Let $\theta = \tan^{-1}x$, so $\tan\theta = x$ and $y = \sec\theta$.

$\frac{dy}{dx} = \sec\theta\tan\theta\cdot\frac{d\theta}{dx} = \sec\theta\tan\theta\cdot\frac{1}{1+x^2}$

Since $\tan\theta=x$ and $\sec\theta=\sqrt{1+x^2}$:

$\frac{dy}{dx} = \sqrt{1+x^2}\cdot x\cdot\frac{1}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$

At $x=1$: $\frac{dy}{dx} = \frac{1}{\sqrt{2}}$