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EEJ MAIN Mathematics QUESTION #6027
Question 1

For $x\in\left(0,\dfrac{1}{4}\right)$, the derivative of $\tan^{-1}\!\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)$ equals $\sqrt{x}\cdot g(x)$. Find $g(x)$.

  • $\dfrac{9}{1+9x^3}$✔️
  • $\dfrac{3x\sqrt{x}}{1-9x^3}$
  • $\dfrac{3x}{1-9x^3}$
  • $\dfrac{3}{1+9x^3}$
Correct Answer Explanation

Write $\frac{6x\sqrt{x}}{1-9x^3} = \frac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}=\frac{2t}{1-t^2}$ where $t=3x^{3/2}$.

So $\tan^{-1}\!\left(\frac{2t}{1-t^2}\right)=2\tan^{-1}(t)=2\tan^{-1}(3x^{3/2})$

Differentiate: $\frac{d}{dx}[2\tan^{-1}(3x^{3/2})] = \frac{2}{1+9x^3}\cdot 3\cdot\frac{3}{2}x^{1/2} = \frac{9\sqrt{x}}{1+9x^3}$

Since this equals $\sqrt{x}\cdot g(x)$: $g(x)=\dfrac{9}{1+9x^3}$