How do you find the derivative of #g(t)=t^2 2^t#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Salvatore I. 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# Answer link Related questions How do I find #f'(x)# for #f(x)=5^x# ? How do I find #f'(x)# for #f(x)=3^-x# ? How do I find #f'(x)# for #f(x)=x^2*10^(2x)# ? How do I find #f'(x)# for #f(x)=4^sqrt(x)# ? What is the derivative of #f(x)=b^x# ? What is the derivative of 10^x? How do you find the derivative of #x^(2x)#? How do you find the derivative of #f(x)=pi^cosx#? How do you find the derivative of #y=(sinx)^(x^3)#? How do you find the derivative of #y=ln(1+e^(2x))#? See all questions in Differentiating Exponential Functions with Other Bases Impact of this question 3544 views around the world You can reuse this answer Creative Commons License