COMEDK · Maths · 23. Inverse Trigonometric Functions
If \(y = \tan^{-1}\left(\dfrac{\sqrt{1+x^3} + \sqrt{1-x^3}}{\sqrt{1+x^3} - \sqrt{1-x^3}}\right)\) then \(\dfrac{dy}{dx} =\)
- A \(\dfrac{6x^2}{\sqrt{1-x^6}}\)
- B \(-\dfrac{6x^2}{\sqrt{1-x^6}}\)
- C \(\dfrac{3x^2}{\sqrt{1-x^6}}\)
- D \(-\dfrac{3x^2}{2\sqrt{1-x^6}}\)
Answer & Solution
Correct Answer
(D) \(-\dfrac{3x^2}{2\sqrt{1-x^6}}\)
Step-by-step Solution
Detailed explanation
Substitute \(x^3 = \cos 2\theta\). Then \(\sqrt{1+x^3} = \sqrt{1+\cos 2\theta} = \sqrt{2}\cos\theta\) And \(\sqrt{1-x^3} = \sqrt{1-\cos 2\theta} = \sqrt{2}\sin\theta\)…
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