Proving that if $y=\operatorname{arcsec}(x)$ then $\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$ I'm trying to prove this formula however, I cannot seem to figure out how to single out the $x$ and remove its power. I would very much appreciate your help towards this.
$$y=\operatorname{arcsec}(x) \implies\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$$
Here's my working out, my reasoning is wrong somewhere, although I cannot find where I went wrong:
$$y = \operatorname{arcsec}; \quad\sec(y)=x$$
Then taking the derivative of both sides:
$$\sec(y)\tan(y)\frac{dy}{dx}=1; \frac{dy}{dx}=\frac{1}{\sec(y)\tan(y)}=\frac{1}{\sqrt{1-x^2}(x^2+1)}$$
How do I simplify the last equation?
 A: What you can show is that
$$
\frac{d}{{dx}}\sec ^{ - 1} (x) = \frac{1}{{x\sqrt {x^2  - 1} }}
$$
for $x>0$. Indeed, by the formula for the derivative of the inverse function, we find
\begin{align*}
\frac{1}{{\left[ {\frac{d}{{dt}}\sec t} \right]_{t = \sec ^{ - 1} (x)} }} & = \left[ {\frac{{\cos ^2 t}}{{\sin t}}} \right]_{t = \sec ^{ - 1} (x)}  = \left[ {\frac{1}{{\left| {\sec t} \right|\sqrt {\sec ^2 t - 1} }}} \right]_{t = \sec ^{ - 1} (x)} \\& = \frac{1}{{\left| x \right|\sqrt {x^2  - 1} }} = \frac{1}{{x\sqrt {x^2  - 1} }}
\end{align*}
for $x>0$.
A: Assuming that you are trying to show that for $x>0$ $$\frac{d}{dx} \operatorname{arcsec}(x) = \frac{1}{x\sqrt{x^2-1}},$$ we can proceed as follows.  Just as you did, we start with $$y = \operatorname{arcsec}(x) \implies \sec(y) = x.\tag{1}$$  This then gives us $$\frac{dy}{dx}=\frac{1}{\sec(y)\tan(y)} = \frac{1}{\sec(\operatorname{arcsec}(x))\tan(\operatorname{arcsec}(x))}.$$  Then, we have $\sec(\operatorname{arcsec}(x)) = x$, and for $\tan(\operatorname{arcsec}(x))$ we consider the Pythagorean identity $$\tan^{2}(y) = \sec^{2}(y) - 1 \implies \tan(y) = \sqrt{\sec^{2}(y) - 1}.$$  Now, using $(1)$, we can write $$\tan(\operatorname{arcsec}(x)) = \sqrt{\sec^{2}(\operatorname{arcsec}(x)) - 1} = \sqrt{x^{2} - 1}.$$  All together, then, we get $$\frac{d}{dx} \operatorname{arcsec}(x) = \frac{1}{x\sqrt{x^2-1}}.$$
