જો $y = \tan^{-1} \sqrt{\frac{a - x}{a + x}}$ હોય,તો $\frac{dy}{dx} = $

  • A
    $-\frac{1}{2\sqrt{a^2 - x^2}}$
  • B
    $\frac{1}{2\sqrt{a^2 - x^2}}$
  • C
    $\frac{1}{2\sqrt{a^2 + x^2}}$
  • D
    આમાંથી કોઈ નહીં

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Similar Questions

જો $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$ અને $x \in \left( {0,\frac{\pi }{2}} \right)$ હોય,તો $\frac{{dy}}{{dx}}$ ની કિંમત શોધો.

ધારો કે $y=f(x)=\sin ^3\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^2+1\right)^{\frac{3}{2}}\right)\right)$. તો,$x =1$ આગળ,

ધારો કે $f : R \rightarrow R$ એક વિકલનીય વિધેય છે અને $f(1) = 4$ છે. તો $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ નું મૂલ્ય શોધો.

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

જો $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$ હોય,તો $\frac{dy}{dx} = $

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