JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If for \(x \in \left( {0,\frac{1}{4}} \right)\) , the derivative of \({\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)\) is \(\sqrt x \cdot g\left( x \right)\) then \(g\left( x \right)\) equals:
- A \(\frac{3}{{1 + 9{x^3}}}\)
- B \(\frac{9}{{1 + 9{x^3}}}\)
- C \(\frac{{3x\sqrt x }}{{1 - 9{x^3}}}\)
- D \(\;\frac{3}{{1 - 9{x^3}}}\)
Answer & Solution
Correct Answer
(B) \(\frac{9}{{1 + 9{x^3}}}\)
Step-by-step Solution
Detailed explanation
Let \(F\left( x \right) = {\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)\) where \(x \in \left( {0,\frac{1}{4}} \right)\).…
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