# What is the derivative of #f(x)=e^(2x) ln(x+2)#?

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To find the derivative of ( f(x) = e^{2x} \ln(x+2) ), you can use the product rule.

The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product ( u(x) \cdot v(x) ) with respect to ( x ) is ( u'(x) \cdot v(x) + u(x) \cdot v'(x) ).

Let ( u(x) = e^{2x} ) and ( v(x) = \ln(x+2) ).

Now, find the derivatives of ( u(x) ) and ( v(x) ):

( u'(x) = 2e^{2x} ) (using the chain rule)

( v'(x) = \frac{1}{x+2} ) (using the derivative of natural logarithm)

Now apply the product rule:

( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) )

( f'(x) = 2e^{2x} \ln(x+2) + e^{2x} \frac{1}{x+2} )

So, the derivative of ( f(x) = e^{2x} \ln(x+2) ) is ( f'(x) = 2e^{2x} \ln(x+2) + \frac{e^{2x}}{x+2} ).

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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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