यदि $y=x^{\log x}+(\log x)^x, x>1$ है, तो $\left(\frac{d y}{d x}\right)_{x=e}=$

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $3$

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Similar Questions

यदि $y = \frac{x^2}{(x - 1)(x - 2)(x - 3)} + \frac{2x}{(x - 2)(x - 3)} + \frac{3}{x - 3} + 1$ है,तो $\frac{xy'}{y}$ का मान क्या होगा? (जहाँ $y' = \frac{dy}{dx}$)

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

यदि $y = ((x+1)(4x+1)(9x+1) \ldots (n^2x+1))^2$ है,तो $x=0$ पर $\frac{dy}{dx}$ का मान ज्ञात कीजिए।

यदि $h(x) = x^{x^x}$ है, तो $x = 1$ पर $\frac{h^{\prime}(x)}{h(x)}$ का मान क्या होगा?

यदि $f(x) = x^{\operatorname{Sec}^{-1} x}$ है,तो $f^{\prime}(2) =$

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