To find the derivative of the function given in the image:

$F(t) = 2 \sin(3t + \tan(t))$

we'll proceed as follows:

Step 1: Differentiate Using the Chain Rule

Since $F(t) = 2 \sin(3t + \tan(t))$, we can apply the chain rule. Let: $u = 3t + \tan(t)$ then, $F(t) = 2 \sin(u)$

The derivative of $F(t)$ with respect to $t$ is: $F'(t) = 2 \cos(u) \cdot \frac{du}{dt}$

Step 2: Differentiate $u = 3t + \tan(t)$

Now, differentiate $u = 3t + \tan(t)$ with respect to $t$: $\frac{du}{dt} = 3 + \sec^2(t)$

Step 3: Substitute $u$ and $\frac{du}{dt}$ into $F'(t)$

Now, substitute $u = 3t + \tan(t)$ and $\frac{du}{dt} = 3 + \sec^2(t)$ into the derivative: $F'(t) = 2 \cos(3t + \tan(t)) \cdot (3 + \sec^2(t))$

Final Answer

The derivative of $F(t)$ is: $F'(t) = 2 \cos(3t + \tan(t)) \cdot (3 + \sec^2(t))$

Would you like more details or have further questions on this solution?

1. What are other examples of trigonometric functions and their derivatives?
2. How does the chain rule work in more complex functions?
3. Can you integrate this derivative back to the original function?
4. How would this change if we had a cosine function instead of sine?
5. What happens if we add another trigonometric function inside the sine function?

Tip: Remember, the chain rule is essential when differentiating composite functions like $\sin(3t + \tan(t))$.