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

1 Answer

Nov 8, 2016

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

Explanation:

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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