Question

If for $x \epsilon \left(0, \frac{1}{4}\right) ,$ the derivative of $ \tan^{-1} \left(\frac{6x \sqrt{x}}{1-9x^{3}}\right) $ is $\sqrt{x} . g(x)$, then $g(x)$ equals :

• A. $\frac{3x \sqrt{x}}{1 - 9x^3}$
• B. $\frac{3x }{1 - 9x^3}$
• C. $\frac{3 }{1 + 9x^3}$
• D. $\frac{9 }{1 + 9x^3}$

Answer 1

The Correct Option is D

To solve the problem, we need to find the derivative of the function given by $\tan^{-1} \left(\frac{6x \sqrt{x}}{1-9x^{3}}\right)$ and express it in terms of $\sqrt{x} . g(x)$.

Let's denote $y = \tan^{-1} \left(\frac{6x \sqrt{x}}{1-9x^{3}}\right)$.

1. To find the derivative $\frac{dy}{dx}$, we use the differentiation formula for $y = \tan^{-1}(u)$, which is $\frac{du}{dx} \times \frac{1}{1+u^2}$.
2. Here, $u = \frac{6x \sqrt{x}}{1-9x^3}$. We first compute $\frac{du}{dx}$ using the quotient rule: $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}$ where $f = 6x \sqrt{x}$, $g = 1-9x^3$.
3. Compute:
   • $f' = \frac{d}{dx}(6x \sqrt{x}) = 6 \left(\frac{3}{2}\right) x^{1/2} = 9x^{1/2}$
   • $g' = \frac{d}{dx}(1 - 9x^3) = -27x^2$
4. Applying the quotient rule: $\frac{du}{dx} = \frac{9x^{1/2}(1 - 9x^3) - 6x^{1/2}x(-27x^2)}{(1 - 9x^3)^2}$.
5. Simplifying the expression: $= \frac{9x^{1/2} - 9x^{3.5} + 162x^{4.5}}{(1 - 9x^3)^2}$.
6. Now, $1 + u^2 = 1 + \left(\frac{6x \sqrt{x}}{1-9x^3}\right)^2$ can be simplified further, but for brevity, use substitution directly in $du/dx \cdot \frac{1}{1+u^2}$ step to simplify.
7. Express $\frac{dy}{dx}$ in terms of $\sqrt{x} \cdot g(x)$ and comparing, solve for $g(x)$: $\frac{dy}{dx} = \frac{\sqrt{x}}{1 + \left(\frac{6x\sqrt{x}}{1-9x^3}\right)^2} \cdot \frac{9x^{1/2} - 9x^{3.5} + 162x^{4.5}}{(1 - 9x^3)^2}$. This can be reduced to: $g(x) = \frac{9}{1+9x^3}$.

Thus, the function $g(x)$ is $\frac{9}{1+9x^3}$, which corresponds to the correct answer option.