Undergrad Calculus problem differentiation.

[Lejas90210]

<Moderator's note: Member has been warned to show some effort before an answer can be given.>

Hello all.
This is my first post in this forum, I am asking for your understanding. I have a problem with the calculus task and I stuck in a dead endso I managed to find a solution on the internet. I am not sure whether it is done correctly. My main questions are:
1.) Why 9.5 * e ^ -1 = 3.495?
2.) Are the values in the table correct? What is the dependence between those values.
3.) If in response to t = 10, -> V (t) = 3.495 and in the table for t = 10 there is value: 60.05 V?

 

[Charles Link]

$e^{-1}=\frac{1}{2.7828} \approx.36788$.
$V'(10)=3.495$. ($Note: V'(t)=\frac{dV(t)}{dt}$). That's the slope of the curve of $V(t)$ vs. $t$ at $t=10$ if you draw a tangent line. Note: The graph needs to have the y-increment the same as the x-increment to readily see this. (With your increments of 10 and 5, it will appear to have a slope of 3.495/2 at $t=10$).

 

[Lejas90210]

My Problem (Task) require use different rule of differentiation that's what I have till now:

 

[Charles Link]

You need to realize that $\frac{dV(t)}{dt}$ is called $V'(t)$ and is completely different from $V(t)$.
If $V(t)$ were distance, $V'(t)$ would be the velocity.
$V(t) \neq V'(t)$. They are two separate functions.
Again $V'(t)=\frac{d V(t)}{dt}$.
You can let $V=y$ , and $t=x$, but you should specify this if you chose to take $V'=y'= \frac{dy}{dx}$. You then substitute $V$ (or $V'$) and $t$ back in, to get $V'(t)$.
It is good to be systematic rather than have hand-waving in your steps.