Correct Answer - Option 3 :
\(\rm 3a\over a^2 +x^2\)
Concept:
\(\rm d\over dx\)f(x)g(x) = f'(x)g(x) + f(x)g'(x)
\(\rm d\over dx\) sin-1 x = \(\rm 1\over\sqrt{1 - x^2}\)
\(\rm d\over dx\) tan-1 x = \(\rm 1\over1 + x^2\)
Calculation:
Given y = \(\rm \tan^{-1}\left[3a^2x-x^3\over a(a^2-3x^2)\right]\)
Let x = a tan θ
y = \(\rm \tan^{-1}\left[3a^2a\tanθ-(a\tanθ)^3\over a[a^2-3(atanθ)^2]\right]\)
Taking a3 common from both denominator and numerator
y = \(\rm \tan^{-1}\left[3\tanθ-\tan^3θ\over (1-3\tan^2θ)\right]\)
y = tan-1[tan 3θ]
y = 3θ = 3 tan-1(\(\rm x\over a\))
Now \(\rm {dy\over dx} = {d\over dx}[3\tan^{-1}{\left(x\over a\right)}]\)
\(\rm {dy\over dx} = {3\over 1+{x^2\over a^2}}\times{1\over a}\)
\(\rm {dy\over dx} = {3a\over a^2+x^2}\)