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If $y=\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$, then $\dfrac{dy}{dx}$ is:
A$\frac{1}{2\sqrt{1-x^2}}$
B$\frac{1}{\sqrt{1-x^2}}$
C$\frac{1}{1+x^2}$
D$\frac{-1}{2\sqrt{1-x^2}}$
Answer & Solution
Correct answer: A. $\frac{1}{2\sqrt{1-x^2}}$
1. Rationalise the bracket by multiplying top and bottom by $\sqrt{1+x}+\sqrt{1-x}$... or set $x=\cos 2\theta$.
2. Rationalising gives $\dfrac{1-\sqrt{1-x^2}}{x}$, so $y=\tan^{-1}\!\left(\dfrac{1-\sqrt{1-x^2}}{x}\right)$.
3. With the substitution $x=\cos 2\theta$ the expression collapses to $y=\dfrac{\pi}{4}-\theta$.
4. Since $\theta=\tfrac12\cos^{-1}x$, differentiating gives $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{1-x^2}}$.
5. The factor $\tfrac12$ is what the half-angle substitution contributes; dropping it leaves $\dfrac{1}{\sqrt{1-x^2}}$.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 10: Differentiation_
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