# What is the derivative of f(x)=(cos^-1(x))/x ?

Jul 28, 2014

$f ' \left(x\right) = - \frac{1}{x \sqrt{1 - {x}^{2}}} - \frac{{\cos}^{-} 1 x}{x} ^ 2$

Using Quotient Rule, which is

$y = f \frac{x}{g} \left(x\right)$, then y'=(f'(x)g(x)−f(x)g'(x))/(g(x))^2

Applying this for given problem, which is $f \left(x\right) = \frac{{\cos}^{-} 1 x}{x}$

$f ' \left(x\right) = \frac{\left({\cos}^{-} 1 x\right) ' \left(x\right) - \left({\cos}^{-} 1 x\right) \left(x\right) '}{x} ^ 2$

$f ' \left(x\right) = \frac{- \frac{1}{\sqrt{1 - {x}^{2}}} \cdot x - {\cos}^{-} 1 x}{x} ^ 2$

$f ' \left(x\right) = - \frac{1}{x \sqrt{1 - {x}^{2}}} - \frac{{\cos}^{-} 1 x}{x} ^ 2$, where $- 1$<$x$<$1$