Look at an inverse function below. $y = \text{cosec}^{-1}(4x^4); \ |4x^4| > 1$. Find $\frac{dy}{dx}$.

$-\frac{4x^3}{|x^4|\sqrt{16x^8-1}}$

The correct answer is Option (1) → $-\frac{4x^3}{|x^4|\sqrt{16x^8-1}}$ ##

Given, $y = \text{cosec}^{-1}(4x^4); \ |4x^4| > 1$

The derivative of $y = \text{cosec}^{-1}(u)$ is:

$\frac{dy}{du} = -\frac{1}{|u| \sqrt{u^2 - 1}}$

Let $u = 4x^4$

$\frac{du}{dx} = 16x^3$

Substitute $u$ into $\frac{dy}{du}$: For $u = 4x^4$:

$\frac{dy}{du} = -\frac{1}{4|x^4| \sqrt{16x^8 - 1}}$

Apply the chain rule:

$\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$

$=-\frac{1}{4|x^4|\sqrt{16x^8-1}}.16x^3$

$\frac{dy}{dx}=-\frac{4x^3}{|x^4|\sqrt{16x^8-1}}$