Question

I am having a problem how he got to (d/dt 80)=0 , d/dt t to the power of 2= 2t and t=4 h'=-40/sec. session 3 Instantaneous speed.

Answer 1

In Clip 3 of Session 3, the notes titled Physical Interpretation of Derivatives

explains
If you’ve had calculus before, you’re probably able to find the derivative of the polynomial \( 80−5t^2 \) on your own. If not, you’ll have to take a few things on faith here.

However, you might want to review the first lecture, because it introduces the definition of a derivative.
\[ f'(x)= \lim_ {\Delta x -> 0} \frac{f\left( x+\Delta x\right) - f(x)}{\Delta x}\]

You can use that definition to find the derivative of a constant:
f(x)= 80 (the constant is 80 in this case)
f(x+ Δx) is still 80 (f(x) is 80 for all values of x)
if you put those values in the definition, you get 0/ Δx which is zero as
Δx approaches 0 (*remember Δx never actually becomes zero, it just gets very close*)

Also, the first lecture tells you the derivative of a function is connected to the slope (of a tangent line) to the function at a specific point. For the equation
y= 80 (a horizontal line), the slope is 0 (everywhere), and this leads to the derivative = 0, which matches with using the definition of the derivative.

The derivative of \( -5t^2\) uses the power rule.

Again, in the notes they say
...we know from last class that the derivative of \( t^2\text{ is }2t^{2−1} =2t \)

Look at clip 4 from Session 2, \( Example\ 2.\ f(x)= x^2 \)
He is showing how to derive the general formula for taking the derivative of x^n
You could try to do the simpler case, just x^2:
\[ \lim_ {\Delta x -> 0} \frac{\left( x+\Delta x\right)^2 - x^2}{\Delta x}\]
square the x+Δx, simplify, and then let Δx->0 . you should get 2x

use that result to find the derivative of
\[ h(t)= -5t^2 \\ h'(t)= -5 d(t^2) \\ h'(t)= -5\cdot 2 \cdot t^{2-1} \\ h'(t)=-10t \]

notice we can factor out the coefficient -5 before taking the derivative.
We can use the definition of the derivative to show it's ok