MathematicsIf y=sec⁡Tan−1⁡x, then dydx at x=1 is 1k then k=

If $y = \sec ({Tan}^{- 1} x), then \frac{dy}{dx} at x = 1 is$ $\frac{1}{\sqrt{k}} t h e n k =$

Solution:

$y = \sec ({Tan}^{- 1} x) = \sqrt{\sec^{2} {Tan}^{- 1} x} = \sqrt{1 + {Tan}^{2} {Tan}^{- 1} x} = \sqrt{1 + x^{2}}$

$\frac{dy}{dx} = \frac{2 x}{2 \sqrt{1 + x^{2}}} = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow {(\frac{dy}{dx})}_{x = 1} = \frac{1}{\sqrt{2}}$

Related content

Area of Isosceles Triangle

Pythagoras Theorem

Triangle Formulae

Perimeter of Triangle Formula

Volume of Cone Formula

Matrices and Determinants_mathematics

Critical Points Solved Examples

Type of relations_mathematics