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 $\frac{d f}{d g}=$

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 $\frac{d f}{d g}=$, $\frac{3}{2}$, $\frac{1+x^2+2 x^3}{\left(1-3 x^2\right)^2}$, $\frac{2}{3}$, $\frac{x^2+x^3}{\left(1+x^2\right)\left(1-3 x^2\right)}$

MARKS App · Accepted Answer

Given, $\cos (f(x))=\frac{1-x^2}{1+x^2}$ Put $\quad x=\tan \theta$ $\begin{aligned} \cos (f(x)) & =\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} \ \cos (f(x)) & =\cos 2 \theta \ f(x) & =2 \theta=2 \tan ^{-1} x \ \tan (g(x)) & =\frac{3 x-x^3}{1-3 x^2} \end{aligned}$ Put $\quad x=\tan \theta$ $\begin{aligned} & \tan (g(x))=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta} \ & \tan (g(x))=\tan 3 \theta \ & \therefore \quad g(x)=3 \theta=3 \tan ^{-1} x \ & \therefore \quad \frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{2}{3} \ & \therefore \quad \frac{d f}{d g}=\frac{2}{3} \end{aligned}$ Hence, answer is (c).

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