Higher-Order Derivatives

Learning Outcomes

Explain the meaning of a higher-order derivative

The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of $y=f(x)$ can be expressed in any of the following forms:

f^{″} (x), f^{‴} (x), f^{(4)} (x), \dots, f^{(n)} (x)

$f''(x), \, f'''(x), \, f^{(4)}(x), \cdots ,f^{(n)}(x)$

y^{″}, y^{‴}, y^{(4)}, \dots, y^{(n)}

$y'', \, y''', \, y^{(4)}, \cdots ,y^{(n)}$

\frac{d^{2} y}{d x^{2}}, \frac{d^{3} y}{d x^{3}}, \frac{d^{4} y}{d x^{4}}, \dots, \frac{d^{n} y}{d x^{n}}

$\dfrac{d^2y}{dx^2}, \, \dfrac{d^3y}{dx^3}, \, \dfrac{d^4y}{dx^4}, \cdots,\dfrac{d^ny}{dx^n}$

It is interesting to note that the notation for $\frac{d^2y}{dx^2}$ may be viewed as an attempt to express $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ more compactly. Analogously,

$\frac{d}{dx}\left(\frac{d}{dx}\left(\frac{dy}{dx}\right)\right)=\frac{d}{dx}\left(\frac{d^2y}{dx^2}\right)=\frac{d^3y}{dx^3}$

Example: Finding a Second Derivative

For $f(x)=2x^2-3x+1$ , find $f''(x)$ .

Show Solution

First find $f^{\prime}(x)$ .

\begin{array}{lllll} f^{'} (x) & = lim_{h \to 0} \frac{(2 (x + h)^{2} - 3 (x + h) + 1) - (2 x^{2} - 3 x + 1)}{h} & \begin{array}{l} Substitute f (x) = 2 x^{2} - 3 x + 1 \\ and \\ f (x + h) = 2 (x + h)^{2} - 3 (x + h) + 1 \\ into f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} . \end{array} \\ = lim_{h \to 0} \frac{4 x h + h^{2} - 3 h}{h} & Simplify the numerator. \\ = lim_{h \to 0} (4 x + h - 3) & \begin{array}{l} Factor out the h in the numerator \\ and cancel with the h in the \\ denominator. \end{array} \\ = 4 x - 3 & Take the limit. \end{array}

$\begin{array}{lllll}f^{\prime}(x) & =\underset{h\to 0}{\lim}\frac{(2(x+h)^2-3(x+h)+1)-(2x^2-3x+1)}{h} & & & \begin{array}{l}\text{Substitute} \, f(x)=2x^2-3x+1 \\ \text{and} \\ f(x+h)=2(x+h)^2-3(x+h)+1 \\ \text{into} \, f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h}. \end{array} \\ & =\underset{h\to 0}{\lim}\frac{4xh+h^2-3h}{h} & & & \text{Simplify the numerator.} \\ & =\underset{h\to 0}{\lim}(4x+h-3) & & & \begin{array}{l}\text{Factor out the} \, h \, \text{in the numerator} \\ \text{and cancel with the} \, h \, \text{in the} \\ \text{denominator.} \end{array} \\ & =4x-3 & & & \text{Take the limit.} \end{array}$

Next, find $f''(x)$ by taking the derivative of $f^{\prime}(x)=4x-3$ .

\begin{array}{lllll} f^{″} (x) & = lim_{h \to 0} \frac{f^{'} (x + h) - f^{'} (x)}{h} & \begin{array}{l} Use f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} with f^{'} (x) in \\ place of f (x) . \end{array} \\ = lim_{h \to 0} \frac{(4 (x + h) - 3) - (4 x - 3)}{h} & \begin{array}{l} Substitute f^{'} (x + h) = 4 (x + h) - 3 and \\ f^{'} (x) = 4 x - 3. \end{array} \\ = lim_{h \to 0} 4 & Simplify. \\ = 4 & Take the limit. \end{array}

$\begin{array}{lllll} f''(x)& =\underset{h\to 0}{\lim}\frac{f^{\prime}(x+h)-f^{\prime}(x)}{h} & & & \begin{array}{l}\text{Use} \, f^{\prime}(x)=\underset{h\to 0}{\lim}\frac{f(x+h)-f(x)}{h} \, \text{with} \, f^{\prime}(x) \, \text{in} \\ \text{place of} \, f(x). \end{array} \\ & =\underset{h\to 0}{\lim}\frac{(4(x+h)-3)-(4x-3)}{h} & & & \begin{array}{l}\text{Substitute} \, f^{\prime}(x+h)=4(x+h)-3 \, \text{and} \\ f^{\prime}(x)=4x-3. \end{array} \\ & =\underset{h\to 0}{\lim}4 & & & \text{Simplify.} \\ & =4 & & & \text{Take the limit.} \end{array}$

Try It

Find $f''(x)$ for $f(x)=x^2$ .

Hint

We found $f^{\prime}(x)=2x$ in a previous checkpoint. Use the definition to find the derivative of $f^{\prime}(x)$

Show Solution

$f''(x)=2$

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

Example: Finding Acceleration

The position of a particle along a coordinate axis at time $t$ (in seconds) is given by $s(t)=3t^2-4t+1$ (in meters). Find the function that describes its acceleration at time $t$ .

Show Solution

Since $v(t)=s^{\prime}(t)$ and $a(t)=v^{\prime}(t)=s''(t)$ , we begin by finding the derivative of $s(t)$ :

\begin{array}{ll} s^{'} (t) & = lim_{h \to 0} \frac{s (t + h) - s (t)}{h} \\ = lim_{h \to 0} \frac{3 (t + h)^{2} - 4 (t + h) + 1 - (3 t^{2} - 4 t + 1)}{h} \\ = 6 t - 4 \end{array}

$\begin{array}{ll}s^{\prime}(t) & =\underset{h\to 0}{\lim}\frac{s(t+h)-s(t)}{h} \\ & =\underset{h\to 0}{\lim}\frac{3(t+h)^2-4(t+h)+1-(3t^2-4t+1)}{h} \\ & =6t-4 \end{array}$

Next,

\begin{array}{ll} s^{″} (t) & = lim_{h \to 0} \frac{s^{'} (t + h) - s^{'} (t)}{h} \\ = lim_{h \to 0} \frac{6 (t + h) - 4 - (6 t - 4)}{h} \\ = 6 \end{array}

$\begin{array}{ll} s''(t) & =\underset{h\to 0}{\lim}\frac{s^{\prime}(t+h)-s^{\prime}(t)}{h} \\ & =\underset{h\to 0}{\lim}\frac{6(t+h)-4-(6t-4)}{h} \\ & =6 \end{array}$

Thus, $a=6 \, \text{m/s}^2$ .

Watch the following video to see the worked solution to Example: Finding Acceleration.

Closed Captioning and Transcript Information for Video

Try It

For $s(t)=t^3$ , find $a(t)$ .

Show Solution

$a(t)=6t$

Hint

Use the previous examples as a guide.

Module 3: Derivatives