`f(t) = t^(3/2)log_2sqrt(t+1)` Find the derivative of the function

We shall use:

Product rule

`(u cdot v)'=u' cdot v+u cdot v'`

Chain rule

`(u(v))'=u'(v) cdot v'`

`u` and `v` are both functions of arbitrary variable over which we are differentiating. 

First we apply product rule.

`f'(t)=(t^(3/2))' log_2 sqrt(t+1)+t^(3/2)(log_2 sqrt(t+1))'=`

Now we apply the chain rule on the second term.

 `3/2 t^(1/2)log_2 sqrt(t+1)+t^(3/2) 1/(sqrt(t+1)ln 2)cdot1/(2sqrt(t+1))=`

Now we simplify the obtained derivative to get the final result.

`3/2 sqrt(t)log_2 sqrt(t+1)+t^(3/2)/(2(t+1)ln2)`