Suppose that f(2)=3 and f′(2)=5. Let g(x)=x^3f(x). Evaluate g′(2). Your final answer should be a single number.
g = x^3 f
g' = x^3 f' + 3x^2 f
g'(2) = 8f'(2) + 12f(2)
= 8*5 + 12*3
= 76
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To evaluate g′(2), we need to find the derivative of g(x) with respect to x and then substitute x=2 into the derivative.
First, let's find the derivative of g(x) using the product rule:
g'(x) = (x^3) * f'(x) + 3x^2 * f(x)
Now, substitute x=2 into g'(x):
g'(2) = (2^3) * f'(2) + 3(2^2) * f(2)
= 8 * 5 + 3 * 4 * 3
= 40 + 36
= 76
Therefore, g′(2) = 76.
To evaluate g'(2), we can use the product rule of differentiation. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by [(u'(x))(v(x)) + (u(x))(v'(x))].
In this case, we have g(x) = x^3f(x), where f(x) is some function that we do not know explicitly. We need to find g'(2), which is the derivative of g(x) evaluated at x=2.
Using the product rule, we have:
g'(x) = [(x^3)(f'(x)) + (3x^2)(f(x))]
To find g'(2), we need to evaluate the expression g'(x) at x=2. Plugging in x=2, we get:
g'(2) = [(2^3)(f'(2)) + (3(2^2))(f(2))]
Given that f(2) = 3 and f'(2) = 5, we can substitute these values into the expression to get:
g'(2) = [(2^3)(5) + (3(2^2))(3)]
= [8(5) + 12(3)]
= [40 + 36]
= 76
Therefore, g'(2) = 76.