If $$y = \sec(\tan^{-1}x)$$, then $$\frac{dy}{dx}$$ at $$x = 1$$ is equal to

We are given the function $$y=\sec\!\bigl(\tan^{-1}x\bigr)$$ and we wish to find its derivative with respect to $$x$$ and then evaluate that derivative at $$x=1$$.

First, let us introduce a new variable for clarity. We set $$\theta=\tan^{-1}x.$$ By definition of the inverse tangent, this means $$\tan\theta = x.$$

Now we re-express $$y$$ in terms of $$\theta$$:

$$y=\sec\theta.$$

To differentiate $$y$$ with respect to $$x$$ we must apply the Chain Rule. The Chain Rule states that if $$y=f(u)$$ and $$u=g(x)$$, then $$\frac{dy}{dx}= \frac{dy}{du}\cdot\frac{du}{dx}.$$

Here, $$f(u)=\sec u$$ and $$u=g(x)=\theta=\tan^{-1}x$$. We differentiate step by step:

• Derivative of the outer function.
We recall the standard derivative formula $$\frac{d}{du}\bigl[\sec u\bigr]=\sec u\,\tan u.$$

• Derivative of the inner function.
For the inverse tangent we use the formula $$\frac{d}{dx}\bigl[\tan^{-1}x\bigr]=\frac{1}{1+x^{2}}.$$

Putting these pieces together via the Chain Rule, we obtain

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

This derivative is expressed in terms of $$\theta$$, but we would like it entirely in terms of $$x$$. To do so, we convert $$\sec\theta$$ and $$\tan\theta$$ into algebraic expressions of $$x$$.

Because $$\tan\theta = x$$, we picture a right-angled triangle where the side opposite $$\theta$$ is $$x$$ and the side adjacent to $$\theta$$ is $$1$$. Using the Pythagorean theorem, the hypotenuse is

$$\sqrt{x^{2}+1}.$$

From this triangle we read off:

• $$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{x}{1}=x.$$

• $$\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{\sqrt{x^{2}+1}}{1} = \sqrt{x^{2}+1}.$$

Substituting these back into the derivative gives

$$\frac{dy}{dx} = \frac{\bigl(\sqrt{x^{2}+1}\bigr)\,(x)}{1+x^{2}} = \frac{x}{\sqrt{x^{2}+1}}.$$

Our next task is to evaluate this expression at $$x=1$$. We substitute $$x=1$$ into the derivative:

$$\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{\sqrt{1^{2}+1}} = \frac{1}{\sqrt{2}}.$$

Thus the numerical value of the derivative at $$x=1$$ is $$\dfrac{1}{\sqrt{2}}$$.

Examining the options, we see that $$\dfrac{1}{\sqrt{2}}$$ corresponds to Option C.

Hence, the correct answer is Option C.