Question

Differentiate $y = \sin^{-1}(3x - 4x^3)$ with respect to $x$, for $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$.

Hint:

When differentiating inverse trigonometric functions, apply the chain rule carefully to account for both the inverse function and its argument.

Answer 1

The task is to find the derivative of the function \( y = \sin^{-1}(3x - 4x^3) \). Let \( u = 3x - 4x^3 \). Applying the chain rule, the derivative of \( \sin^{-1} u \) with respect to \( x \) is given by \( \frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \). First, we compute the derivative of \( u \) with respect to \( x \): \( u = 3x - 4x^3 \), so \( \frac{du}{dx} = 3 - 12x^2 \). Substituting this into the chain rule formula, we obtain the derivative of \( y \) with respect to \( x \): \( \frac{dy}{dx} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2) \). This represents the required derivative.