What is the derivative of $f (x) = \frac{{cos}^{- 1} (x)}{x}$ $f(x)=(cos^-1(x))/x$ ?

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Answer 1 · 2014-07-28T23:14:27

$f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2$

Using Quotient Rule, which is

$y=f(x)/g(x)$ , then $y'=(f'(x)g(x)−f(x)g'(x))/(g(x))^2$

Applying this for given problem, which is $f(x)=(cos^-1x)/x$

$f'(x)=((cos^-1x)'(x)-(cos^-1x)(x)')/x^2$

$f'(x)=(-1/sqrt(1-x^2)*x-cos^-1x)/x^2$

$f'(x)=-1/(xsqrt(1-x^2))-(cos^-1x)/x^2$ , where $-1$ < $x$ < $1$

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