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

Derivative of arcsin x

Derivative $f’$ of function $f(x)=\arcsin{x}$ is: \(\forall x \in ]–1, 1[ ,\quad f'(x) = \dfrac{1}{\sqrt{1-x^2}}\)


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

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

Using result of derivative of inverse functions, we have:


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



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

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

So, we have:

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

We have

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


by definition

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

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

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

Now, we have:

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

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

\[\forall x \in [-1,1], \quad \arcsin x \in [-\frac{\pi}{2},\frac{\pi}{2}]\]

since it is the inverse function of $\sin:[-\dfrac{\pi}{2},\dfrac{\pi}{2}] \to [-1,1]$.

Since angle $\arcsin x \in \displaystyle [-\frac{\pi}{2},\frac{\pi}{2}]$, then cosine of this angle $\cos (\arcsin x)$ is greater than or equal to zero. Then the only possible solution is :

\[\cos (\arcsin x) = + \sqrt{1 - x^2}\]

We conclude that:

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