# What is the instantaneous rate of change of #f(x)=ln(4x^2+2x ) # at #x=-1 #?

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To find the instantaneous rate of change of the function ( f(x) = \ln(4x^2 + 2x) ) at ( x = -1 ), we can use the derivative of the function.

The derivative of ( \ln(u) ) is ( \frac{1}{u} \cdot u' ), where ( u ) is the function inside the logarithm.

So, let ( u = 4x^2 + 2x ). Then, ( u' = (8x + 2) ).

Using the chain rule, the derivative of ( f(x) ) is ( \frac{1}{4x^2 + 2x} \cdot (8x + 2) ).

Now, evaluate this derivative at ( x = -1 ):

[ f'(-1) = \frac{1}{4(-1)^2 + 2(-1)} \cdot (8(-1) + 2) ] [ = \frac{1}{4 - 2} \cdot (-8 + 2) ] [ = \frac{1}{2} \cdot (-6) ] [ = -3 ]

So, the instantaneous rate of change of ( f(x) ) at ( x = -1 ) is ( -3 ).

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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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