If y=\sin ^{-1}({x^{2}+1}/{\\sqrt{x^{4}+3 x^{2}-1}})(x>0), then \frac{d y}{d x}=

Correct Answer is : x2−1x4+3x2+1

NIMCET 2021 Question Paper with solutions

Question : 37 of 120

Marks: +1, -0

y=\sin ^{-1}({x^{2}+1}/{\√{x^{4}+3 x^{2}-1}})(x>0)

, then

{d y}/{d x}=

Solution:

y=\sin ^{-1}({x^{2}+1}/{\√{x^{4}+3 x^{2}-1}})

{d y}/{d x}={1}/{\√{1-({x^{2}+1}/{\√{x^{4}+3 x^{2}+1}})^{2}}} \×

{2 x \√{x^{4}+3 x^{2}+1}}/{(\√{x^{4}+3 x^{2}+1})^{2}}

\× {(x^{2}+1)(4 x^{3}+6 x)}/{2 \√{x^{4}+3 x^{2}+1}}(

By chain rule)

\⇒{\√{x^{4}+3 x^{2}+1}}/{\√{(x^{4}+3 x^{2}+1) -(x^{2}+1)^{2}}} \

×{x[2(x^{4}+3 x^{2}+1)-(x^{2}+1)(2 x^{2}+3)}/{(x^{4}+3 x^{2}+1)(\√{x^{4}+3 x^{2}+1})}

\⇒ {1}/{\√{x^{4}+3 x^{2}+1-(x^{4}+3 x^{2}+1)}} \

×{x[2 x^{4}+6 x^{2}+2-2 x^{4}-3 x^{2}-2 x^{2}-3]}/{(x^{4}+3 x^{2}-1)}

⇒{1}/{\√{x^{2}}} \×{x[x^{2}-1]}/{x^{4}+3 x^{2}+1}

\⇒{x^{2}-1}/{x^{4}+3 x^{2}+1}

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