## 3.7 Higher Order Derivatives

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 \(f\) is denoted by \(f”(x)\) or \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\), which is commonly abbreviated to \(\dfrac{d^{2}y}{dx^{2}}.\) Thus \(f^{\prime\prime}=(f’)’\). The second derivative is also indicated by \(y”\) 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 \(f'(x)\): \[y”=f”(x)=\frac{d^{2}y}{dx^{2}}=\frac{d^{2}f}{dx^{2}}=6x-10.\]

In a similar fashion, we can define the third derivative as the derivative of the second derivative. It is denoted by \[y”’=f”'(x)=\frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}};\] 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 \(f\) is said to be

**\(n\) times differentiable**at \(x_{0}\).

**Example 3.15**. If \(y=\sin
x\), find \(y^{(4)}.\)

**Solution**

\[\begin{aligned} y & =\sin x\\ y’ & =\frac{d}{dx}\sin x=\cos x\\ y^{\prime\prime} & =\dfrac{d}{dx}\cos x=-\sin x\\ y^{\prime\prime\prime} & =\dfrac{d}{dx}(-\sin x)=-\dfrac{d}{dx}\sin x=-\cos x\\ y^{(4)} & =\frac{d}{dx}(-\cos x)=-\frac{d}{dx}\cos x=-(-\sin x)=\sin x\end{aligned}\] Therefore, \(y^{(4)}=y=\sin x\).

**Example 3.16**. Determine \(a,b\), and \(c\) such that \(f”(x)\) exists everywhere if \[f(x)=\begin{cases}
x^{3} & \text{when }x\leq1\\
ax^{2}+bx+c & \text{when }x>1
\end{cases}.\]

**Solution**

Because \(f(x)x^{3}\)We have three
unknowns: \(a,b,c\) and hence we need
three equations. For \(f\) to have a
second derivative at \(x=1\), we
need:

(1) \(f\) to be continuous at \(x=1\),

(2) \(f\) to have a derivative at \(x=1\) or \(f’_{-}(1)=f’_{+}(1)\), and

(3) \(f_{+}”(1)=f_{-}”(1)\).

Now

(1) The continuity of \(f\) at \(x=1\) implies \[f(1)=1^{3}=a\cdot1^{2}+b\cdot1+c.\tag{i}\]
(2) \(f\) has a derivative at \(x=1\). Thus \[\begin{aligned}
f’_{-}(1) & =f’_{+}(1)\\
\left.3x^{2}\right|_{x=1} & =\left.2ax+b\right|_{x=1}\\
3 & =2a+b.\tag{ii}\end{aligned}\] (3) \(f\) has a second derivative at \(x=1\). Thus \[\begin{aligned}
f”_{-}(1) & =f”_{+}(1)\\
\left.6x\right|_{x=1} & =\left.2a\right|_{x=1}\\
6 & =2a\tag{iii }\end{aligned}\] From (i), (ii), and (iii),
we conclude \[a=3,\qquad
b=-3,\quad\text{and}\quad c=1.\]

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