## 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