how to solve following up pgt mathematics previous years question

Question : Solve following up pgt mathematics previous years questionSolution :given function \[y=\tan^{-1}{\left(\frac{3a^2x-x^3}{a\left(a^2-3x^2\right)}\right)\ }\] to find derivative of y \[\frac{dy}{dx}=\frac{1}{1+\frac{\left(3a^2x-x^3\right)^2}{a^2\left(a^2-3x^2\right)^2}\ }\frac{d}{dx}\left(\frac{3a^2x-x^3}{a\left(a^2-3x^2\right)}\right)\] simplifying this we get \[=\frac{1}{1+\frac{\left(3a^2x-x^3\right)^2}{a^2\left(a^2-3x^2\right)^2}\ }\left(\frac{3a^2-3x^2}{a(a^2-3x^2)}+\frac{6x(3a^2x-x^3)}{a{(a^2-3x^2)}^2}\right)\] after simplification, we obtain \[\frac{dy}{dx}=\frac{3a}{a^2+x^2}\] is the required derivative of y. Hence correct…

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