Limit Continuity And Differentiability Ques 106

Answer:

Correct Answer: 106.(d)

Solution:

106. Let $y=\tan ^{-1} \frac{6 x \sqrt{x}}{1-9 x^{3}}=\tan ^{-1} \frac{2 \cdot\left(3 x^{3 / 2}\right)}{1-\left(3 x^{3 / 2}\right)^{2}}$

$$ \begin{alignedat} & =2 \tan ^{-1}\left(3 x^{3 / 2}\right) \because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ \therefore \quad \frac{d y}{d x} & =2 \cdot \frac{1}{1+\left(3 x^{3 / 2}\right)^{2}} \cdot 3 \times \frac{3}{2}x^{1 / 2}=\frac{9}{1+9 x^{3}} \cdot \sqrt{x} \\ \therefore \quad g(x) & =\frac{9}{1+9 x^{3}} \end{aligned} $$