# What is the derivative of #f(x)=e^(4x)*ln(1-x)# ?

Explanation:

In general Product Rule is,

Plugging these in product rule definition yields,

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To find the derivative of ( f(x) = e^{4x} \cdot \ln(1-x) ), you can use the product rule of differentiation, which states that if ( u ) and ( v ) are functions of ( x ), then the derivative of ( u \cdot v ) with respect to ( x ) is ( u'v + uv' ). Applying this rule:

( f(x) = e^{4x} \cdot \ln(1-x) )

Let ( u = e^{4x} ) and ( v = \ln(1-x) ).

Then, ( u' = 4e^{4x} ) (derivative of ( e^{4x} )) and ( v' = \frac{-1}{1-x} ) (derivative of ( \ln(1-x) )).

Now, applying the product rule:

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

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

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

So, the derivative of ( f(x) = e^{4x} \cdot \ln(1-x) ) is ( f'(x) = 4e^{4x} \cdot \ln(1-x) - \frac{e^{4x}}{1-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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