y = √(3 - t^8) dy = - (8t^7 / (3∛(3 - t^8))) dx

Explanation:

Step1: Identify the outer - inner functions

Let $u = 3 - t^{8}$, so $y=\sqrt{u}=u^{\frac{1}{2}}$.

Step2: Differentiate the outer function

The derivative of $y = u^{\frac{1}{2}}$ with respect to $u$ is $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

Step3: Differentiate the inner function

The derivative of $u = 3 - t^{8}$ with respect to $t$ is $\frac{du}{dt}=-8t^{7}$.

Step4: Apply the chain - rule

By the chain - rule $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dt}$: $\frac{dy}{dt}=\frac{1}{2\sqrt{3 - t^{8}}}\cdot(-8t^{7})=-\frac{4t^{7}}{\sqrt{3 - t^{8}}}$. And $dy =-\frac{4t^{7}}{\sqrt{3 - t^{8}}}dt$.

Answer:

$dy =-\frac{4t^{7}}{\sqrt{3 - t^{8}}}dt$