Question #86235

2 Answers
Nov 8, 2017

Look below

Explanation:

Rewrite it as arccos(1/x)

now use the chain rule

= -\frac{1}{sqrt(1-(1/x)^2}) times \frac{d}{dx}[1/x]

=d/dx[1/x]=\frac{d/dx[x]}{x^2}

now subsitute the 1 with the function

= -\frac{\frac{d/dx[x]}{x^2}}{sqrt(1-(1/x)^2})

move the x^2 to the denominator

= \frac{1}{sqrt(1-(1/x)^2)x^2

Nov 9, 2017

1/(|x|sqrt(x^2-1))

Explanation:

let

y=cos^(-1)(1/x)

=>1/x=cosy

differentiate wrt" "x

-1/x^2=(dy)/(dx)(-siny)

(dy)/(dx)=1/(x^2siny)

=1/(x^2(1-cos^2y)

=1/(x^2sqrt((1-(1/x))^2)

=1/((x^2)(sqrt((x^2-1)/x^2)))

=1/(x^2/xsqrt(x^2-1))

=1/(|x|sqrt(x^2-1))