If $f\left(x\right)={\sin }^{-1}\left(3x-4x^3\right)$, then the value of $f^{\text{'}}\left(0\right)$ is

The given function is $f\left(x\right)={\sin }^{-1}\left(3x-4x^3\right)$.

It is known that $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$.

Put $x=\sin \theta$ in the given function.

$$\begin{gathered} f\left(x\right)={\sin }^{-1}\left(3x-4x^3\right) \\ f\left(x\right)={\sin }^{-1}\left(3\sin \theta -4{\left(\sin \theta \right)}^3\right) \\ f\left(x\right)={\sin }^{-1}\left(3\sin \theta -4{\sin }^3\theta \right) \\ f\left(x\right)={\sin }^{-1}\left(\sin 3\theta \right) \\ f\left(x\right)=3\theta \\ f\left(x\right)=3{\sin }^{-1}x \end{gathered}$$

Differentiating the function,

$$\begin{gathered} f\text{'}\left(x\right)=3\times \frac{1}{\sqrt{1-x^2}} \\ f\text{'}\left(x\right)=\frac{3}{\sqrt{1-x^2}} \end{gathered}$$

At $x=0$,

$$\begin{gathered} f\text{'}\left(0\right)=\frac{3}{\sqrt{1-0}} \\ f\text{'}\left(0\right)=3 \end{gathered}$$

Hence, the value of $f\text{'}\left(0\right)$ is $3$.