To find the derivative of the given function:

$y = \frac{3 \tan(t)}{4 + 4 \sec(t)}$

We'll use the quotient rule. The quotient rule states that if you have a function of the form $y = \frac{f(t)}{g(t)}$, then the derivative is given by:

$\frac{dy}{dt} = \frac{f'(t) g(t) - f(t) g'(t)}{[g(t)]^2}$

In this case:

• $f(t) = 3 \tan(t)$
• $g(t) = 4 + 4 \sec(t)$

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

1. The derivative of $f(t) = 3 \tan(t)$ is: $f'(t) = 3 \sec^2(t)$

2. The derivative of $g(t) = 4 + 4 \sec(t)$ is: $g'(t) = 4 \sec(t) \tan(t)$

Step 2: Apply the quotient rule

Now, we apply the quotient rule:

$\frac{dy}{dt} = \frac{f'(t) g(t) - f(t) g'(t)}{[g(t)]^2}$

Substitute $f(t)$, $f'(t)$, $g(t)$, and $g'(t)$ into the formula:

$\frac{dy}{dt} = \frac{(3 \sec^2(t))(4 + 4 \sec(t)) - (3 \tan(t))(4 \sec(t) \tan(t))}{(4 + 4 \sec(t))^2}$

Step 3: Simplify the expression

1. Simplify the numerator: $(3 \sec^2(t))(4 + 4 \sec(t)) = 12 \sec^2(t) + 12 \sec^3(t)$ $(3 \tan(t))(4 \sec(t) \tan(t)) = 12 \sec(t) \tan^2(t)$

So the numerator becomes: $12 \sec^2(t) + 12 \sec^3(t) - 12 \sec(t) \tan^2(t)$

2. The denominator is: $(4 + 4 \sec(t))^2 = 16(1 + \sec(t))^2$

Thus, the derivative is: $\frac{dy}{dt} = \frac{12 \sec^2(t) + 12 \sec^3(t) - 12 \sec(t) \tan^2(t)}{16(1 + \sec(t))^2}$

Final Answer:

$\frac{dy}{dt} = \frac{12 \sec^2(t) + 12 \sec^3(t) - 12 \sec(t) \tan^2(t)}{16(1 + \sec(t))^2}$

Would you like more details on the steps or any clarifications?

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Here are some related questions you might find interesting:

1. What is the quotient rule and how does it apply to functions with trigonometric expressions?
2. How do you simplify expressions involving secant and tangent functions?
3. Can you find the second derivative of the function $y = \frac{3 \tan(t)}{4 + 4 \sec(t)}$?
4. How does the chain rule work with trigonometric functions like $\sec(t)$ and $\tan(t)$?
5. What happens if we apply L