QVQ: 更睿智地看世界

Let’s tackle this problem step by step. So, we have two functions, f(x) and g(x), and their derivatives f’(x) and g’(x), tabulated for specific values of x. We’re asked to find h’(2), where h(x) is the product of f(x) and g(x). That means h(x) = f(x) * g(x).

First, I need to recall the product rule for derivatives. The product rule tells us how to find the derivative of the product of two functions. It states that if you have two differentiable functions u(x) and v(x), then the derivative of their product is given by:

(u*v)’(x) = u’(x)*v(x) + u(x)*v’(x)

So, in this case, u(x) is f(x) and v(x) is g(x). Therefore, h’(x) = f’(x)*g(x) + f(x)*g’(x).

Now, I need to find h’(2), which means I need to evaluate this derivative at x = 2.

Looking at the table:

• At x = 2, f(2) = -3

• f’(2) = -8

• g(2) = 1

• g’(2) = 7

Plugging these values into the product rule formula:

h’(2) = f’(2)*g(2) + f(2)*g’(2)

= (-8)(1) + (-3)(7)

= -8 + (-21)

= -29

Wait a minute, let me double-check that. Maybe I should consider if there’s any other way to approach this problem.

Another approach could be to consider the properties of derivatives and see if there’s a different way to compute h’(2). But I think the product rule is the most straightforward method here.

Let me verify the values from the table again:

• f(2) = -3

• f’(2) = -8

• g(2) = 1

• g’(2) = 7

These seem correct as per the table.

Alternatively, perhaps I should consider if the functions f(x) and g(x) have any special properties or if there’s a pattern I’m missing. For example, is g(x) a constant function? Looking at the table:

• g(-2) = -11

• g(-1) = -2

• g(0) = -1

• g(1) = -2

• g(2) = 1

It doesn’t seem like g(x) is constant, as the values change. Similarly, f(x) is changing as well.

Maybe I should think about the units if this were a real-world problem, but since it’s an abstract mathematical problem, that might not help.

Alternatively, perhaps I should consider the geometric interpretation of the derivative. The derivative h’(2) represents the instantaneous rate of change of h(x) at x = 2. But again, that might not help in calculating the numerical value.

Let me try plugging in the values again to make sure I didn’t make a miscalculation.

h’(2) = f’(2)*g(2) + f(2)*g’(2)

= (-8)(1) + (-3)(7)

= -8 + (-21)

= -29

No, that seems correct.

Wait, maybe I should consider if the table has any units or if there’s a scaling factor I’m missing, but the problem doesn’t indicate any units, so I assume the values are just numerical.

Alternatively, perhaps there’s a typo in the problem or the table, but based on the given information, the calculation seems straightforward.

Let me consider if there’s a way to cross-verify this result. Suppose I were to consider the function h(x) = f(x)*g(x), and I want to find h’(2).

Another way to think about it is to consider the definition of the derivative:

h’(2) = lim[h -> 0] [h(2 + h) - h(2)] / h

But without knowing the explicit forms of f(x) and g(x), this approach isn’t practical because I can’t compute h(2 + h) for arbitrary h.

Therefore, relying on the product rule seems like the most appropriate method here.

So, my final answer is h’(2) = -29.

Final Answer

$$ \boxed{-29} $$