# How do you differentiate #g(x) =(2x+1)^(5/2)cos(5x) # using the product rule?

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To differentiate the function g(x) = (2x + 1)^(5/2)cos(5x) using the product rule, we apply the formula: (uv)' = u'v + uv'.

Let u = (2x + 1)^(5/2) and v = cos(5x).

Then, u' = (5/2)(2x + 1)^(3/2) and v' = -5sin(5x).

Using the product rule, the derivative of g(x) is:

g'(x) = u'v + uv'

g'(x) = [(5/2)(2x + 1)^(3/2)]cos(5x) + (2x + 1)^(5/2)(-5sin(5x))

So, g'(x) = (5/2)(2x + 1)^(3/2)cos(5x) - 5(2x + 1)^(5/2)sin(5x).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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