Question:
Differentiate:
\[
\frac{\tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right)}{x} \quad \text{w.r.t.} \quad \cos^{-1}(2x\sqrt{1 - x^2}), \quad x \in \left(\frac{1}{\sqrt{2}}, 1\right)
\]
Differentiate:
\[
\frac{\tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right)}{x} \quad \text{w.r.t.} \quad \cos^{-1}(2x\sqrt{1 - x^2}), \quad x \in \left(\frac{1}{\sqrt{2}}, 1\right)
\]
Show Hint
In problems involving inverse trigonometric functions, use substitution to simplify the differentiation, and don't forget to apply the product and chain rules.
Updated On: Mar 8, 2026
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Solution and Explanation
Let \( y = \frac{\tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right)}{x} \)
We need to differentiate this with respect to \( x \), and the given function involves both inverse trigonometric and trigonometric functions.
Using standard differentiation rules for inverse functions and product rule, the solution becomes:
\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\tan^{-1}\left(\frac{\sqrt{1 - x^2}}{x}\right)}{x} \right)
\]
The detailed differentiation involves simplifying the terms and applying the chain rule for the inverse trigonometric part, yielding the final derivative expression.
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