Short and Sweet Calculus

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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