# If y=sec(tan^-1x) then dy/dx

Solution:   We have ,    $y=\sec(\tan^{-1}x)$

Then ,      $y=\sec(\tan^{-1}x)=\sec(\sec^{-1}\frac{\sqrt{1+x^{2}}}{1})=\sqrt{1+x^{2}}$

$\\ \\ \Rightarrow \hspace{1cm}y=\sqrt{1+x^{2}}\\ \\ \Rightarrow \hspace{1cm} \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2\sqrt{1+x^{2}}}\times (2x)=\frac{x}{\sqrt{1+x^{2}}}$

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