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Find the derivative of \(\rm sin^{-1}\left ( \frac{2x}{1+x^2} \right )\) with respect to x ?
1. \(\rm \frac{1}{\left ( 1+x^2 \right )} \)
2.  \(\rm \frac{2}{\left ( 1+x^2 \right )} \)
3. \(\rm \frac{-2}{\left ( 1+x^2 \right )} \)
4. \(\rm \frac{2}{\left ( 1+x \right )} \)

1 Answer

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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.

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