Question

If for x ∈ $\left(0,\frac{1}{4}\right)$, the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^{3}}\right)$ is √x.g(x), then g(x) equals:
(a) $\frac{3}{1+9x^{3}}$
(b) $\frac{9}{1+9x^{3}}$
(c) $\frac{3x\sqrt{3}}{1-9x^{3}}$
(d) $\frac{3x}{1-9x^{3}}$

Answer 1

Hello mate ☺

Option C is correct one

Hope it helps u...❤

Answer 2

If for x ∈ $\left(0,\frac{1}{4}\right)$, the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^{3}}\right)$ is √x.g(x), then g(x) equals:

(a) $\frac{3}{1+9x^{3}}$

(b) $\frac{9}{1+9x^{3}}$

(c) $\frac{3x\sqrt{3}}{1-9x^{3}}$✔

(d) $\frac{3x}{1-9x^{3}}$