Limit Continuity And Differentiability Ques 2
- 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
(2017 Main)
(a) $\frac{9}{1+9 x^3}$
(b) $\frac{3 x \sqrt{x}}{1-9 x^3}$
(c) $\frac{3 x}{1-9 x^3}$
(D) $\frac{3 }{1+9 x^3}$
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Answer:
Correct Answer: 2.(a)
Solution: (a) Let $y=\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)=\tan ^{-1}\left[\frac{2 \cdot\left(3 x^{3 / 2}\right)}{1-\left(3 x^{3 / 2}\right)^2}\right]$
$=2 \tan ^{-1}\left(3 x^{3 / 2}\right) \quad \left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right] $
$\therefore \quad \frac{d y}{d x} =2 \cdot \frac{1}{1+\left(3 x^{3 / 2}\right)^2}+3 \times \frac{3}{2}(x)^{1 / 2}=\frac{9}{1+9 x^3} \cdot \sqrt{x} $
$\therefore \quad g(x) =\frac{9}{1+9 x^3}$
BETA
