g(t) = t^2*2^t Find the derivative of the function

The derivative of a function f at a point x is denoted as  y' = f'(x)
There are basic properties and formula we can apply to simplify a function such as the  Product Rule provides the formula:
d/dx(u*v) = u' *v + u * v'
 For the problem:g(t) = t^2* 2^t we let :
u = t^2  
v = 2^t
Now we want to find the derivative of each function.
Recall the power rule for derivatives:d/dx(u^n)=n*u^(n-1) du/dx  
So, for u = t^2 , 
u' = 2t
Recall that for differentiating exponential functions: 
d/dx(a^u) =a^u* ln(a)*du/dx  where  a!=1 .
With the function v = 2^t , we get v' = 2^t*ln(2) *1 = 2^tln(2)
We now have:
u = t^2
u' = 2t
v=2^t
v' =2^tln(2)
Then following the Product Rule:d/(dx)(u*v) = u' *v + u * v', we get:
g'(t) = 2t*2^t + t^2* 2^tln(2)
g'(t) = t2^(t+1) + t^2 2^tln(2 )