જો $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$ હોય,તો $\frac{dy}{dx}$ શોધો,જ્યાં $0 < x < \frac{1}{\sqrt{2}}$.

  • A
    $\frac{2}{\sqrt{1-x^2}}$
  • B
    $\frac{-2}{\sqrt{1-x^2}}$
  • C
    $\frac{1}{\sqrt{1-x^2}}$
  • D
    $\frac{-1}{\sqrt{1-x^2}}$

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Similar Questions

જો $\cos^{-1} x + \cos^{-1} y = 2\pi$ હોય,તો $\sin^{-1} x + \sin^{-1} y$ ની કિંમત શોધો.

જો $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$ હોય,તો $x + y + z$ ની કિંમત શું થાય?

$\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ ની કિંમત શોધો.

$\sin [\cot ^{ - 1}(\cos \tan ^{ - 1}x)] =$

$\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ નો ઉકેલ શોધો.

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