Inverse Trig Derivative Here's a simple problem that I'm not getting.
y = arcsec (1/x), where o < x < 1.
What is the derivative?
y' = 1 / { x^-1 * (x^(-2) - 1)^(1/2) } * -x^-2
y' = -x / { x^2 * (x^(-2) - 1)^(1/2) }
y' = -1 / { x * (x^(-2) - 1)^(1/2) }
That's what I'm getting.
But the answer should be y' = -1 / {1-x^2}^(1/2)
That's very different.

$$
y =arcsec(\frac{1}{x})\Rightarrow \frac{1}{x}=\sec y=\frac{1}{\cos y}\Rightarrow x=\cos y.
$$
i.e
$$
x(y)=\cos y\Rightarrow x'(y)=-\sin y
$$
$$
x'(y)=\frac{1}{y'(x)}\Rightarrow y'(x)=\frac{1}{x'(y)}
$$
$$
y'(x)=-\frac{1}{\sin y}=-\frac{1}{\sqrt{1-\cos^2 y}}=-\frac{1}{\sqrt{1-x^2}}
$$
$$
y' =arcsec(\frac{1}{x})=-\frac{1}{\sqrt{1-x^2}}
$$
