How do you find the derivative of #g(t)=t^2 2^t#?

Answer 1

#g'(t)=2^(t+1)t+ 2^(t)t^2ln2#

Let's rewrite it #g(t)=t^2e^(tln2)# Now we can derive it
#g'(t)=2t*e^(tln2)+t^2ln2e^(tln2)# that can be rewritten as #g'(t)=e^(tln2)*(2t+t^2ln2))=2^(t+1)*t+ 2^t*t^2ln2#
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Answer 2

To find the derivative of ( g(t) = t^2 \cdot 2^t ), you can use the product rule, which states that the derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

So, applying the product rule to ( g(t) = t^2 \cdot 2^t ), we get:

[ g'(t) = 2t \cdot 2^t + t^2 \cdot \ln(2) \cdot 2^t ]

This can be simplified to:

[ g'(t) = 2t \cdot 2^t + t^2 \cdot 2^t \cdot \ln(2) ]

[ g'(t) = 2^t(2t + t^2 \ln(2)) ]

So, the derivative of ( g(t) ) is ( 2^t(2t + t^2 \ln(2)) ).

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Answer from HIX Tutor

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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