To derive the differentiation formula for \(\frac{d}{d x}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}\), we will start by defining \(y=\arcsin x\). This means that \(x=\sin y\)

Now, we'll differentiate both sides of the equation \(x=\sin y\) with respect to \(x\), applying implicit differentiation. Remember that we'll need to use the chain rule when differentiating the term involving \(y\) :

\[

\frac{d}{d x}(x)=\frac{d}{d x}(\sin y)

\]

Differentiating both sides, we get:

\[

1=(\cos y) \frac{d y}{d x}

\]

Now, we need to express \(\cos y\) in terms of \(x\). To do this, we use the Pythagorean identity:

\[

\sin ^2 y+\cos ^2 y=1

\]

Since we know that \(x=\sin y\), we can substitute \(x\) for \(\sin y\) :

\[

x^2+\cos ^2 y=1

\]

Now, solve for \(\cos y\) :

\[

\cos y=\sqrt{1-x^2}

\]

Substitute this expression for \(\cos y\) back into our original equation:

\[

1=\left(\sqrt{1-x^2}\right) \frac{d y}{d x}

\]

Now, solve for \(\frac{d y}{d x}\) :

\[

\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}

\]

Since \(y=\arcsin x\), we can express the result as:

\[

\frac{d}{d x}[\arcsin x]=\frac{1}{\sqrt{1-x^2}}

\]

And we've derived the differentiation formula for the arcsine function.