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

We are asked to differentiate the function: \[ y = \sin^{-1}(3x - 4x^3). \] Let $u = 3x - 4x^3$. Using the chain rule, the derivative of $\sin^{-1} u$ is: \[ \frac{d}{dx} \sin^{-1} u = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}. \] First, compute $\frac{du}{dx}$: \[ u = 3x - 4x^3 \quad \Rightarrow \quad \frac{du}{dx} = 3 - 12x^2. \] Now, apply the chain rule: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2). \] This is the required derivative.