3.5: Higher-Order Derivatives

Higher-Order Derivatives

Recall the definition of the derivative:

The derivative of a function $f$ with respect to $x$ is the function $f'$ where \[f'(x) = \lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\] The domain of $f'$ is the set of all $x$ for which the limit exists.

What if, in the definition, we had $f'(x+h) - f'(x)$ instead? This gives us the derivative of the derivative, or the second derivative.

To find the second derivative, first find $f'(x)$, then find the derivative of $f'(x)$.

In general, this extends to the $n$th derivative. Here are common notations for multiple derivatives:

  1. $f'(x), f''(x), f'''(x), \cdots, f^{(n)}(x)$
  2. $D_xf(x), D^2_xf(x), D^3_xf(x), \cdots, D^(n)_xf(x)$
  3. $y', y'', y''', \cdots, y^{(n)}$
  4. $\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}, \dfrac{d^3y}{dx^3}, \cdots, \dfrac{d^ny}{dx^n}$

Find the derivatives of all orders of \[f(x) = x^5 - 3x^4 + 4x^3 - 2x^2 + x - 8\]

Sometimes we need to use the product rule in the second derivative.

Find the second derivative of \[y = (2x^2 + 3)^{3/2}\]

We saw in Lecture 2.6 if $f(t)$ gives the position of an object moving in a straight line $t$ seconds after rest, then $f'(t)$ gives the velocity of the object at time $t$.

The rate of change of velocity is acceleration. This means $f''(t)$ gives the acceleration of the object at time $t$.

Suppose a maglev's position at time $t$ is determined by the function \[f(t) = 4t^2 \qquad 0 \leq t \leq 10\] What is the maglev's acceleration at time $t$?
Suppose a ball is thrown straight up. The height of the ball is measured by \[f(t) = -16t^2 + 24t + 120\] where $t$ is measured in seconds and $f(t)$ is measured in feet. Find the velocity and acceleration of the ball at time $t$.

The Consumer Price Index (CPI) is a weighted average of a market basket of various goods and services. For example, clothing, food,, and energy are all contained in the market basket. CPI is a rough estimate of how much consumers are spending.

If $I(t)$ is the CPI at time $t$, then inflation rate at $t = c$ is \[\dfrac{\text{rate of change of CPI at } t = c}{\text{CPI at } t = c} = \dfrac{I'(c)}{I(c)}\] so if $I'(t)$ were positive, then consumers are paying more for goods and services (inflation).

Consider $I''(t)$. Intuitively, it is telling us the rate of change of inflation (the $I(t)$ term can be ignored, it is a positive number). If $I''(t) < 0$, this is telling us the rate of change of inflation is negative.

It is possible for $I'(t) > 0$ and $I''(t) < 0$. This means we are experiencing inflation but the rate at which inflation is growing is slowing down.

Suppose \[I(t) = -0.2t^3 + 3t^2 + 100 \qquad 0 \leq t \leq 9\] is the CPI of an economy.
  1. Find the inflation rate at $t = 6$.
  2. Show that the rate at which inflation is growing is slowing down.