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For x ∈ ( 0, 1/4), if the derivative of tan^-1 ( 6x √x/ 1- 9x^3) is √x g (x), then g(x) equals

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For x ∈ ( 0, 1/4), if the derivative of tan-1 ( 6x √x/ 1- 9x3) is √x g (x), then g(x) equals

(a) 9/(1+9x3 )

(b) 3x√x/(1-9x3 ) 

(c) 3x/(1-9x3) 

(d) 3/(1+9x3)

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The correct option is (a) 9/(1+9x3 )

Explanation:

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