find the derivative of (Atcos(4sqrt3t)+Btsin(4sqrt3t)

To find the derivative of the function \( f(t) = At\cos(4\sqrt{3}t) + Bt\sin(4\sqrt{3}t) \), we will apply the product rule, which states:

\[
\frac{d}{dt} [uv] = u'v + uv'
\]

Here, \( u = t \) and \( v = \cos(4\sqrt{3}t) \) or \( v = \sin(4\sqrt{3}t) \) for each respective term.

### Step 1: Differentiate \( At\cos(4\sqrt{3}t) \)
Using the product rule:
\[
\frac{d}{dt} [At\cos(4\sqrt{3}t)] = A \left( \frac{d}{dt}(t) \cdot \cos(4\sqrt{3}t) + t \cdot \frac{d}{dt}(\cos(4\sqrt{3}t)) \right)
\]

- \( \frac{d}{dt}(t) = 1 \)
- \( \frac{d}{dt}[\cos(4\sqrt{3}t)] = -4\sqrt{3} \sin(4\sqrt{3}t) \)

So, the derivative of the first term is:
\[
A(1 \cdot \cos(4\sqrt{3}t) + t \cdot (-4\sqrt{3} \sin(4\sqrt{3}t)))
\]
\[
= A\cos(4\sqrt{3}t) - 4A\sqrt{3}t\sin(4\sqrt{3}t)
\]

### Step 2: Differentiate \( Bt\sin(4\sqrt{3}t) \)
Similarly, applying the product rule to the second term:
\[
\frac{d}{dt} [Bt\sin(4\sqrt{3}t)] = B \left( \frac{d}{dt}(t) \cdot \sin(4\sqrt{3}t) + t \cdot \frac{d}{dt}(\sin(4\sqrt{3}t)) \right)
\]

- \( \frac{d}{dt}(t) = 1 \)
- \( \frac{d}{dt}[\sin(4\sqrt{3}t)] = 4\sqrt{3} \cos(4\sqrt{3}t) \)

Thus, the derivative of the second term is:
\[
B(1 \cdot \sin(4\sqrt{3}t) + t \cdot (4\sqrt{3} \cos(4\sqrt{3}t)))
\]
\[
= B\sin(4\sqrt{3}t) + 4B\sqrt{3}t\cos(4\sqrt{3}t)
\]

### Step 3: Combine the results
Now, we combine the two derivatives:
\[
\frac{d}{dt} [At\cos(4\sqrt{3}t) + Bt\sin(4\sqrt{3}t)] = A\cos(4\sqrt{3}t) - 4A\sqrt{3}t\sin(4\sqrt{3}t) + B\sin(4\sqrt{3}t) + 4B\sqrt{3}t\cos(4\sqrt{3}t)
\]

This is the derivative of the given function.

Would you like more details or further explanations?

### 5 Related Questions:
1. How do we use the chain rule for trigonometric functions?
2. Can you differentiate functions with more complex trigonometric expressions?
3. What is the product rule and when is it applied in differentiation?
4. How do we handle implicit differentiation with trigonometric functions?
5. How can derivatives be used to find the rate of change in oscillating systems?

### Tip:
When differentiating functions involving products of a variable and trigonometric functions, always remember to apply the product rule carefully, as both the variable and the trigonometric part must be treated separately.

Find the derivative of (At cos(4 sqrt(3) t) + B t sin(4 sqrt(3) t))

Find the derivative of the function At cos(4 sqrt(3) t) + B t sin(4 sqrt(3) t) using the product rule and trigonometric differentiation. This solution provides a detailed step-by-step approach, including key calculus concepts such as the chain rule and product rule, suitable for students in Grades 11-12.

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