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

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

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

ExpandSolution:
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 $t=4$

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

Return to Topics Page

This page were made by Tri Phan