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 \(\frac{d f}{d g}=\)
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 \(\frac{d f}{d g}=\)
- \(\frac{3}{2}\)
- \(\frac{1+x^2+2 x^3}{\left(1-3 x^2\right)^2}\)
- \(\frac{2}{3}\)
- \(\frac{x^2+x^3}{\left(1+x^2\right)\left(1-3 x^2\right)}\)
Solution
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).
Asked in: AP EAMCET 2019 (23 Apr Shift 1)
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