ધારો કે $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,જ્યાં $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$ છે. તો $\frac{d}{d(\tan \theta)}(f(\theta))$ નું મૂલ્ય શોધો.

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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$\sqrt{x}$ ની સાપેક્ષમાં $\tan^{-1}\sqrt{x}$ નું વિકલન સહગુણક શું છે?

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{1+6x^2} \right) \right) = $ . . . . . .

$\frac{d}{dx} \left\{ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right\} = $

જો $f(x) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$ હોય,તો $f^{\prime}(e)$ ની કિંમત શોધો.

જો $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$ હોય,તો $\frac{dy}{dx} = $

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