# How do you differentiate #g(x) = sqrt(x-3)e^(4x)cos(7x)# using the product rule?

Let us break down this problem into manageable chunks by differentiating each function independently and entering our solution at the end of the problem.

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To differentiate ( g(x) = \sqrt{x-3} e^{4x} \cos(7x) ) using the product rule, follow these steps:

- Identify the functions being multiplied together: ( f(x) = \sqrt{x-3} e^{4x} ) and ( h(x) = \cos(7x) ).
- Apply the product rule: ( g'(x) = f'(x)h(x) + f(x)h'(x) ).
- Find the derivatives of ( f(x) ) and ( h(x) ):
- ( f'(x) = \frac{1}{2\sqrt{x-3}} e^{4x} + \sqrt{x-3} \cdot 4e^{4x} ).
- ( h'(x) = -7\sin(7x) ).

- Plug the derivatives into the product rule formula and simplify.

The derivative of ( g(x) ) using the product rule is:

[ g'(x) = \left(\frac{1}{2\sqrt{x-3}} e^{4x} + \sqrt{x-3} \cdot 4e^{4x}\right) \cos(7x) + \sqrt{x-3} e^{4x} \left(-7\sin(7x)\right) ]

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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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