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Let's break down the problem and find the derivative \( h'(1) \) given the function \( h(x) = f(2x)(g(x))^2 \).

### Step 1: Differentiate \( h(x) \)
The function \( h(x) \) is defined as:
\[
h(x) = f(2x)(g(x))^2
\]
To find \( h'(x) \), we'll use the product rule, which states that if \( h(x) = u(x) \cdot v(x) \), then:
\[
h'(x) = u'(x)v(x) + u(x)v'(x)
\]
Here, let:
- \( u(x) = f(2x) \)
- \( v(x) = (g(x))^2 \)

So, \( h(x) = u(x) \cdot v(x) \).

#### Differentiate \( u(x) = f(2x) \)
Using the chain rule:
\[
u'(x) = \frac{d}{dx}f(2x) = f'(2x) \cdot 2
\]

#### Differentiate \( v(x) = (g(x))^2 \)
Using the chain rule and the power rule:
\[
v'(x) = 2g(x) \cdot g'(x)
\]

### Step 2: Combine using the product rule
Now, substitute the derivatives into the product rule:
\[
h'(x) = u'(x)v(x) + u(x)v'(x)
\]
\[
h'(x) = 2f'(2x) \cdot (g(x))^2 + f(2x) \cdot 2g(x)g'(x)
\]
\[
h'(x) = 2f'(2x)(g(x))^2 + 2f(2x)g(x)g'(x)
\]

### Step 3: Evaluate at \( x = 1 \)
We need to find \( h'(1) \):
\[
h'(1) = 2f'(2 \cdot 1) \cdot (g(1))^2 + 2f(2 \cdot 1) \cdot g(1) \cdot g'(1)
\]
This simplifies to:
\[
h'(1) = 2f'(2) \cdot (g(1))^2 + 2f(2) \cdot g(1) \cdot g'(1)
\]

### Step 4: Use the provided values
From the table:
- \( g(1) = 35 \)
- \( g'(1) = -2 \)

From the graph for \( f'(x) \):
- \( f'(2) = -1 \) (observing the graph at \( x = 2 \))

We also know:
- \( f(2) = 4 \) (since \( f'(x) = 0 \) for \( x < 2 \), and it appears \( f(1) = 3 \), so \( f(2) = 3 + 1 \cdot (2-1) = 4 \)).

Substitute these values into the equation:
\[
h'(1) = 2(-1) \cdot (35)^2 + 2 \cdot 4 \cdot 35 \cdot (-2)
\]
\[
h'(1) = -2 \cdot 1225 - 560
\]
\[
h'(1) = -2450 - 560
\]
\[
h'(1) = -3010
\]

### Final Answer:
The derivative \( h'(1) \) is \( -3010 \).

Would you like more details, or do you have any questions?

### Relative Questions:
1. How would the derivative change if \( f(x) \) had a different initial value?
2. What if \( g(x) \) was a different function, how would that affect the derivative?
3. Could the chain rule be applied differently to find \( h'(x) \)?
4. What are the implications of \( h(x) \) having such a large derivative value at \( x = 1 \)?
5. How does the product rule assist in differentiating products of functions?

### Tip:
Always check the conditions provided in the problem to ensure you're using the correct values when substituting into derived equations.

Let h(x) = f(2x)(g(x))^2 where f(x) is a function such that f(1) = 3 and the graph of f'(x) is given. Using this information, determine h'(1), giving full details of how you have found your answer.

Learn how to find the derivative h'(1) using the product rule and chain rule. The problem involves a function h(x) = f(2x)(g(x))^2 with given values of g(x) and g'(x), and a graph of f'(x). Detailed solution provided.

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