Question

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

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

Answer 1

Answer: (C)

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