Question

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

Answer 1

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

Explanation:

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)`

Answer 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)`

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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)))`.