વિધેય $y^{x} = x^{y}$ માટે $\frac{dy}{dx}$ શોધો.

  • A
    $\frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$
  • B
    $\frac{x}{y} \left( \frac{y - x \log y}{x - y \log x} \right)$
  • C
    $\frac{y}{x} \left( \frac{x - y \log x}{y - x \log y} \right)$
  • D
    $\frac{x}{y} \left( \frac{x - y \log x}{y - x \log y} \right)$

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જો $\sqrt{x} + \sqrt{y} = \sqrt{xy}$ હોય,તો $\frac{dy}{dx} = $

જો $x+y=\tan ^{-1} y$ અને $\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}$ હોય,તો $f(y)$ ની કિંમત શોધો.

જો $y = \sqrt{(1 - x)(1 + x)}$ હોય,તો

જો $f(x) = (\cos x)(\cos 2x) \ldots (\cos nx)$ હોય, તો $f^{\prime}(x) + \sum_{r=1}^n (r \tan rx) f(x)$ ની કિંમત શોધો.

જો $y = \sqrt[3]{\tan x + y}$ હોય, તો $\frac{dy}{dx} =$ શું થાય?

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