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.