To find the derivative of the function P(x) = (x^2 + 1)(2x - 3), we will use the product rule. The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Let's denote:
f(x) = x^2 + 1
g(x) = 2x - 3

The derivatives of these functions are:
f'(x) = 2x
g'(x) = 2

Now, we can apply the product rule:
P'(x) = f(x)g'(x) + g(x)f'(x)
P'(x) = (x^2 + 1)*2 + (2x - 3)*2x
P'(x) = 2x^2 + 2 + 4x^2 - 6x
P'(x) = 6x^2 - 6x + 2

None of the provided options match this result. There might be a mistake in the options or in the question itself.

Question

To find the derivative of the function P(x) = (x^2 + 1)(2x - 3), we will use the product rule. The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Let's denote:
f(x) = x^2 + 1
g(x) = 2x - 3

The derivatives of these functions are:
f'(x) = 2x
g'(x) = 2

Now, we can apply the product rule:
P'(x) = f(x)g'(x) + g(x)f'(x)
P'(x) = (x^2 + 1)*2 + (2x - 3)*2x
P'(x) = 2x^2 + 2 + 4x^2 - 6x
P'(x) = 6x^2 - 6x + 2

None of the provided options match this result. There might be a mistake in the options or in the question itself.

Knowee AI · Accepted Answer

To find the derivative of the function P(x) = (x^2 + 1)(2x - 3), we will use the product rule. The product rule states that the derivative of two functions multiplied together is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Let's denote:
f(x) = x^2 + 1
g(x) = 2x - 3

The derivatives of these functions are:
f'(x) = 2x
g'(x) = 2

Now, we can apply the product rule:
P'(x) = f(x)g'(x) + g(x)f'(x)
P'(x) = (x^2 + 1)*2 + (2x - 3)*2x
P'(x) = 2x^2 + 2 + 4x^2 - 6x
P'(x) = 6x^2 - 6x + 2

None of the provided options match this result. There might be a mistake in the options or in the question itself.

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