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Ask a math problem with high second derivative.
Divide f (1+△t)-f (1) by △t to get the average rate of change, and then take the average rate of change when △t approaches 0 to get the instantaneous rate of change when t= 1, which is the derivative, that is, when t=65438.

Or:

h'(t)=(-4.9t? )'+(6.5t)'+( 10)'

H '(t)=-9.8 tons +6.5

Then substitute t= 1 to get the instantaneous velocity at t= 1.

The second method is to solve it directly with derivative formula.