If y=^{-1}≤ft({x²+1}{{x^{4}+3x²-1}})(x>0), then dy/dx=

Correct Answer is : x2−1x4+3x2+1{x²-1}{x^{4}+3x²+1}x4+3x2+1x2−1

Correct Answer is : x2−1x4+3x2+1{x²-1}{x^{4}+3x²+1}x4+3x2+1x2−1

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NIMCET 2021 Question Paper with solutions

Question : 37 of 120

Marks: +1, -0

y=\sin^{-1}\left(\frac{x^{2}+1}{\sqrt{x^{4}+3x^{2}-1}}\right)(x>0)

\frac{dy}{dx}=

Solution:

y=\sin^{-1}\left(\frac{x^{2}+1}{\sqrt{x^{4}+3x^{2}-1}}\right)

\frac{dy}{dx}=\frac{1}{\sqrt{1-\left(\frac{x^{2}+1}{\sqrt{x^{4}+3x^{2}+1}}\right)^{2}}}\times

\frac{2x\sqrt{x^{4}+3x^{2}+1}}{\left(\sqrt{x^{4}+3x^{2}+1}\right)^{2}}

\times \frac{(x^{2}+1)(4x^{3}+6x)}{2\sqrt{x^{4}+3x^{2}+1}}

By chain rule)

\Rightarrow \frac{\sqrt{x^{4}+3x^{2}+1}}{\sqrt{(x^{4}+3x^{2}+1)-(x^{2}+1)^{2}}}

\times \frac{x\left[2(x^{4}+3x^{2}+1)-(x^{2}+1)(2x^{2}+3)\right]}{(x^{4}+3x^{2}+1)\left(\sqrt{x^{4}+3x^{2}+1}\right)}

\Rightarrow \frac{1}{\sqrt{x^{4}+3x^{2}+1-(x^{4}+3x^{2}+1)}}

\times \frac{x\left[2x^{4}+6x^{2}+2-2x^{4}-3x^{2}-2x^{2}-3\right]}{(x^{4}+3x^{2}-1)}

\Rightarrow \frac{1}{\sqrt{x^{2}}}\times \frac{x\left[x^{2}-1\right]}{x^{4}+3x^{2}+1}

\Rightarrow \frac{x^{2}-1}{x^{4}+3x^{2}+1}

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