The Derivative as a Function: Learn It 4

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The Derivative as a Function: Learn It 4

Higher-Order Derivatives

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,

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

higher-order derivatives

Higher-order derivatives are the derivatives of a function taken multiple times. The first derivative is the rate of change of the function, the second derivative is the rate of change of the first derivative, and so on.

Notation: Common notations for higher-order derivatives include:

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}}

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

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

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

\begin{matrix} f^{'} (x) & = lim h \to 0 \frac{(2 (x + h)^{2} - 3 (x + h) + 1) - (2 x^{2} - 3 x + 1)}{h} & \begin{matrix} 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{matrix} = 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{matrix} Factor out the h in the numerator and cancel with the h in the denominator. \end{matrix} = 4 x - 3 & Take the limit. \end{matrix}

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

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

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

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

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 $t$ .

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

\begin{matrix} 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{matrix}

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

Next,

\begin{matrix} 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{matrix}

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

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

Watch the following video to see the worked solution to this example.

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