Question

y=sec(tan^-1x) then dy/dx at x=2​

Answer 1

Step-by-step explanation:

We have,

$y = \sec( \tan^{ - 1} (x) )$

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

$\implies \: \rm \: y = \sqrt{1 + {x}^{2} }$

$\implies \: \rm \: \frac{dy}{dx} = \frac{2x}{2\sqrt{1 + {x}^{2} } }$

$\implies \: \rm \: \frac{dy}{dx} = \frac{x}{\sqrt{1 + {x}^{2} } }$

$\implies \: \rm \: \frac{dy}{dx}\bigg|_{x=2} = \frac{2}{\sqrt{1 + {2}^{2} } }=\frac{2}{5}$