Choose the correct alternative in the following:

Let $f(x)=\sec \left(\tan ^{-1} x\right)$

Let $\theta=\tan ^{-1} x$

$\frac{d \theta}{d x}=\frac{1}{1+x^{2}}$ .....(1)

Now $\theta=\tan ^{-1} x$

$=x=\tan \theta$

$=\sqrt{1+x^{2}}=\sec \theta \because \sec ^{2} \theta-\tan ^{2} \theta=1$

Putting values, we get

$=\sec \theta \cdot \tan \theta \cdot \frac{1}{1+\mathrm{x}^{2}}$

$=\sqrt{1+\mathrm{x}^{2}} \cdot \mathrm{x} \cdot \frac{1}{\left(1+\mathrm{x}^{2}\right)}$

$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}$