{-1}\left(a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\right)\\ & =& \frac{1}{1+{\left(x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}^2}\frac{1}{3}{x}^{\frac{1}{3}-1}\\ & =& \frac{1}{3\left(1+x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}{x}^{-\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ & =& \frac{1}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(1+x^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right)}\end{array}$

Hence, the correct option is A.