To differentiate the function $F(t) = \frac{t^2}{\sqrt{t^3 + 1}}$, we apply the quotient rule and the chain rule.

Quotient Rule:

If $F(t) = \frac{f(t)}{g(t)}$, then: $F'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}.$

Here, $f(t) = t^2$ and $g(t) = \sqrt{t^3 + 1} = (t^3 + 1)^{1/2}$.

Step 1: Differentiate $f(t)$ and $g(t)$

1. $f(t) = t^2 \implies f'(t) = 2t$.
2. $g(t) = (t^3 + 1)^{1/2}$.
   • By the chain rule, $g'(t) = \frac{1}{2}(t^3 + 1)^{-1/2} \cdot 3t^2 = \frac{3t^2}{2\sqrt{t^3 + 1}}$.

Step 2: Apply the Quotient Rule

$F'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}.$

Substituting $f(t)$, $f'(t)$, $g(t)$, and $g'(t)$: $F'(t) = \frac{(2t)(\sqrt{t^3 + 1}) - (t^2)\left(\frac{3t^2}{2\sqrt{t^3 + 1}}\right)}{(\sqrt{t^3 + 1})^2}.$

Step 3: Simplify

1. The denominator becomes: $(\sqrt{t^3 + 1})^2 = t^3 + 1.$

2. Simplify the numerator: $\text{Numerator} = 2t\sqrt{t^3 + 1} - \frac{3t^4}{2\sqrt{t^3 + 1}}.$

   Combine terms over a common denominator $\sqrt{t^3 + 1}$: $\text{Numerator} = \frac{2t(t^3 + 1) - 3t^4}{2\sqrt{t^3 + 1}}.$

   Expand $2t(t^3 + 1)$: $\text{Numerator} = \frac{2t^4 + 2t - 3t^4}{2\sqrt{t^3 + 1}} = \frac{-t^4 + 2t}{2\sqrt{t^3 + 1}}.$

3. The final derivative is: $F'(t) = \frac{-t^4 + 2t}{2(t^3 + 1)\sqrt{t^3 + 1}}.$

Final Answer:

$F'(t) = \frac{-t^4 + 2t}{2(t^3 + 1)\sqrt{t^3 + 1}}.$

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Would you like further clarification or step-by-step elaboration?

Related Questions:

1. How does the chain rule apply to differentiating radicals?
2. Can this function be rewritten in a simpler equivalent form?
3. How would this differentiation change if $t^3 + 1$ were replaced by $t^3 - 1$?
4. What are the critical points of $F(t)$?
5. How does $F'(t)$ behave as $t \to \infty$?

Tip:

When differentiating fractions or radicals, carefully track the application of the quotient and chain rules to avoid algebraic mistakes.