Question bank/JEE Main/Mathematics

If y = sec$\left( \tan^{- 1}x \right)$, and value
of $\frac{dy}{dx}\$at x = 1 is equal to
$\frac{\lambda}{3\sqrt{2}}$ then $\lambda$ is equal to

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Let tan^-1$x\ = \theta$ $\Rightarrow x = tan\ \theta\$
$\Rightarrow \ y = sec\ \theta\$ $\Rightarrow$ $y = \sqrt{1 + x^{2}}$
$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^{2}}} \cdot 2x\$ at  x = 1
$\Rightarrow$ $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx