Mathematics - Differentiation Question with Solution | TestHub

Let  $3x^{\frac{3}{2}}=y$, then $y\in \left(0,\frac{3}{8}\right)$

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

$\Rightarrow 2{\tan }^{-1}y=2{\tan }^{-1}\left(3x^{\frac{3}{2}}\right)$

Now $\Rightarrow \frac{d}{dx}2{\tan }^{-1}\left(3x^{\frac{3}{2}}\right)=\frac{2}{1+9x^3}\cdot 3\cdot \frac{3}{2}\cdot x^{\frac{1}{2}}$

$\Rightarrow \frac{d}{dx}2{\tan }^{-1}\left(3x^{\frac{3}{2}}\right)=\frac{9}{1+9x^3} \cdot \sqrt{x}$

$\Rightarrow \sqrt{x}.g\left(x\right)=\frac{9}{1+9x^3} \cdot \sqrt{x}$

$\Rightarrow g\left(x\right)=\frac{9}{1+9x^3}$