Question

The derivative of ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ w.r.t. ${\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)$ is

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Suppose we substitute $x$ to be $\tan t$ in ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, we get ${\cos }^{-1}\left(\cos 2t\right)=2t=2{\tan }^{-1}x$

Similarly, when substituted $x=\tan t$ in ${\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right)$, we get ${\cot }^{-1}\left(\frac{1-3{\tan }^{2}t}{3\tan t-{\tan }^{3}t}\right)={\cot }^{-1}\left(\cot 3t\right)=3t=3{\tan }^{-1}x$

Now, derivative of $2{\tan }^{-1}x$ w.r.t. $3{\tan }^{-1}x$ is $\frac{2}{3}$