The image has been opened. Let's extract the mathematical problem and solve it.

Given:

$h(x) = 3f(x) - 2g(x) - 5 \cos(x) - 3$ You are asked to find $h'(0)$ given:

• $f(0) = 3$, $g'(0) = 7$
• $f$ and $g$ are differentiable functions.

We start by differentiating $h(x)$: $h'(x) = 3f'(x) - 2g'(x) - 5(-\sin(x))$ Now, substitute $x = 0$: $h'(0) = 3f'(0) - 2g'(0) + 5\sin(0)$ Since $\sin(0) = 0$, this simplifies to: $h'(0) = 3f'(0) - 2g'(0)$ Substitute $g'(0) = 7$: $h'(0) = 3f'(0) - 14$

To get the exact value of $h'(0)$, we would need $f'(0)$, which isn't given explicitly in the problem. However, this is the expression for $h'(0)$.

Would you like to proceed with additional details or calculations? Here are five related questions to deepen your understanding:

1. What is the significance of knowing both $f'(0)$ and $g'(0)$ in determining $h'(0)$?
2. How would the derivative of $h(x)$ change if $f(x)$ or $g(x)$ were not differentiable?
3. Why does the sine function $\sin(0) = 0$ simplify the problem?
4. How would the problem change if the constants in front of $f(x)$ and $g(x)$ were different?
5. Can the problem be extended to higher-order derivatives of $h(x)$?

Tip:

When differentiating composite functions, remember to apply the chain rule if any of the functions have more complex structures (like compositions).