# How do you differentiate #f(x)= e^(2x)* (x^2 - 4) *ln x# using the product rule?

#To differentiate the product of 3 functions

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To differentiate the function ( f(x) = e^{2x} \cdot (x^2 - 4) \cdot \ln(x) ) using the product rule, follow these steps:

- Identify the functions ( u ) and ( v ) where ( f(x) = u(x) \cdot v(x) ).
- Apply the product rule: ( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) ).
- Differentiate each function separately.
- Substitute the derivatives and the original functions into the product rule formula.

The derivative of ( e^{2x} ) with respect to ( x ) is ( 2e^{2x} ). The derivative of ( x^2 - 4 ) with respect to ( x ) is ( 2x ). The derivative of ( \ln(x) ) with respect to ( x ) is ( \frac{1}{x} ).

Applying the product rule formula:

[ f'(x) = (2e^{2x}) \cdot (x^2 - 4) \cdot \ln(x) + e^{2x} \cdot (2x) \cdot \ln(x) + e^{2x} \cdot (x^2 - 4) \cdot \frac{1}{x} ]

So, the derivative of the function ( f(x) ) with respect to ( x ) using the product rule is:

[ f'(x) = 2e^{2x}(x^2 - 4) \ln(x) + 2xe^{2x} \ln(x) + \frac{e^{2x}(x^2 - 4)}{x} ]

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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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