Lecture 18: Higher-Order Derivatives

1.1 Definition of Multi-Order Derivatives

Representation of Multi-Order Derivatives

$(y^{\prime})^{\prime}=y^2=\frac{d^2y}{dx^2}\quad\quad y^{m}\quad\quad y^{(4)}\quad\quad y^{(n)}=\frac{d^ny}{dx^n}$

#####Definition: Advancedderivatives

Description: Specifically, the definition of an n-th order derivative is:

Definition 1: $f^{(n)}(x_0)=\lim_{\Delta x\to0}\frac{f^{(n-1)}(x_0+\Delta x)-f^{(n-1)}(x_0)}{\Delta x}$

Definition 2: $f^{(n)}(x_0)=\lim_{x\to x_0}\frac{f^{(n-1)}(x)-f^{(n-1)}(x_0)}{x-x_0}$

Explanation

• Conclusion:
• If function $f(x)$ is n-th differentiable at x, then in some neighborhood of x, $f(x)$ must have all derivatives below n-th order**;
• Concept:
• If $f^{(n)}(x)$ is continuous on interval $I$, $f(x)$ is said to be n-th-order continuously differentiable on $I$
• The n-th derivative exists, and it is continuous;

1.2 Common Ways to Find the n-th Order Derivative

1.2.1 Common Conclusions

Common Ways to Find n-th Order Derivatives

• Index
• $(e^{x})^{(n)}=e^{x}$
• Trigonometric function
• $(\sin x)^{(n)}=\sin(x+n\frac\pi2)$
• $(\cos x)^{(n)}=\cos(x+n\frac{\pi}{2})$
• In function
• $(\ln(1+x))^{(n)}=(-1)^{n-1}\frac{(n-1)!}{(1+x)^n}$
• Addition
• $(u\pm v)^{(n)}=u^{(n)}\pm v^{(n)}$
• Multiplication
• Leibniz formula
• $(uv)^{(n)}=\sum_{k=0}^n\binom{n}{k}u^{(k)}v^{(n-k)}$

Example Question: $\text{ Let }f(x)=\frac1{x^2-1}\text{ , please }f^{(n)}(x)$

Approach: Since addition or multiplication formulas cannot be used directly, the denominator needs to be factored and split:

• $\begin{aligned}f(x)&=\frac{1}{(x-1)(x+1)}=\frac{1}{2}\frac{(x+1)-(x-1)}{(x-1)(x+1)}\\&=\frac{1}{2}\left[\frac{1}{x-1}-\frac{1}{x+1}\right]\end{aligned}$

After the split, use the n-th order derivative of addition:

• $f^{(n)}(x)=\frac12\left[(\frac1{x-1})^{(n)}-(\frac1{x+1})^{(n)}\right]$
• Then let's analyze one first: $(\frac1{x-1})^{(n)}=(-1)^nn!(x-1)^{-(n+1)}$
• $f^{(n)}(x)=\frac{1}{2}\left[\frac{(-1)^{n}n!}{(x-1)^{n+1}}-\frac{(-1)^{n}n!}{(x+1)^{n+1}}\right].$

1.2.2 Higher-Order Differentiation of Composite Functions

Example: For example, find the derivative of $y=sin3x$;

• Here, sinx has a corresponding formula for the higher-order derivative derivative but compounds 3x;
• In this case, you can set u = 3x and then use the principle of differentiation with composite functions to propose the compound part caused by u, then substitute the original higher-order derivative formula;

1.3 Frequently Tested Question Types

Question Type: Higherorderderivativestofindthedifferentiation

PART 1: Problem-solving methods

Basic Approach

• 1. If you have a formula, substitute it;
• Find the n-th order derivative $f^{(n)}(x)$
• 2. If there is no formula, substitute the first and second order derivatives to find general patterns;
• Find the n-th order derivative $f^{(n)}(x)$
• 3. Expand using Taylor's formula
• Specify the point: $f^{(n)}(x_0)$ Develop Taylor at this point

PART 2: Typical Example Problems

Example Question: $\text{ Let the function be }\quad y=\frac1{2x+3},\quad\text{ then }y^{(n)}(0)=$?

• Analysis
• Analysis
• Method 1: Find patterns
• $\begin{aligned}&y^{\prime}=(-1)(2 x+3)^{-2}\cdot 2 \\&y^{\prime\prime}=(-1)(-2)(2 x+3)^{-3}\cdot 2^{2} \\&y^{(n)}=(-1)^{n}n! (2 x+3)^{-(n+1)}. 2^{n}\end{aligned}$
• Then substitute the derivative from point 0 to order n;
• Method 2: Use Taylor
• $f(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\cdots$
• Question Type: Higherorderderivativestofindthedifferentiation