Question

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

Answer 1

`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`