Difference between revisions of "Math 22 Higher-Order Derivative"

Higher-Order Derivatives

The "standard" derivative $f'(x)$ is called the first derivative of $f(x)$ . The derivative of $f'(x)$ is the second derivative of$f(x)$ , denoted by $f''(x).$ By continuing this process, we obtain higher-order derivative of $f(x)$ .

Note: The 3rd derivative of $f(x)$ is $f'''(x)$ . However, we simply denote the $n^{th}$ derivative as $f^{(n)}(x)$ for $n\geq 4$ Example: Find the first four derivative of

1) $f(x)=x^{4}+5x^{3}-2x^{2}+6$ Solution:
$f'(x)=4x^{3}+15x^{2}-4x$ $f''(x)=12x^{2}+30x-4$ $f'''(x)=24x+30$ $f^{(4)}(x)=24$ 2) $f(x)=(x^{3}+1)(x^{2}+3)$ Solution:
It is better to rewrite $f(x)=(x^{3}+1)(x^{2}+3)=x^{5}+3x^{3}+x^{2}+3$ Then, $f'(x)=5x^{4}+9x^{3}+2x$ $f''(x)=20x^{3}+27x^{2}+2$ $f'''(x)=60x^{2}+54x$ $f^{(4)}(x)=120x+54$ Acceleration

If $f(x)$ is the position function, then $f'(x)$ is the velocity function and $f''(x)$ is the acceleration function.

Word-Problem Example: A ball is thrown upward from the top of a $200$ -foot cliff. The initial velocity of the ball is $32$ feet per second. The position function is $f(t)=-16t^{2}+32t+200$ where $t$ is measured in seconds. Find the height, velocity, and acceleration of the ball at $t=4$ Solution:
$f(t)=-16t^{2}+32t+200$ (Position function)
$f'(t)=-32t+32$ (Velocity function)
$f''(x)=-32$ (Acceleration function)
So, when $t=4$ , from the functions above, we can have:
${\text{Height = }}f(4)=-16(4^{2})+32(4)+200=72$ ${\text{Velocity = }}f'(4)=-32(4)+32=-96$ ${\text{Acceleration = }}f''(4)=-32$ 