વિધેયનું વિકલન શોધો: $\frac{d}{dx} \{(\sin x)^{\log x}\}$

  • A
    $(\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \cot x \right]$
  • B
    $(\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \cot x \log x \right]$
  • C
    $(\sin x)^{\log x} \left[ \frac{1}{x} \log \sin x + \log x \right]$
  • D
    આમાંથી કોઈ નહીં

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Similar Questions

વિધાન $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
કારણ $(R)$: $\frac{d}{d x}\left(\frac{u v}{w}\right)=\frac{u v}{w}\left[\frac{u^{\prime}}{u}+\frac{v^{\prime}}{v}-\frac{w^{\prime}}{w}\right]$

જો $y = \frac{e^{2x} \cos x}{x \sin x}$ હોય,તો $\frac{dy}{dx} = $

જો $y = \frac{2(x - \sin x)^{3/2}}{\sqrt{x}}$ હોય,તો $\frac{dy}{dx} = $

જો $f(x) = \frac{e^{-x} \sin x}{\log_e x}$ અને $f'(x) = f(x) \cdot g(x)$ હોય, તો $g'(e) =$

જો $y = 2x^{3x}$ હોય,તો $x = 1$ આગળ $\frac{dy}{dx}$ ની કિંમત શોધો.

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