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If y = \(\rm \tan^{-1}\left[3a^2x-x^3\over a(a^2-3x^2)\right]\), then find out \(\rm dy\over dx\)
1. \(\rm a\over a^2 +x^2\)
2. \(\rm 2a\over a^2 +x^2\)
3. \(\rm 3a\over a^2 +x^2\)
4. \(\rm a\over a^2 - x^2\)

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Best answer
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 \(\rm {dy\over dx} = {d\over dx}[3\tan^{-1}{\left(x\over a\right)}]\)

\(\rm {dy\over dx} = {3\over 1+{x^2\over a^2}}\times{1\over a}\)

\(\rm {dy\over dx} = {3a\over a^2+x^2}\)

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