`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` )  `