Correct Answer - Option 2 :
\(\rm \frac{2}{\left ( 1+x^2 \right )} \)
Concept:
Derivative of sin-1 x with respect to x is \(\frac{1}{\sqrt{1-x^2}}\)
Quotient rule:
\(\rm \left ( \frac{u}{v} \right )'= \frac{vu'-uv'}{v^2}\)
Calculation:
Let \(\rm y=sin^{-1}\left ( \frac{2x}{1+x^2} \right )\)
We differentiate the function y with respect to x
As we know that, the derivative of sin-1 x with respect to x is \(\frac{1}{\sqrt{1-x^2}}\)
⇒ \(\rm y' = \frac{1}{\sqrt{1- \left ( \frac{2x}{1+x^2} \right )^2 }}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{2x}{1+x^2} \right )\)
As we know that,
\(\rm \left ( \frac{u}{v} \right )'= \frac{vu'-uv'}{v^2}\)
⇒ \(\rm y' = \frac{1+x^2}{\sqrt{(1+x^2)^2- 4x^2 }}\times \frac{(1+x^2)2-2x(0+2x)}{\left ( 1+x^2 \right )^2} \)
⇒ \(\rm y' = \frac{1+x^2}{\sqrt{1+x^4+2x^2- 4x^2 }}\times \frac{(2+2x^2-4x^2)}{\left ( 1+x^2 \right )^2}\)
⇒ \(\rm y' = \frac{1+x^2}{\sqrt{1+x^4- 2x^2 }}\times \frac{(2-2x^2)}{\left ( 1+x^2 \right )^2} \)
⇒ \(\rm y' = \frac{1+x^2}{\sqrt{(1- x^2)^2 }}\times \frac{2(1-x^2)}{\left ( 1+x^2 \right )^2} \)
⇒ \(\rm y' = \frac{1+x^2}{{(1- x^2) }}\times \frac{2(1-x^2)}{\left ( 1+x^2 \right )^2} \)
⇒ \(\rm y' = \frac{2}{\left ( 1+x^2 \right )} \)
Hence, option 2 is correct.