Higher Order Derivatives

nth Derivatives If the derivative of $y = f(x)$ which is $y^{\prime} = f^{\prime}(x)$ is differentiable, then we can think of $(y^{\prime})^{\prime}$. We call this second derivative of $y = f(x)$ and write

$$y^{\prime\prime}, f^{\prime\prime}(x), \frac{d^{2}y}{dx^{2}}, D^{2}f(x)$$

If this second derivative is differentiable again, we can think of the 3rd derivatives. In this way, we can define nth derivative. If $f^{(n)}(x)$ exists, we say $f(x)$ is n times differentiable

Symbols

3rd derivative is denoted by $f'''(x)$ or $\frac{d^3 y}{dx^3}$. But the 4th derivative is not denoted by $f''''(x)$. Instead $f^{4. }(x)$ is used.

NOTE If $f^{(n)}(x)$ is continuous, we say $f(x)$ is class $C^{n}$ . Also, if for all $n$, $f^{(n)}(x)$exist. Then $f(x)$ is called infinitely differentiable or class $C^{\infty}$.

Infinitely Differentiable $\sin{x}, \cos{x}, e^x$ are infinitely differentiable on $(-\infty,\infty)$.

Properties of Higher Order Derivatives

Theorem 2..6   Suppose that $f(x)$ and $g(x)$ are in class $C^{n}$ and $c$ is constant. Then we have the following.

$${1. \{f(x) \pm g(x)\}^{(n)} = f^{(n)}(x) \pm g^{(n)}(x)}$$

$${2. \{cf(x)\}^{(n)} = cf^{(n)}(x)}$$

$${3. \{f(x)g(x)\}^{(n)} = \sum_{i=0}^{n}\binom{n}{i}f^{(n-i)}(x)g^{(i)}(x)}$$

$${4. \frac{dt}{dx} = c \mbox{implies} \frac{d^n f(t)}{dx^n} = c^n \frac{d^n f(t)}{dt^n}}$$

NOTE The theorem 3. is called general Leibnitz rule. $${\binom{n}{i} = \frac{n!}{i!(n-i)!}}$$

Proof of 3. Use induction on $n$. For $n=1$, we have

$$(f(x)g(x))^{\prime} = f^{\prime}(x)g(x) + f(x)g^{\prime}(x)$$

Next assume this theorem holds for $n=k$ and consider for $n=k+1$.

$$(f(x)g(x))^{(k+1)} = [(f(x)g(x))^{(k)}]^{\prime} = [ \sum_{i=0}^{k}\binom{k}{i}f^{(k-i)}(x)g^{(i)}(x)]^{\prime}$$

$$= \sum_{i=0}^{k}\binom{k}{i}f^{(k+1-i)}(x)g^{(i)}(x) + \sum_{i=0}^{k}\binom{k}{i}f^{(k-i)}(x)g^{(i+1)}(x)$$

$$= \sum_{i=0}^{k}\binom{k}{i}f^{(k+1-i)}(x)g^{(i)}(x) + \sum_{i=1}^{k+1}\binom{k}{i-1}f^{(k+1-i)}(x)g^{(i)}(x)$$

$$= f^{(k+1)}(x)g(x) + \sum_{i=1}^{k}(\binom{k}{i} + \binom{k}{i-1})f^{(k+1-i)}(x)g^{(i)}(x) + f(x)g^{(k+1)}(x)$$

$$= \sum_{i=0}^{k+1}\binom{k+1}{i}f^{(k+1-i)}(x)g^{(i)}(x).$$

Thus we have

Explanation

We show why $\binom{k}{i} + \binom{k}{i-1} = \binom{k+1}{i}$ holds. We study outcomes of the case when you draw 2 balls out of box containing 5 balls. Suppose you put some mark on one of the balls. Then the number of outcomes for drawing two balls out of 4 balls is $\binom{4}{2}$. Next, the number of outcomes for drawing two balls with one is marked is $\binom{4}{1}$. Thus, $\binom{4}{1} + \binom{4}{2} = \binom{5}{2}$.

$$\{f(x)g(x)\}^{(n)} = \sum_{i=0}^{n}\binom{n}{i}f^{(n-i)}(x)g^{(i)}(x)\ensuremath{ \blacksquare}$$

Proof of 4. Suppose $${\frac{dt}{dx} = c}$$. Then $${\frac{df(t)}{dx} = \frac{df(t)}{dt}\cdot \frac{dt}{dx} = c \frac{df(t)}{dt}}$$. Thus true for $n=1$. Now assume true for $n-1$ and consider for $n$.

$$\frac{d^n f(t)}{dx^n} = \frac{d(\frac{d^{n-1}f(t)}{dx^{n-1}})}{dx} = \frac{d(c^{n-1}\frac... ...}})}{dx} = c^{n-1}\frac{d(\frac{d^{n-1}f(t)}{dt^{n-1}})}{dt}\cdot \frac{dt}{dx}$$

$$= c^n \frac{d^n f(t)}{dt^n}\ensuremath{ \blacksquare}$$

Example 2..10   Find the nth derivative of the following functions.
$${1. f(x) = \sin{x}}$$ $${2. f(x) = \sin{2x}}$$ $${3. f(x) = \frac{1}{1 - x}}$$

Derivatives of $\sin,\cos$ If you start at $\sin x$, then rotate clockwise in the picture, now you get derivative of $\sin x$. Now note that if you add $\pi/2$ to $\sin{x}$, then you get $\cos{x}$. From this observation, $\sin(x+\pi/2) = \cos{x} = (\sin{x})'$

Figure 2.4:

SOLUTION 1.
$$\begin{array}{ll} f^{\prime}(x) &= \cos{x} = \sin{(x + \frac... ...\prime}(x) &= -\cos{x} = \sin{(x + \frac{3\pi}{2})} \end{array}$$
Then we can show $${f^{(n)}{x} = \sin{(x + \frac{n \pi}{2})}}$$ by induction $\blacksquare$
2. Let $t = 2x$. Then $${\frac{dt}{dx} = 2}$$. Thus,

$$\frac{d^n (\sin{2x})}{dx^n} = 2^n \frac{d^n(\sin{t})}{dt^n} = 2^n \sin(t + \frac{n \pi}{2}) = 2^n \sin(2x + \frac{n \pi}{2}).$$

3.

$$f^{\prime}(x) = \frac{-(1-x)'}{(1-x)^2} = \frac{1}{(1-x)^2} = (1-x)^{-2}$$

$$f^{\prime\prime}(x) = -2(1-x)^{-3}(-1) = 2(1-x)^{-3}$$

$$f^{\prime\prime\prime}(x) = -6(1-x)^{-4}(-1) = 6(1-x)^{-4}$$

Thus we can show $${f^{(n)}{x} = \frac{n!}{(1-x)^{n+1}}}$$ by induction $\blacksquare$

Basic Formula $$\begin{array}{l} (\sin{x})^{(n)} = \sin(x + \frac{n\pi}{2})\\... ...)\\ (\frac{1}{1-x})^{(n)} = \frac{n!}{(1-x)^{n+1}} \end{array}$$

Exercise 2..10   Find the nth derivative of the following functions.
$${1. \frac{1}{1-x^2}}$$ $${2. h(x) = x^3 e^{-x}}$$ $${3. \frac{x^3}{1-x}}$$

SOLUTION 1. Using the partial fraction, to write

$$\frac{1}{1 - x^{2}} = \frac{1}{2}(\frac{1}{1+x} + \frac{1}{1-x})$$

Then find $${(\frac{1}{1+x})^{(n)}}$$ and $${(\frac{1}{1-x})^{(n)}}$$.

$$(\frac{1}{1-x})^{(n)} = n!(1 - x)^{-(n+1)} = \frac{n!}{(1-x)^{n+1}}$$

Next consider $\frac{1}{1+x} = \frac{1}{1 - (-x)}$. Put $t = -x$. Then $\frac{dt}{dx} = -1$. Thus,

$$(\frac{1}{1+x})^{(n)} = \frac{(-1)^n n!}{(1+x)^{n+1}} .$$

Therefore,

$$\left(\frac{1}{1 - x^{2}}\right)^{(n)} = \frac{1}{2}\left((-1)^{n}n!(1 + x)^{-(n+1)} + n!(1 - x)^{-(n+1)}\right) \ensuremath{ \blacksquare}$$

2. Note that $(x^3)^{(4)} = 0$. Thus we let $g(x) = x^3$ and $f(x) = e^{-x}$ and use general Leibnitz rule,

$$h^{(n)}(x) = (f(x)g(x))^{(n)} = \sum_{k=0}^{n} \binom{n}{k} (f(x))^{(n-k)} (g(x))^{(k)}$$

$$= (-1)^n e^{-x}(x^3) + n((-1)^{n-1} e^{-x}(3x^2))$$

$$+ \frac{n(n-1)}{2}((-1)^{n-2}e^{-x}(6x)) + \frac{n(n-1)(n-2)}{3!}((-1)^{n-3}e^{-x}6. )$$

$$= (-1)^n e^{-x}(x^3 -3nx^2 + 3n(n-1)x - n(n-1)(n-2) \ensuremath{ \blacksquare}$$

Recall the partial fraction. Write $\frac{1}{1-x^2}$ as $\frac{1}{1-x^2} = \frac{A}{1+x} + \frac{B}{1-x}$. Then clear the denominator to have $1 = A(1-x) + B(1+x) = (-A+B)x + A+B$. Now this equation must be true for all $x$. Thus we have $\left\{\begin{array}{l} -A+B = 0\\ A+B = 0. \end{array}\right.$ From this, we have $A = \frac{1}{2}, B = \frac{1}{2}$.

Note that $n$th derivative of $x^3$ and $e^{-x}$ are $(x^{3})^{(n)} = 0 n > 3$ and $(e^{-x})^{(n)} = (-1)^{n}e^{-x}$. Then, in the Leibnitz theorem, we take $g(x) = x^3$.

The nth derivative of $x^3$ is 0 for $n > 3$, and the nth derivative of $(1-x)^{-1}$ is $((1-x)^{-1})^{(n)} = n!(1-x)^{-(n+1)}$. Thus we can use general Leibnitz rule. But it is usually not good way to solve.

3. Since the degree of the numerator $= 3 \geq 1 =$ the degree of the denominator, divide the numerator by the denominator.

$$\frac{x^3}{1-x} = -x^2 -x - 1 + \frac{1}{1-x}.$$

Since $(\frac{1}{1-x})^{(n)} = n!(1-x)^{-(n+1)} = \frac{n!}{(1-x)^{n+1}}$, we have

$$\big(\frac{x^3}{1-x}\big)^{(n)} = \left\{\begin{array}{ll} -2x - ... ...\frac{n!}{(1-x)^{n+1}} & n \geq 3 \end{array}\right.\ensuremath{ \blacksquare}$$

Exercise A

1.

When the motion of an object is given by the following equation, find the position, veclocity, and acceleration at $t_{0} = 0$

(a) $${y(t) = 4 + 3t - t^{2}}$$ (b) $${y(t) = t^{3} - 6t}$$ (c) $${y(t) = \frac{18}{t+2}}$$

2.

Find the second derivative of the following functions:

(a) $${f(x) = \sqrt{x^{2} + 1}}$$ (b) $${f(x) = x\log{x}}$$ (c) $${f(x) = e^{x} \sin{x}}$$

Exercise B

1.

Show the following formulas are true.

(a) $${(\sin{x})^{(n)} = \sin{(x + \frac{n \pi}{2})}}$$ (b) $${(\cos{x})^{(n)} = \cos{(x + \frac{n \pi}{2})}}$$

(c) $${\left[(1 + x)^{\alpha}\right]^{(n)} = \alpha(\alpha -1)\cdots(\alpha - n + 1)(1+x)^{\alpha - n}}$$

2.

Find the $n$-th derivative of the following functions.

(a) $${f(x) = \frac{x^{3}}{1 - x}}$$ (b) $${f(x) = x^{2} \sin{x}}$$ (c) $${f(x) = e^{x} \sin{x}}$$