Question

If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \frac{-1}{\sqrt{3}}\)

(a) 3

(b) \(\frac{3}{1+x}\)

(c) –\(\frac{3}{1+x^2}\)

(d) \(\frac{3}{1+x^2}\)

I had been asked this question in an internship interview.

I would like to ask this question from Differentiability in division Continuity and Differentiability of Mathematics – Class 12

Answer 1

The correct choice is (d) \(\frac{3}{1+x^2}\)

For explanation I would say: Given function is \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2})\),

Now taking RHS and substituting x = tang in it and then we get,

y = \(tan^{-1}(\frac{3tang-tang^3}{1-3tang^2})\), Now it becomes the expansion of the function tan3g,

Hence the given function becomes y= tan^-1(tan3g). Which is equal to 3g, now substituting the value of g= tan^-1x, now after differentiating both sides we get the answer \(\frac{3}{1+x^2}\).