MCQ
यदि $y=\sec \left(\tan ^{-1} x\right)$ हो, तो $\frac{d y}{d x}=$ $x y \quad-x$
  • $\frac{x y}{\sqrt{1+x^2}}$
  • B
    $\frac{-x}{\sqrt{1+x^2}}$
  • C
    $\frac{x}{\sqrt{1-x^2}}$
  • D
    कोई नहीं

Answer

Correct option: A.
$\frac{x y}{\sqrt{1+x^2}}$
(A)

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