To find the derivative of the given expression:

$s = \frac{4t + 2}{\sqrt{t^2 + 6}}$

we apply the quotient rule, which states:

$\frac{d}{dt} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}$

Here:

• $u = 4t + 2$ and $v = \sqrt{t^2 + 6} = (t^2 + 6)^{1/2}$

Step 1: Differentiate $u$ and $v$

1. $u = 4t + 2 \implies u' = 4$
2. $v = (t^2 + 6)^{1/2} \implies v' = \frac{1}{2}(t^2 + 6)^{-1/2} \cdot 2t = \frac{t}{\sqrt{t^2 + 6}}$

Step 2: Apply the quotient rule

Substitute into the quotient rule:

$\frac{ds}{dt} = \frac{\sqrt{t^2 + 6} \cdot 4 - (4t + 2) \cdot \frac{t}{\sqrt{t^2 + 6}}}{t^2 + 6}$

Step 3: Simplify the numerator

The numerator is:

$\sqrt{t^2 + 6} \cdot 4 - (4t + 2) \cdot \frac{t}{\sqrt{t^2 + 6}}$

Rewrite it under a common denominator:

$\frac{4(t^2 + 6) - (4t^2 + 2t)}{\sqrt{t^2 + 6}}$

Simplify the terms:

$4(t^2 + 6) - (4t^2 + 2t) = 4t^2 + 24 - 4t^2 - 2t = 24 - 2t$

So the numerator becomes:

$\frac{24 - 2t}{\sqrt{t^2 + 6}}$

Step 4: Combine everything

The derivative is:

$\frac{ds}{dt} = \frac{\frac{24 - 2t}{\sqrt{t^2 + 6}}}{t^2 + 6} = \frac{24 - 2t}{(t^2 + 6)^{3/2}}$

Final Answer:

$\frac{ds}{dt} = \frac{24 - 2t}{(t^2 + 6)^{3/2}}$

Would you like me to go into further details or explore another aspect of this problem?

Questions for further exploration:

1. How does the quotient rule work in other scenarios with trigonometric functions?
2. Can you find the critical points of $s(t)$ by setting $\frac{ds}{dt} = 0$?
3. What is the behavior of $\frac{ds}{dt}$ as $t \to \infty$?
4. How does the chain rule contribute to the derivative of $\sqrt{t^2 + 6}$?
5. Can you use logarithmic differentiation to solve the same problem?

Tip:

When solving complex derivatives, always simplify individual components (numerator and denominator) before combining!