Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

Higher-order Derivatives

As we indicated in Definition 4.5, the derivative $f^{\prime }\left(x\right)$ of a function $f\left(x\right)$ is a function that we may be able to differentiate. Then, the derivative of the function $f^{\prime }\left(x\right)$ is called the second derivative function of $f$, and we write $f^{″}\left(x\right)$. [Note that $f^{\prime }\left(x\right)$ is referred to as the first derivative of $f$.] Notation must be established for this second derivative.

For a real number $a$,

${\displaystyle f^{″}\left(a\right)={\displaystyle \frac{d^2}{d{x}^2}}f\left(x\right){|}_{x=a}={\displaystyle \frac{d^{2}f}{d{x}^2}}{|}_{x=a}=D_x^{2}f\left(a\right)\text{,}}$

and for the second derivative function,

${\displaystyle f^{″}\left(x\right)={\displaystyle \frac{d^2}{d{x}^2}}f\left(x\right)={\displaystyle \frac{d^{2}f}{d{x}^2}}=D_x^{2}f\left(x\right)\text{.}}$

If $y=f\left(x\right)\text{,}$ then

${\displaystyle f^{″}\left(a\right)={\displaystyle \frac{d^{2}y}{d{x}^2}}{|}_{x=a}\text{ and }{f}^{″}\left(x\right)={\displaystyle \frac{d^{2}y}{d{x}^2}}\text{.}}$

Observe that this notation makes sense, because we are saying that

${\displaystyle f^{″}\left(x\right)={\displaystyle \frac{d}{dx}}{f}^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{df}{dx}}\right)={\displaystyle \frac{d^{2}f}{d{x}^2}}\text{.}}$

Example 4.64. The first derivative of $sec x$ is $sec x tan x\text{.}$ So, the second derivative of $sec x$ is

${\displaystyle \begin{array}{ll}\hfill {\displaystyle \frac{d}{dx}}\sec x\tan x& =\sec x{\text{ sec}}^{2}x+\sec x\tan x\tan x\hfill \\ \hfill & ={\sec }^{3}x+\sec x{\tan }^{2}x\text{,}\hfill \end{array}}$

and we write

${\displaystyle {\displaystyle \frac{d^2}{d{x}^2}}sec x={sec}^{\mathrm{3 }}x+sec x{ tan}^{\mathrm{2 }}x\text{.}}$

Example 4.65.

${\displaystyle {\displaystyle \frac{d^2}{d{x}^2}}\sqrt{x}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{d}{dx}}{x}^{1∕2}\right)={\displaystyle \frac{d}{dx}}{\displaystyle \frac{x^{-1∕2}}{2}}=-{\displaystyle \frac{x^{-3∕2}}{4}}=-{\displaystyle \frac{1}{4x^{3∕2}}}\text{.}}$

By analogy, the third derivative of $f$ is the derivative of the second derivative $f^{″}$ and we write $f^{‴}\text{.}$ Hence,

${\displaystyle f^{‴}\left(x\right)={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{d^{2}f}{d{x}^2}}\right)={\displaystyle \frac{d^{3}f}{d{x}^3}}=D_x^{3}f\left(x\right)\text{.}}$

In the same manner, we can continue with this process and obtain the fourth, fifth and any other higher-order derivative. In general, the $n$-th derivative of $y=f\left(x\right)$ is

${\displaystyle y^n=f^{\left(n\right)}\left(x\right)={\displaystyle \frac{d^{n}f}{d{x}^n}}={\displaystyle \frac{d^{n}y}{d{x}^n}}=D_x^{n}f\left(x\right)\text{ for }n\ge 4\text{.}}$