How do you find the instantaneous rate of change of the function #d=200t * 2^(-t)# when t=1?
I found:
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To find the instantaneous rate of change of the function (d = 200t \cdot 2^{-t}) when (t = 1), we need to find the derivative of the function with respect to (t), and then evaluate it at (t = 1).
First, we find the derivative of (d) with respect to (t), denoted as (\frac{d}{dt}): [ \frac{d}{dt}(d) = \frac{d}{dt}(200t \cdot 2^{-t}) ]
Using the product rule and the chain rule, we get: [ \frac{d}{dt}(d) = 200 \cdot 2^{-t} - 200t \cdot 2^{-t} \cdot \ln(2) ]
Now, we evaluate this derivative at (t = 1): [ \frac{d}{dt}(d) \Bigg|_{t=1} = 200 \cdot 2^{-1} - 200 \cdot 1 \cdot 2^{-1} \cdot \ln(2) ]
Solving this expression gives us the instantaneous rate of change of the function (d) when (t = 1).
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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.
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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