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 $a=3,\qquad b=-3,\quad\text{and}\quad c=1.$