Differentiation

P3 Textbook page 123 to 145

Part 1 - Operation rules 运算法则

1.The Chain Rule

👼 Basics:

\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\\[1em] \text{If }y=(f(x))^n\text{ , then }\frac{dy}{dx}=n(f(x))^{n-1}f^{\prime}(x)\\[1em] \text {If } y=f(g(x)) \text { , then } \frac{d y}{d x}=f^{\prime}(g(x)) g^{\prime}(x)

Where $y$ and $u$ are two functions of $x$

$y$ 和 $u$ 是关于 $x$ 的两个方程

The second and third expressions are just different ways to express the first one

第二和第三条表达，其实和第一条一样, 只是写法不同，下一个 Example 会说明

💡 Example 1-1

y=(2x^4+x)^5\qquad \frac{dy}{dx}=? \\[2em] \text{Let } u=2x^4+x\\[2ex] \therefore y=u^5\\[2ex] \frac{du}{dx}=8x^3+1\\[3ex] \frac{dy}{du}=5u^4\\[3ex] \because \frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}\\[3ex] \begin{aligned} \therefore \frac{dy}{dx} &=5u^4(8x^3+1)\\[2ex] &=5(2x^4+x)^4(8x^3+1) \end{aligned}

There’s no need to open the bracket(further simplification/reduction is not needed)

到这一步就不用再继续化简了

🛹 Trick

It’s actually very easy, you just have to think that this is a function in side

a function, like a smaller function wrapped in a bigger one So you need to

differentiate the bigger one then multiply the differentiation of the inside

wrapped function

你只需要把他们看成, 一个方程, 套了另一个方程. 那么我们需要先 differentiate 外面

的方程, 再乘以套在里面方程的 derivative 就好了

🚧 Note

We can also solve the equation by using the second and third formula in the

above Basics sections We can let $y=f(x)=3x^4+5$ , and we need to find the

derivative of $y^5=(f(x))^5$

Or we can let $g(x)=3x^4+5$ , $f(x)=x^5$ , and find the derivative of

$f(g(x))=(3x^4+5)^5$ (If you don’t understand this part, you need to revisit

Function)

如果要用 Basic 里面后两个表达式来解这一题

这里可以理解成 $y=f(x)=3x^4+5$ , 而现在要求 $y^5=(f(x))^5$ 的 derivative，

或理解成 $g(x)=3x^4+5$ , $f(x)=x^5$ , 而现在要求 $f(g(x))=(3x^4+5)^5$ 的 derivative

（如果你看不懂这里，那么你需要复习 Function）

You can practice Exercise 6C on page 130

2.The Product Rule

👼 Basics:

\text {If } y=uv \text { , then } \frac{\mathrm{d} y}{\mathrm{~d} x}=u \frac{\mathrm{d} v}{\mathrm{~d} x}+v \frac{\mathrm{d} u}{\mathrm{~d} x}

Where $v$ and $v$ are two functions of $x$

💡 Example 1-2

\text{Given that }f(x)=x^2\sqrt{3x-1}\qquad f^{\prime}(x)=?\\[2em] \text{Let } u=x^{2} \text{ and } v=\sqrt{3 x-1}=(3 x-1)^{\frac{1}{2}}\\[2ex] \therefore \frac{d u}{d x}=2x\qquad \frac{d v}{d x}=3 \cdot \frac{1}{2}(3 x-1)^{-\frac{1}{2}}\\[3ex] \because g \frac{d y}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\\[3ex] \begin{aligned} \therefore f^{\prime}(x)&=x^{2} \cdot \frac{3}{2}(3 x-1)^{-\frac{1}{2}}+\sqrt{3 x-1} \cdot 2 x\\[3ex] &=\frac{3 x^{2}+12 x^{2}-4 x}{2 \sqrt{3 x-1}} \\[3ex] &=\frac{15 x^{2}-4 x}{2 \sqrt{3 x-1}}=\frac{x(15 x-4)}{2 \sqrt{3 x-1}} \end{aligned}

You can practice Exercise 6D on page 133

3.The Quotient Rule

👼 Basics:

\text{If }y=\frac uv \text{ , then }\frac{dy}{dx}={{v\frac{du}{dx}-u\frac{dv}{dx}}\over v^2}\\[2em] \text{If }f(x)=\frac{g(x)}{h{(x)}}\text{ , then }f^{\prime}(x)=\frac{h(x) g^{\prime}(x)-g(x) h^{\prime}(x)}{(h(x))^{2}}

Where $u$ and $v$ are two functions of $x$

The second expression is just different ways to express the first one

同 chain rule, 第二个表达式只是第一则的另一种表达方式

✅ Quotient rule is just a special case of the product rule, $y=uv$ , where $v$ is now $v^{-1}$ ，it can be deduced by the product rule

Quotient rule 只是 product rule 的一种特殊形式，并且可以被 product rule 推导出来

只要把 $y=uv$ 中的 $v$ 改写成 $v^{-1}$ 就可以推导

💡 Example 1-3

\text{Given that }y=\frac {x}{2x+5}\qquad y^{\prime}(x)=?\\[2em] \text{Let } u=x \text{ and } v=2x+5\\[3ex] \therefore \frac{du}{dx}=1\qquad \frac{dv}{dx}=2\\[3ex] \because \frac{dy}{dx}={{v\frac{du}{dx}-u\frac{dv}{dx}}\over v^2}\\[3ex] \begin{aligned} \therefore y^{\prime}&=\frac{(2x+5)\cdot 1-x\cdot 2}{(2x+5)^2}\\[3ex] &=\frac{5}{(2x+5)^2} \end{aligned}

You can practice Exercise 6E on page 135

Part 2 - Trigs & Logs & Exponentials 三角函数等微分

1. Beginner 新手入门

👼 Basics:

y=\sin kx\qquad y^{\prime}=k\cos kx \\[2ex] y=\cos kx \qquad y^{\prime}=-k\sin kx \\[2ex] y=e^{kx} \qquad y^{\prime}=k\ e^{kx}\\[2ex] y=ln\ kx \qquad y^{\prime}=k \ \frac 1x \\[2ex]

⚠️ Proof the derivative of $sin\ x$ and $cos\ x$ by fisrt principal

第一原理证明三角函数的微分

First you need to know the small angle approximation(saa) for $sin$ and $cos$

首先你需要知道 small angle approximation (saa)

(saa = when x approaches 0, the value of $sin\ x \text{ and }cos\ x$ )

(saa = 当 $x$ 趋近于 0 时, $sin\ x$ 和 $cos\ x$ 的值)

$sin\ x\approx x$ , $cos\ x\approx1-\frac1 {2}x^2$ , $x$ is in radians

You don’t need to know how to prove saa, it is not required by the exam

你不需要知道如何证明 saa, 考试不要求掌握

(But if you are really interested in it, click here to see a video explanation)

(如果你对证明很感兴趣, 点这里查看视频证明)

Now by using saa we can find the value of following limits

现在通过 saa 我们可以推导出一下极限的值

\lim _{h \rightarrow 0} \frac{\sin h}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=1\\[2em] \lim _{h \rightarrow 0} \frac{\cos h-1}{h}=\lim _{h \rightarrow 0} \frac{1-\frac{1}{2} h^{2}-1}{h}=\lim \_{h \rightarrow 0}\left(-\frac{1}{2} h\right)=0

You don’t have to memorise the limit, if the relevant topic is tested, the question will provide the hint of the limit for you

以上两个极限求值不用记忆, 考试如果考到需要运用的题目, 会直接靠诉你以上两个条件

Now Let’s prove the derivative! 现在让我们来证明三角函数的微分！

\text{Let} \quad f(x)=\sin x\\[2em] \begin{align*} f^{\prime}(x) &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\[3ex] &=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}\\[3ex] &=\lim _{h \rightarrow 0} \frac{\sin x \cos h+\cos x \sin h-\sin x}{h} \\[3ex] &=\lim _{h \rightarrow 0} \left(\frac{cos h-1}{h}\right)\sin x + \left(\frac{\sin h}{h}\right)\cos x \end{align*}\\[3ex] \text{Since }\quad \frac{\cos h-1}{h} \rightarrow 0 \quad \text{ and } \quad \frac{\sin h}{h} \rightarrow 1 \\[3ex] \text{The expression inside the limit tends to}\\[3ex] 0 \times \sin x+1 \times \cos x\\[3ex] \text{So } \lim \_{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}=\cos x \\[3ex] \text{Hence the derivative of } \sin x \text{ is } \cos x.

Try the derivative of $\cos x$ yourself. The answer is on page 124 of P3 text book.

尝试自己推导 $\cos x$ 的微分，答案在 P3 教科书的第 124 页.

🔥 If you don’t have the textbook, hit us on Wechat, we will send you for free!

💡 Example 2-1

\text{If }y=4\cos 4x \qquad y^{\prime}=? \\[2em] \because y=\cos kx \qquad y^{\prime}=-k\sin kx\\[2ex] \therefore \ y^{\prime}=-4\sin 4x \cdot 4=-16\sin 4x

Easy, right? It’s just $\cos x$ with $4x$ plus the chain rule!

You can practice Exercise 6A on page 125

💡 Example 2-2

\text{If }y=4e^{2x} \qquad y^{\prime}=? \\[2em] \because \ y=e^{kx} \qquad y^{\prime}=k\ e^{kx}\\[2ex] \therefore \ y^{\prime}=4e^{2x}\cdot 2 = 8e^{2x}

💡 Example 2-3

\text{If }y=4ln2x \qquad y^{\prime}=? \\[2em] \because y=ln\ kx \qquad y^{\prime}=k\frac 1x\\[2ex] \therefore \ y^{\prime}=4 \cdot \frac 1{2x} \cdot 2= \frac{4}{x}

🕳️ Something trickier

y=a^x \qquad y^{\prime}=? \\[2em] \text{First we need to convert }a^x \text{to somthing with base } e^? \\[2ex] \begin{align*} y&=a^x\\ &=e^{\ln a^x}\\ &=e^{x\ln a} \end{align*}

⚠️This is because:

Let’s assume $e^b=a^x$

Then $\log_e{a^x}=b$

$\therefore\ a^x=e^b= e^{log_ea^x}=e^{\ln a^x}$

If you don’t understand this, you need to revisit logarithms

Now we can continue our differentiation:

\begin{align*} y^{\prime}&=e^{x\ln a}\cdot \ln a\\[1ex] &=a^x\ln a \end{align*}

⚠️You don’t have memorize this, it will be on the formula sheet

But I highly recommend you to practice it

Remember $\ln a$ is just a constant so the derivative of $x\ln a$ is just $\ln a$

⛳ Example 2-4

\text{If }y=3e^{3x}+2^{3x} \qquad y^{\prime}=? \\[2em] y=3e^{3x}+e^{3x\ln 2} \\[2ex] \ y^{\prime}=3e^{3x}\cdot\ 3\ +2^{3x}\ \cdot\ 3\ln 2

You can practice Exercise 6B on page 127

2. Intermediate 中级进阶

👼 Basics

\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\\[3ex] y=\arcsin x \qquad \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}\\[3ex] y=\arccos x \qquad \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}\\[3ex] y=\arctan x \qquad \frac{dy}{dx}=\frac{1}{1+x^{2}}

$arc$ is the inverse function of the trigonometric functions

If $\sin x=b$ , then $\arcsin b =x$

💡 Example 2-5

y=\arcsin x \qquad y^{\prime}=? \\[2em] \therefore\ \sin y=x\\[2ex] \frac{dx}{dy}=\cos y\\[3ex] \sin^2y+\cos^2y=1\\[2ex] \frac{dx}{dy}=\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}\\[3ex] \therefore\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}

Try prove the rest by yourself. The examples are on P3 textbook page 139 to140

❓ Your turn

y=\sin^5x \qquad y^{\prime}=?

Hint

$\sin^5x=(\sin x)^5=\left(f(x)\right)^5$ Use the product rule

🗝️Answer key

y^{\prime}=5\ \cdot\ \sin ^4 x\ \cdot\ \cos x

❓ A Harder one

y=x^{2} \tan \frac{1}{2} x+\tan \left(x-\frac{1}{2}\right) \qquad

Hint

$\tan x=\frac{\sin x}{\cos x}$ , use the quotient rule and the first part of the equation is product rule ( $y=uv$ )

🗝️Answer key

y=x^{2} \tan 2x+\tan \left(x-\frac{1}{2}\right)\qquad y^{\prime}=?\\[2em] u=x^2 \qquad v=\tan 2x \\[3ex] u^{\prime}=2x\\[3ex] \begin{align*} v^{\prime}=\left(\frac {\sin 2x}{ \cos 2x}\right)^{\prime} &={\cos 2x \cdot 2 \cdot \cos 2x \ -\ (-\sin 2x)\cdot 2 \cdot \sin 2x \over \cos^2 2x}\\[3ex] &={2\cos^2 2x \ + \ 2\sin^2 2x\over \cos^2 2x}\\[3ex] &={2(\cos^2 2x +\sin^2 2x)\over \ cos^2 2x}\\[3ex] &=\frac 2{\cos^2 2x}\\[3ex] &=2\sec^2 2x \end{align*} \\[3ex] \text{Now we know:}\\[3ex] \tan kx =k\sec kx\\[3ex] \therefore \text{The second part of the quation is:}\\[2ex] \left( \tan \left(x-\frac{1}{2}\right) \right)^{\prime}=\sec^2 \left( {x- \frac 12} \right) \cdot 1\\[3ex] \begin{align*} \therefore \ y^{\prime} &=u^{\prime}v+v^{\prime}u+\sec^2 \left( {x- \frac 12} \right)\\[3ex] &=2x \cdot \tan 2x +\ 2\sec^2 2x \cdot x^2 \ + \sec^2 \left( {x- \frac 12} \right) \end{align*}

When you finish the question, or you read the answer key, you will find than we can also derivative $\tan x$ and other trigonometric equations.

Following are the cheat sheet for the derivative of further trigonometric functions. But we high recommend you be familiar with them.

You don’t have to memorize these. These will be on provided on the formula book during exams.

当你做完上面这道题, 或者读完解析, 你会发现我们还可以 derivative $\tan x$ 等其他三角函数

以下是三角函数微分的公式, 你不用记住他们, 考试的时候会随公示表提供. 但我们强烈建议你熟悉这些公式.

Formula Book

👼 Further trig derivatives

\begin{aligned}&y=\tan k x \qquad \frac{dy}{dx}=k \sec ^{2} k x \\[3ex] &y=\operatorname{cosec} k x \quad \frac{{d} y}{dx}=-k \operatorname{cosec} k x \cdot \cot k x \\[3ex] &y=\sec k x \qquad \frac{\mathrm{d} y}{{d} x}=k \sec k x \cdot \tan k x \\[3ex] &y=\cot k x \qquad \frac{{d} y}{{d} x}=-k \operatorname{cosec}^{2} k x\end{aligned}

Try prove other trig functions than $\tan x$ , the examples are in page 137 to 138 尝试自己推倒一下其他方程的 derivative, 例子在书上 137 to 138

You can practice Exercise 6F on page 140

🎆 Congrats 👏! This is the end of the chapter, complete the exercise between page 123 and 143 on the text book. You can skip some if they are too simple, but we highly recommend 💪 you to finish all the questions in the “Chapter review” section (Page 142). If you don’t have the textbook or the answers to the practice questions, 🔥 hit us on Discord we will send you for free 🙌🏼.

恭喜㊗️!本章结束了, 请完成教科书 123 至 143 之前的作业. 如果你觉得简单, 可以跳过一些练习. 但是我们强烈 💪 你完成”Chapter review”部分的习题 (142 页). 如果你没有教材, 或者练习题的答案, 🔥 加我们的 Discord 或者微信, 我们会免费发送给你 🙌🏼.

We know it is a big chapter, so start practicing💨💨💨 The more you practice, the better you get ‼️