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

In general,

#d/dxcos^-1x=-1/sqrt(1-x^2)#

Here's how we obtain this common derivative:

#y=cos^-1x -> x=cosy# from the definition of an inverse function.

Differentiate both sides of #x=cosy.#

This will entail using Implicit Differentiation on the right side:

#d/dx(x)=d/dxcosy#

#1=-dy/dxsiny#

Solve for #dy/dx#:

#dy/dx=-1/siny#

We need to get rid of the #siny.#

We previously said #y=cos^-1x#. So,

#dy/dx=-1/sin(cos^-1x)#

Now, recall the identity

#sin^2x+cos^2x=1#

In the identity, replace #x# with #cos^-1x:#

#sin^2(cos^-1x)+cos^2(cos^-1x)=1#

#cos^2(cos^-1x)=(cos(cos^-1x))^2=x^2#

#sin^2(cos^-1x)+x^2=1#

#sin^2(cos^-1x)=1-x^2#

#sin(cos^-1x)=sqrt(1-x^2)#

Thus,

#dy/dx=-1/sqrt(1-x^2)#

#f(x)=cos^-1(x)" "=>" "cos(f(x))=x#

Take the derivative of both sides. Use the chain rule on the left.

#-sin(f(x))*f'(x)=1#

#=>" "f'(x)=(-1)/sin(f(x))=(-1)/sqrt(1-cos^2(f(x)))#

The last step came from the identity #sin^2(theta)+cos^2(theta)=1#, which is restated as #sin(theta)=sqrt(1-cos^2(theta))#. We should also remember that #cos(f(x))=x#, which we saw in the first line, so finally:

#f'(x)=(-1)/sqrt(1-x^2)#

Note about domain: the domain of #cos^-1(x)# is #0lt=xlt=pi#. Note that on this interval, #sin(x)gt=0#. This allows us to only take the positive root when we say that #sin(f(x))=sqrt(1-cos^2(f(x)))#.