Determine if the function $f(x)=\sqrt[3]{x}$ is differentiable at $x=-8$. If not, identify why.
a.) $f(x)=\sqrt[3]{x}$ is not differentiable at $x=-8$ because $f(-8)$ is not defined.
b.) $f(x)=\sqrt[3]{x}$ is not differentiable at $x=-8$ because $f^{\prime}(-8)$ is not defined.
c.) $f(x)=\sqrt[3]{x}$ is not differentiable at $x=-8$ because $\lim _{x \rightarrow-8} \sqrt[3]{x}$ does not exist.
d.) $f(x)=\sqrt[3]{x}$ is differentiable at $x=-8$

Step 1 ：The function \(f(x)=\sqrt[3]{x}\) is defined for all real numbers, including \(x=-8\). Therefore, option a is incorrect.

Step 2 ：The limit as \(x\) approaches \(-8\) also exists, so option c is incorrect.

Step 3 ：We need to check if the derivative of the function at \(x=-8\) exists to determine if the function is differentiable at that point.

Step 4 ：The derivative of \(f(x)=\sqrt[3]{x}\) is \(f'(x)=\frac{1}{3\sqrt[3]{x^2}}\).

Step 5 ：We need to check if this derivative is defined at \(x=-8\).

Step 6 ：The derivative of the function at \(x=-8\) is defined and is equal to \(-\frac{1}{12}\).

Step 7 ：Therefore, the function \(f(x)=\sqrt[3]{x}\) is differentiable at \(x=-8\).

Step 8 ：Final Answer: \(\boxed{\text{(d) } f(x)=\sqrt[3]{x} \text{ is differentiable at } x=-8}\)