Question

How do you find the derivative of $y=tan{h}^{-1}\left(\frac{1}{x}\right)$?

Answer 1

$\frac{dy}{dx}=\frac{1}{1-x^2}$
Explanation:
$y=tan{h}^{-1}\left(\frac{1}{x}\right)\Rightarrow tanhy=\frac{1}{x}$
Differentiating implicitly gives:
$(1-tan{h}^{2}y)\frac{dy}{dx}=-\frac{1}{x^2}$
$\therefore (1-{\left(\frac{1}{x}\right)}^2)\frac{dy}{dx}=-\frac{1}{x^2}$
$\therefore \frac{dy}{dx}=-\frac{1}{\left(x^2\right)(1-{\left(\frac{1}{x}\right)}^2)}$
$\therefore \frac{dy}{dx}=-\frac{1}{\left(x^2\right)(1-\frac{1}{x^2})}$
$\therefore \frac{dy}{dx}=-\frac{1}{x^2-1}$
$\therefore \frac{dy}{dx}=\frac{1}{1-x^2}$