## Simple way to find the derivative of a function (ADVANCE)

In example 1, the method we find the acceleration function is called find the derivative of the velocity function. In Calculus there is a simple way to find the derivative of a function.
Give Functions

`F(x) = ax^n+bx^m + C`

where `C` is a constant.
The derivative of this function is:
`F'(x) = anx^(n-1) + bmx^(m-1)`

Notes: the constant `C` just dropped.
So in example 1 the function of velocity is `v_x = 40 - 5t^2`
`a_x = (dv_x)/(dt) = (-5)(2)t^(2-1) = -10t`
More practice:
`f(x) = 3x^3 + 4x^2 - x +5`
`f'(x) = (3)(3)x^(3-1) + (4)(2)x^(2-1) +(-1)(1)x^(1-1)`
`f'(x) = 9x^2 + 8x -1`
`x = 100 + 20t + 5t^2`
`(dx)/(dt) = 20 +(5)(2)t = 20+ 10t`
Example 2:
The displacement of a particle moving along the `x` axis varies according to the expression `x = 10 - 5t^2+4t^3`, where `x` is in meters and `t` is in seconds. Find displacement, velocity and acceleration at `t=2.0\ s`
Displacement at `t=2.0\ s`:
`x = 10 - 5(2.0)^2 + 4(2.0)^3 = 10-20 + 32 = 22\ m`
Velocity is the derivative of displacement function:
`v = (dx)/(dt) = (-5)(2)t + (4)(3)t^2 = -10t + 12t^2`
Velocity at `t = 2.0\ s`
`v = (-10)(2.0) + 12(2.0)^2 = -20 + 48 = 24\ m//s`
Acceleration is the derivative of velocity function:
`a = (dv)/(dt) = -10 + 24t `
Acceleration at `t = 2.0\ s`
`a = -10 + 24(2.0) = 38\ m//s^2`