Question

If \(y = tan^{-1}[\frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}]\) , then \(\frac{dy}{dx} =\)

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

Hint:

Put x = sin(theta) so the fraction becomes tan(theta - pi/4).

Answer 1

The Correct Option is B

Step 1: Differentiate directly:
Let $u = \dfrac{x - \sqrt{1-x^2}}{x + \sqrt{1-x^2}}$. Then $\dfrac{dy}{dx} = \dfrac{u'}{1 + u^2}$.

Step 2: Compute pieces:
Let $w = \sqrt{1-x^2}$, $w' = -x/w$. $u' = \dfrac{(1 + x/w)(x + w) - (x - w)(1 - x/w)}{(x+w)^2} = \dfrac{2(w + x^2/w)}{(x+w)^2} = \dfrac{2}{w(x+w)^2}$. Also $1 + u^2 = \dfrac{2(x^2+w^2)}{(x+w)^2} = \dfrac{2}{(x+w)^2}$.

Step 3: Ratio:
$\dfrac{dy}{dx} = \dfrac{2/(w(x+w)^2)}{2/(x+w)^2} = \dfrac1w = \dfrac{1}{\sqrt{1-x^2}}$, option (B).

Final Answer:
The derivative is 1/sqrt(1 - x^2). \[ \boxed{\text{(B) }\dfrac{1}{\sqrt{1-x^2}}} \]