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

2 Answers
Apr 26, 2018

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)

Apr 26, 2018

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))).