Question

If $y=ta{n}^{-1}\left(\frac{x-a}{1+xa}\right)$ then $\frac{dy}{dx}=$

• A. $\frac{1}{\sqrt{1+x^2}}$
• B. $\frac{1}{1+x^2}$
• C. $\frac{1}{\sqrt{1-x^2}}$
• D. $\frac{1}{1-x^2}$

Answer 1

Answer: (B)

We have,
$y=ta{n}^{-1}\left(\frac{x-a}{1+ax}\right)$
$\Rightarrow y=ta{n}^{-1}x-ta{n}^{-1}a$
Differentiating w.r.t. x, we get
$\frac{dy}{dx}=\frac{1}{1+x^2}-0$
$=\frac{1}{1+x^2}$