We have learned the meaning of the first derivative of a function. Now we want to know what the second, the third, and the n-th derivatives of a function are defined and how we can calculate them.
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If $s(t)$ is the position of an object moving on a straight line, then the derivative of s is the velocity of the object $v(t)=s'(t)$. The derivative of the velocity is the acceleration of the object $a(t)$. So $a(t)=v'(t)$ or $a(t)=(s’)'(t)$, which is often written simply as $a(t)=s^{\prime\prime}(t)$.We say the acceleration is the second derivative of the position. In physics, acceleration plays an important role as it appears in Newton’s second law $F=ma$.
In general, if we take the derivative of $y=f(x)$, we obtain a new function $f’$ (also denoted by $y’$ or $dy/dx$). We can take the derivative of $f’$ and obtain another function called the second derivative of $f$ (or $y$). The second derivative of $f(x)$ is denoted by \(f^{\prime\prime}(x)\) or \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\), which is commonly abbreviated to \(\dfrac{d^{2}y}{dx^{2}}.\) Thus $f^{\prime\prime}=(f’)^\prime$ or
\[\bbox[#F2F2F2,5px,border:2px solid black]{f^{\prime\prime}(x)=\lim_{\Delta x\to0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}.\tag{a}}\]
The second derivative is also indicated by \(y^{\prime\prime}\) or \(\dfrac{d^{2}f}{dx^{2}}\).
For example, if \(f(x)=x^{3}-5x^{2}+3x-1\), then the first derivative is \[y’=f'(x)=\frac{dy}{dx}=\frac{df}{dx}=3x^{2}-10x+3,\] and the second derivative of \(f\) is the derivative of \(f'(x)\):
\[y^{\prime\prime}=f^{\prime\prime}(x)=\frac{d^{2}y}{dx^{2}}=\frac{d^2f}{dx^2}=6x-10.\]
In a similar fashion, we can define the third derivative as the derivative of the second derivative. It is denoted by
\[y^{\prime\prime\prime}=f^{\prime\prime\prime}(x)=\frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}};\]
the fourth derivative is the derivative of the third derivative, and is denoted by
\[y^{(4)}=f^{(4)}(x)=\frac{d^{4}y}{dx^{4}}=\frac{d^{4}f}{dx^{4}},\] and so on. In general, the \(n\)-th derivative of \(y=f(x)\) is indicated by one of the following symbols:
\[y^{(n)}=f^{(n)}(x)=\frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}.\]
- If $f^{(n)}(x_0)$ exists, then it is said that $f$ is $n$-times differentiable at $x_0$.
- If a function is differentiable, its derivative is not necessarily differentiable. In other words, from the existence of \(f'(x_{0})\), we cannot infer the existence of \(f^{\prime\prime}(x_{0})\). For instance, see the following example.
- In the above example, \(f\) is differentiable (= \(f'(x)\) exists) everywhere. But because \(f'(x)\) is not continuous at \(x=0\), \(f^{\prime\prime}(0)\) does not exist.