Derivative f’ of function f(x)=arccos x is: f’(x) = - 1 / √(1 - x²) for all x in ]-1,1[. To show this result, we use derivative of the inverse function cos x.

Derivative of arccos x

Derivative $f’$ of function $f(x)=\arccos{x}$ is:

\[\forall x \in ]–1, 1[ ,\quad f'(x) = -\frac{1}{\sqrt{1-x^2}}\]


Remember that function $\arcsin$ is the inverse function of $\cos$ :

\[\left(f^{-1} \circ f\right)=\left(\cos \circ \arccos\right)(x)=\cos(\arccos(x))=x\]

Using result of derivative of inverse functions, we have:


Taking : $g^{-1}=f=\arccos$ and $g=f^{-1}=\cos$, we have:



Derivative of cosine , $g(x)=\cos x$ is:

\[\forall x \in \mathbb{R}, \quad g'(x) = - \sin x\]

So, we have:

\[\begin{aligned} f^{\prime}(x)&=\frac{1}{\cos^{\prime}(f(x))}\\ &=-\frac{1}{\sin (f(x))}\\ &=-\frac{1}{\sin (\arccos x)}\\ \end{aligned}\]

We have

\[\forall X \in \mathbb{R}, \quad \cos^2 X + \sin^2 X =1\]


by definition

\[\left(f^{-1} \circ f\right)=\left(\cos \circ \arccos\right)(x)={\color{red}{\cos(\arccos(x))=x}}\]

Taking $X=\arccos x$, it gives:

\[\begin{aligned} 1&=\cos^2 X + \sin^2 X\\ &={\color{red}{\cos^2 (\arccos x)}} + \sin^2 (\arccos x) \\ &={\color{red}{x^2}} + \sin^2 (\arccos x) \\ \end{aligned}\]

Now, we have:

\[\sin^2 (\arccos x) = 1 - x^2 \Longrightarrow \sin (\arccos x) = \pm \sqrt{1 - x^2}\]

Function $\arccos x$ is defined for all $x \in [-1,1]$ and we have

\[\forall x \in [-1,1], \quad \arccos x \in [0, \pi]\]

since it is the inverse function of $\cos:[0, \pi] \to [-1,1]$.

Since angle $\arccos x \in \displaystyle [0, \pi]$, then sine of this angle $\sin (\arccos x)$ is greater than or equal to zero. Then the only possible solution is :

\[\sin (\arccos x) = + \sqrt{1 - x^2}\]

We conclude that:

\[\forall x \in ]–1, 1[ ,\quad f'(x) = - \frac{1}{\sin (\arccos x)} = - \frac{1}{\sqrt{1 - x^2}}\]