r(t)=(t+e^t)(3-[sqrt(t)])
find r'(t)
Recall the product rule: $\displaystyle \big[ f(t)g(t) \big]' = {\color{red}f'(t)}g(t) + f(t){\color{blue}g'(t)}$
Here imagine: $\displaystyle f(t) = t + e^t$ and $\displaystyle g(t) = 3 - \sqrt{t}$
So plugging in we have: $\displaystyle \big[ (t+e^t)(3-\sqrt{t}) \big]' = {\color{red}(t + e^t)'}(3-\sqrt{t}) + (t+e^t){\color{blue}(3 - \sqrt{t})'}$
Now it's just a matter of finding the derivatives of the red and the blue which should be pretty straightforward.
sorry but i dont know latex so i know this will be hard to read but...
r'(t)= (1+e^t)[3-sqrt(t)] + (t+e^t)(1/2t^(-1/2))
its like how would i multiply e^t and t^(1/2) together? and where e^t would need to be multiplied. i just dont know how to simplify further.
With product rule questions, you are usually not expected to simplify that much.
As for your question, you can't multiply $\displaystyle e^t$ and $\displaystyle \sqrt{t}$ together to get anything new so you just simply leave it as $\displaystyle e^t\sqrt{t}$
So, just multiplying out the brackets:
$\displaystyle \begin{array}{rcl}r'(t) & = & (1 + e^t)(3-\sqrt{t}) + (t + e^t )\displaystyle \left(\frac{1}{2\sqrt{t}}\right) \\ & & \\ & = & \displaystyle 3 - \sqrt{t} + 3e^t - e^t\sqrt{t} + {\color{blue}\frac{t}{2\sqrt{t}}} + \frac{e^{t}}{2\sqrt{t}} \end{array}$
You can do a bit of simplifying (simplify the blue and see if you can combine it with another term) or combine like terms another way. Either way your final answer isn't going to look all that much neater