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MHT CET · Maths · Differentiation

For \(x \in\left(0, \frac{1}{4}\right)\), if the derivative of \(\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)\) is \(\sqrt{x} \cdot g(x)\), then \(g(x)\) equals

  1. A \(\frac{9}{1+9 x^3}\)
  2. B \(\frac{3 x}{1-9 x^3}\)
  3. C \(\frac{3 x \sqrt{x}}{1-9 x^3}\)
  4. D \(\frac{3}{1+9 x^3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{9}{1+9 x^3}\)

Step-by-step Solution

Detailed explanation

\(y=\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)=\tan ^{-1}\left\{\frac{2\left(3 x^{\frac{3}{2}}\right)}{1-\left(3 x^{\frac{3}{2}}\right)^2}\right\}=2 \tan ^{-1}\left(3 x^{\frac{3}{2}}\right) \)
\( \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \times \frac{1}{1+9 x^3} \times 3 \times \frac{3}{2} \sqrt{x}=\frac{9}{1+9 x^3} \cdot \sqrt{x} \)
\( \Rightarrow g(x)=\frac{9}{1+9 x^3}\)