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 \(y = f \left(\right. x \left.\right)\) can be expressed in any of the following forms:
\[f^{''} \left(\right. x \left.\right) , f ‴ \left(\right. x \left.\right) , f^{\left(\right. 4 \left.\right)} \left(\right. x \left.\right) ,\ldots , f^{\left(\right. n \left.\right)} \left(\right. x \left.\right)\]
\[y^{''} \left(\right. x \left.\right) , y ‴ \left(\right. x \left.\right) , y^{\left(\right. 4 \left.\right)} \left(\right. x \left.\right) ,\ldots , y^{\left(\right. n \left.\right)} \left(\right. x \left.\right)\]
\[\frac{d^{2} y}{d x^{2}} , \frac{d^{3} y}{d x^{3}} , \frac{d^{4} y}{d x^{4}} ,\ldots , \frac{d^{n} y}{d x^{n}} .\]
It is interesting to note that the notation for \(\frac{d^{2} y}{d x^{2}}\) may be viewed as an attempt to express \(\frac{d}{d x} \left(\right. \frac{d y}{d x} \left.\right)\) more compactly. Analogously, \(\frac{d}{d x} \left(\right. \frac{d}{d x} \left(\right. \frac{d y}{d x} \left.\right) \left.\right) = \frac{d}{d x} \left(\right. \frac{d^{2} y}{d x^{2}} \left.\right) = \frac{d^{3} y}{d x^{3}} .\)
Example
3.15
Finding a Second Derivative
For \(f \left(\right. x \left.\right) = 2 x^{2} - 3 x + 1 ,\) find \(f^{''} \left(\right. x \left.\right) .\)
Solution
First find \(f^{'} \left(\right. x \left.\right) .\)
\[f^{'} \left(\right. x \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{\left(\right. 2 \left(\left(\right. x + h \left.\right)\right)^{2} - 3 \left(\right. x + h \left.\right) + 1 \left.\right) - \left(\right. 2 x^{2} - 3 x + 1 \left.\right)}{h} & & & \begin{matrix}\text{Substitute} f \left(\right. x \left.\right) = 2 x^{2} - 3 x + 1 \\ \text{and} \\ f \left(\right. x + h \left.\right) = 2 \left(\left(\right. x + h \left.\right)\right)^{2} - 3 \left(\right. x + h \left.\right) + 1 \\ \text{into} f^{'} \left(\right. x \left.\right) = \underset{h \rightarrow 0}{\text{lim}} \frac{f \left(\right. x + h \left.\right) - f \left(\right. x \left.\right)}{h} .\end{matrix} \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{4 x h + 2 h^{2} - 3 h}{h} & & & \text{Simplify the numerator}. \\ & = \underset{h \rightarrow 0}{\text{lim}} \left(\right. 4 x + 2 h - 3 \left.\right) & & & \begin{matrix}\text{Factor out the} h \text{in the numerator} \\ \text{and cancel with the} h \text{in the} \\ \text{denominator}.\end{matrix} \\ & = 4 x - 3 & & & \text{Take the limit}.\]
Next, find \(f^{''} \left(\right. x \left.\right)\) by taking the derivative of \(f^{'} \left(\right. x \left.\right) = 4 x - 3 .\)
\[f^{''} \left(\right. x \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{f^{'} \left(\right. x + h \left.\right) - f^{'} \left(\right. x \left.\right)}{h} & & & \begin{matrix}\text{Use} f^{'} \left(\right. x \left.\right) = \underset{h \rightarrow 0}{\text{lim}} \frac{f \left(\right. x + h \left.\right) - f \left(\right. x \left.\right)}{h} \text{with} f^{'} \left(\right. x \left.\right) \text{in} \\ \text{place of} f \left(\right. x \left.\right) .\end{matrix} \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{\left(\right. 4 \left(\right. x + h \left.\right) - 3 \left.\right) - \left(\right. 4 x - 3 \left.\right)}{h} & & & \begin{matrix}\text{Substitute} f^{'} \left(\right. x + h \left.\right) = 4 \left(\right. x + h \left.\right) - 3 \text{and} \\ f^{'} \left(\right. x \left.\right) = 4 x - 3.\end{matrix} \\ & = \underset{h \rightarrow 0}{\text{lim}} 4 & & & \text{Simplify}. \\ & = 4 & & & \text{Take the limit}.\]
Checkpoint
3.9
Find \(f^{''} \left(\right. x \left.\right)\) for \(f \left(\right. x \left.\right) = x^{2} .\)
Example
3.16
Finding Acceleration
The position of a particle along a coordinate axis at time \(t\) (in seconds) is given by \(s \left(\right. t \left.\right) = 3 t^{2} - 4 t + 1\) (in meters). Find the function that describes its acceleration at time \(t .\)
Solution
Since \(v \left(\right. t \left.\right) = s^{'} \left(\right. t \left.\right)\) and \(a \left(\right. t \left.\right) = v^{'} \left(\right. t \left.\right) = s^{''} \left(\right. t \left.\right) ,\) we begin by finding the derivative of \(s \left(\right. t \left.\right) :\)
\[\begin{aligned} s^{'} \left(\right. t \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{s \left(\right. t + h \left.\right) - s \left(\right. t \left.\right)}{h} \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{3 \left(\right. t + h \left.\right)^{2} - 4 \left(\right. t + h \left.\right) + 1 - \left(\right. 3 t^{2} - 4 t + 1 \left.\right)}{h} \\ & = 6 t - 4. \end{aligned}\]
Next,
\[\begin{aligned} s^{''} \left(\right. t \left.\right) & = \underset{h \rightarrow 0}{\text{lim}} \frac{s^{'} \left(\right. t + h \left.\right) - s^{'} \left(\right. t \left.\right)}{h} \\ & = \underset{h \rightarrow 0}{\text{lim}} \frac{6 \left(\right. t + h \left.\right) - 4 - \left(\right. 6 t - 4 \left.\right)}{h} \\ & = 6. \end{aligned}\]
Thus, \(a = 6 \left( \text{m}/\text{s}\right)^{2} .\)
Checkpoint
3.10
For \(s \left(\right. t \left.\right) = t^{3} ,\) find \(a \left(\right. t \left.\right) .\)