Differentiation

P3 Textbook page 123 to 145

Part 1 - Operation rules

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 $x$

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

💡 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)

🛹 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

🚧 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)

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

✅ 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

💡 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

1. Beginner

👼 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

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

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

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

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

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

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

\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

Now Let’s prove the derivative!

\text{Let} \quad f(x) =\sin x\\[2em] \begin{aligned} 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{aligned} \\[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.

🔥 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

👼 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.

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

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 🙌🏼.

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