The equation of motion of an object is s = T ^ 3 + 2T ^ 2-1. When t = 2,

The equation of motion of an object is s = T ^ 3 + 2T ^ 2-1. When t = 2,

V=dS/dt=3t^2+4t=20
a=dV/dt=6t+4=16

The equation of motion of a particle s = sin2t calculates the acceleration of the particle (by derivative)

Y '= 2cos2t (speed)
Y '' = - 4cos2t (acceleration)

The displacement s of a derivative in mathematics changes with time as s = 5-3t square, then the average velocity of the particle in the time period (1,1 + triangle T) is

Average speed v = - 6-3 Δ t

Second derivative problem: Tongji higher mathematics, p103, question 7, assuming that the velocity of the x-axis motion of the particle is DX / dt = f (x), try to find the acceleration of the particle motion The answer is a = D ^ 2x / dt ^ 2, that is, the second-order derivative of velocity, why not f '(x) but f' (x) multiplied by F (x) Why must a = D ^ 2x / dt ^ 2 = D (f (x)) / DX * DX / dt = f '(x) f (x) in the calculation process? Why should DX be introduced here? Is chain derivation used here?

a=d^2x/dt^2=d(dx/dt)dt=d(f(x))/dt=d(f(x))/dx*dx/dt=f'(x)f(x)
Supplement:
Because f 'represents the derivative of its function value to its independent variable, that is, f' (x) = DF / DX
And DF / DT is actually DF (x (T)) / DT, so we have to use the chain derivative
The key is DX / dt = f (x), not DX / dt = f (T)
If f (x) = x ^ 2,
Then a = DF / dt = D (x ^ 2) / dt = 2xdx / DT
And if the original question is DX / dt = f (T) = T ^ 2
Then a = DF / dt = D (T ^ 2) / dt = 2T
I hope it will be clear

The motion equation of the particle is s = 3T + 1 (displacement unit: m, time unit: s). Calculate the velocity when t = 1 and 2 respectively. Write it in derivative, The branch is to

s=3t+1
v=ds/dt=3 m/s
V is a constant, moving at a constant speed
When t = 1,2s, the speed is 3m / s

Y = ㏑ [(1-e ⁿ) / E ⁿ], then find the derivative of Y Y = ㏑ [(1-e ⁿ) / E ⁿ], then find the derivative of Y. please write in detail,

y=ln[(1-e^n)/e^n]=ln(1-e^n)-ln(e^n)=ln(1-e^n)-ny'=dy/dn=[ln(1-e^n)-n]'=ln'(1-e^n)-n'=(1-e^n)'/(1-e^n)-1=-(e^n)/(1-e^n)-1=(-e^n-1+e^n)/(1-e^n)=1/(e^n-1)