Method of Differentiation
Inverse Trigonometric Functions
Premium Question
Grade 12
Question:
If $y = \tan^{-1}\left(\frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}\right)$, then $\frac{dy}{dx}$ equals:
(1) $\frac{-x}{\sqrt{1 - x^4}}$
(2) $\frac{x}{\sqrt{1 - x^4}}$
(3) $\frac{-1}{\sqrt{1 - x^4}}$
(4) $\frac{1}{2\sqrt{1 - x^4}}$
Step-by-Step Solution
Key Concept: Substitute $x^2 = \cos \phi$; the expression inside the inverse tangent then simplifies to $\tan\left(\frac{\pi}{4} + \frac{\phi}{2}\right)$, giving $y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x^2)$.
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Correct Answer: (1)