Question

If $\cos (f(x))=\frac{1-x^{2}}{1+x^{2}}$ and $\tan (g(x))=\frac{3 x-x^{3}}{1-3 x^{2}}$ then $\frac{d f}{d g}=$

• A. $\frac{3}{2}$
• B. $\frac{1+x^{2}+2 x^{3}}{\left(1-3 x^{2}\right)^{2}}$
• C. $\frac{2}{3}$
• D. $\frac{x^{2}+x^{3}}{\left(1+x^{2}\right)\left(1-3 x^{2}\right)}$

Answer 1

Answer: (C)

Given, $ \cos (f(x))=\frac{1-x^{2}}{1+x^{2}} $
Put $ x=\tan\, \theta $
$\cos (f(x))=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} $
$\cos (f(x))=\cos 2 \theta $
$f(x)=2\, \theta=2 \tan ^{-1} \,x$
$\tan (g(x))=\frac{3 x-x^{3}}{1-3 x^{2}}$
Put $ x=\tan \,\theta $
$ \tan (g(x)) =\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta} $
$ \tan (g(x)) =\tan 3 \theta $
$\therefore \, g(x) =3 \theta=3 \tan ^{-1} x $
$\therefore \, \frac{f'(x)}{g'(x)} =\frac{2}{3} $
$\therefore \, \frac{d f}{d g} =\frac{2}{3}$