AP EAMCET · Maths · Differentiation
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 & Solution
Correct Answer
(C) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
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…
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