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

## Higher-Order Derivatives

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


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

Example: Find the first four derivative of

1) ${\displaystyle f(x)=x^{4}+5x^{3}-2x^{2}+6}$

Solution:
${\displaystyle f'(x)=4x^{3}+15x^{2}-4x}$
${\displaystyle f''(x)=12x^{2}+30x-4}$
${\displaystyle f'''(x)=24x+30}$
${\displaystyle f^{(4)}(x)=24}$

2) ${\displaystyle f(x)=(x^{3}+1)(x^{2}+3)}$

Solution:
It is better to rewrite ${\displaystyle f(x)=(x^{3}+1)(x^{2}+3)=x^{5}+3x^{3}+x^{2}+3}$
Then, ${\displaystyle f'(x)=5x^{4}+9x^{3}+2x}$
${\displaystyle f''(x)=20x^{3}+27x^{2}+2}$
${\displaystyle f'''(x)=60x^{2}+54x}$
${\displaystyle f^{(4)}(x)=120x+54}$

## Acceleration

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

Word-Problem Example: A ball is thrown upward from the top of a ${\displaystyle 200}$-foot cliff. The initial velocity of the ball is ${\displaystyle 32}$ feet per second. The position function is ${\displaystyle f(t)=-16t^{2}+32t+200}$ where ${\displaystyle t}$ is measured in seconds. Find the height, velocity, and acceleration of the ball at ${\displaystyle t=4}$

Solution:
${\displaystyle f(t)=-16t^{2}+32t+200}$ (Position function)
${\displaystyle f'(t)=-32t+32}$ (Velocity function)
${\displaystyle f''(x)=-32}$ (Acceleration function)
So, when ${\displaystyle t=4}$, from the functions above, we can have:
${\displaystyle {\text{Height = }}f(4)=-16(4^{2})+32(4)+200=72}$
${\displaystyle {\text{Velocity = }}f'(4)=-32(4)+32=-96}$
${\displaystyle {\text{Acceleration = }}f''(4)=-32}$