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# The Derivative

The Value of g(x)

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## The derivative of (tan-1(6x√x)) / (1-9x^3) is √xg(x) then what is the value of g(x).

Answer:  We need to prove tan-1((6x√x) / (1-9x^3)) = √xg(x)

Now, let us consider y = tan-1((6x√x) / (1-9x^3))

Now dy/ dt = 1 / ( 1+ (36 x^3)/ (1- 9 x^ 3) ^ 2) *[ (1-9 x ^ 3) * (6 * (3/2) x ^ (1/2)) + 6 x^ (3/2) * 27 x^2] / (1 – 9 x^3)^ 2        [Where if y = tan-1 x, then dy/ dx = 1/ (1 + x^2) and d(u/v)/dx = (v du/ dx- u dv/ dx)/ v^2

Hence, dy/ dx = (((1- 9 x^ 3) ^ 2)/ (((1- 9 x^ 3) ^ 2) + 36 x^3)) * [(1- 9 x^ 3) (9 √x) + (6 x √x (27x^2)) / (1- 9 x^ 3) ^ 2)]

Now, dy/ dx = (1- 9 x^ 3) (9 √x) + (6 x √x (27x^2)) / (((1- 9 x^ 3) ^ 2) + 36 x^3)

Now, dy/ dx = √x [{((1- 9 x^ 3) 9) + (6 x √x (27x^2))}/ (((1- 9 x^ 3) ^ 2) + 36 x^3)]

Now, dy/ dx = √x g (x)

Hence, g (x) = [{((1- 9 x^ 3) 9) + (6 x √x (27x^2))}/ (((1- 9 x^ 3) ^ 2) + 36 x^3)]

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