Mathematics

The key is recognizing that $3x - 4x^3$ is the triple-angle formula for sine: $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$. This means $f(x) = \sin^{-1}(\sin(3\theta))$ where $\sin(\theta) = x$, so $f(x) = 3\theta = 3\sin^{-1}(x)$ (on the appropriate domain). Once you simplify $f$ this way, finding $f'(0)$ becomes straightforward using the derivative of inverse sine.

The key is recognizing that $3x - 4x^3$ is the triple-angle formula for sine: $\sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta)$. This means $f(x) = \sin^{-1}(\sin(3\theta))$ where $\sin(\theta) = x$, so $f(x) = 3\theta = 3\sin^{-1}(x)$ (on the appropriate domain). Once you simplify $f$ this way, finding $f'(0)$ becomes straightforward using the derivative of inverse sine.