5. continuity and differentiation — ગણિત ધોરણ 12 સાયન્સ — Question

Put $x = a\tan \theta $

$ \Rightarrow \frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^3}\tan \theta - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right]$

$ = \frac{d}{{dx}}{\tan ^{ - 1}}(\tan 3\theta ) = \frac{d}{{dx}}(3\theta ) = \frac{{3a}}{{{x^2} + {a^2}}}$

If $x = 0,$  then  $\frac{d}{{dx}}{\tan ^{ - 1}}\left[ {\frac{{3{a^2}x - {x^3}}}{{a({a^2} - 3{x^2})}}} \right] = \frac{3}{a}$.