The derivative of sin⁻¹ (2x/1+x²) with respect to cos-1 (1−x²/1+x²) is

The correct option is 1.

Let us find the derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ using substitution.

Let $u={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ and $v={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.

We want to find the derivative of $u$ with respect to $v$, which is given by $\frac{du}{dv}=\frac{du/dx}{dv/dx}$.

To simplify both expressions, let us substitute $x=\tan \theta$, where $\theta ={\tan }^{-1}\left(x\right)$.

First, simplifying $u$:
$u={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$

Using the trigonometric identity $\sin (2\theta )=\frac{2\tan \theta }{1+{\tan }^2\theta }$, we get:
$u={\sin }^{-1}\left(\sin (2\theta )\right)=2\theta$

Substituting back $\theta ={\tan }^{-1}\left(x\right)$, we have:
$u=2{\tan }^{-1}\left(x\right)$

Differentiating $u$ with respect to $x$:
$\frac{du}{dx}=\frac{2}{1+x^2}$

Second, simplifying $v$:
$v={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$

Using the trigonometric identity $\cos (2\theta )=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$, we get:
$v={\cos }^{-1}\left(\cos (2\theta )\right)=2\theta$

Substituting back $\theta ={\tan }^{-1}\left(x\right)$, we have:
$v=2{\tan }^{-1}\left(x\right)$

Differentiating $v$ with respect to $x$:
$\frac{dv}{dx}=\frac{2}{1+x^2}$

Finally, we calculate $\frac{du}{dv}$:
$\frac{du}{dv}=\frac{du/dx}{dv/dx}=\frac{\left({\displaystyle \frac{2}{1+x^2}}\right)}{\left({\displaystyle \frac{2}{1+x^2}}\right)}=1$

Thus, the derivative is equal to 1.