Differentiation Part1

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Notes

Learning Objectives (Syllabus)

use the notations $f'(x), f''(x), \frac{dy}{dx},\frac{{d}^{2}y}{d{x}^{2}}$
use the derivative of ${x}^{n}\,$ (for any rational number $n\,$ ), ${e}^{x},\ln x, \sin x, \cos x, \tan x \,$
carry out differentiation of $kf(x),f(x)\pm g(x),f(x)g(x),\frac{f(x)}{g(x)},\left(f\circ g\right)(x)$
find the first derivative of an implicit function
find the first derivative of a function defined parametrically

Prior Knowledge

Strong basics in index is a MUST in this topic
Strong in algebraic manipulations (factoring, rearranging, fractions, etc) is also a MUST for simplifying

Notation

When we differentiate $y=f(x)\,$ with respect to (w.r.t.) $x\,$ we will obtain the derivative which can be written in any of the following form

$\frac{dy}{dx}$
$f'\left(x\right)$
$\frac{d}{dx}\left[f\left(x\right)\right]$

Basic Formulas

Constants

A) $\frac{d}{dx}c$ $=0\,$ ( $c\,$ is a constant)

Remember : Derivative of ANY constant is zero
Note :
- Which of the following are constants? $1,-50,\frac{1}{2}, \ln 2, \pi, e, \frac{\pi}{100},{e}^{2}+2$ . ALL
- Any other "alphabets" may/may not be constants (it will be stated in the question)

X power n

B) $\frac{d}{dx}{x}^{n}$ $=n{x}^{n-1}\,$

Remember :
Note :
- Terms like $\frac{1}{x},\sqrt{x},\sqrt[3]{x},\frac{1}{\sqrt{x}}$ can be changed to the ${x}^{n}\,$ form before differentiating

Examples

$\frac{d}{dx}\left({x}^{6}\right)$ $=6{x}^{5}\,$ $\frac{d}{dx}\left({x}^{{\color{Red}6}}\right)={\color{Red}6}{x}^{{\color{Blue}5}}$
$\frac{d}{dx}\left({x}^{-10}\right)$ $=-10{x}^{-11}\,$ Be careful, $-10-1\,$ is $-11\,$ not $-9\,$
$\frac{d}{dx}\left(x\right)$
$\frac{d}{dx}\left(\frac{1}{x}\right)$
$\frac{d}{dx}\left(\sqrt{x}\right)$ $=\frac{1}{2\sqrt{x}}$ $\frac{d}{dx}\left(\sqrt{x}\right)=\frac{d}{dx}\left({x}^{{\color{Red}\frac{1}{2}}}\right)={\color{Red}\frac{1}{2}}{x}^{{\color{Blue}-\frac{1}{2}}}=\frac{1}{2\sqrt{x}}$
$\frac{d}{dx}\left(\frac{1}{{x}^{4}}\right)$
$\frac{d}{dx}\left(\frac{1}{\sqrt[3]{{x}^{2}}}\right)$ $=-\frac{2}{3{x}^{\frac{5}{3}}}$ $\frac{d}{dx}\left(\frac{1}{\sqrt[3]{{x}^{2}}}\right)=\frac{d}{dx}\left({x}^{{\color{Red}-\frac{2}{3}}}\right)={\color{Red}-\frac{2}{3}}{x}^{{\color{Blue}-\frac{5}{3}}}=-\frac{2}{3{x}^{\frac{5}{3}}}$

Useful Formulas

These should be memorized

$\frac{d}{dx}\left(x \right)=1$ Thus, $\frac{d}{dx}\left(cx \right)=c$ (if $c\,$ is constant) - refer below
$\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{{x}^{2}}$
$\frac{d}{dx}\sqrt x=\frac{1}{2\sqrt{x}}$

Plus/Minus/Multiply with Constant

C) $\frac{d}{dx}\left[f\left(x\right)\pm g\left(x\right)\right]$ $=\frac{d}{dx}\left[f\left(x\right)\right]\pm\frac{d}{dx}\left[g\left(x\right)\right]$

Remember :

D) $\frac{d}{dx}c\left[f\left(x\right)\right]$ $=c\frac{d}{dx}\left[f\left(x\right)\right]$ ( $c\,$ is a constant)

Remember :

Note :
Note the difference between
- $\frac{d}{dx}c$ $=0\,$
- $\frac{d}{dx}c\left[f\left(x\right)\right]$ $=c\frac{d}{dx}\left[f\left(x\right)\right]$
- differentiate constant $=0\,$
- differentiate constant times function $\rightarrow$ leave the constant outside

Note :

Example

$y=3{x}^{2}+\left(\pi+1\right)x-\frac{2}{x}+{\pi}^{2}-4{a}^{3}+1$ where $a\,$ is a constant.

$\frac{dy}{dx}=6x+\left(\pi+1\right)+\frac{2}{{x}^{2}}$
- Analyze :
- Differentiate $\frac{1}{x}$ is $-\frac{1}{{x}^{2}}$ , thus it becomes ${\color{Red}+}\frac{2}{{x}^{2}}$
- Thus, $\frac{dy}{dx}=6x+\left(\pi+1\right)+\frac{2}{{x}^{2}}$

Writing the Working

When writing the working out, always remember the function and its derivative

Example 1 :
- Correct? WRONG $6x^{2}\,$ is definitely NOT EQUAL to $12x\,$ .
- It MUST be written as $\begin{align}y &=6x^{2} \\ \frac{dy}{dx}& = 12x \end{align}$

Example 2 :
- How should we correct it? $\begin{align} y &=\sqrt{x}\\ &= {x}^{\frac{1}{2}}\\ \frac{dy}{dx}&=-\frac{1}{2}{x}^{-\frac{1}{2}}\\ &=-\frac{1}{2\sqrt{x}} \end{align}$
- Notice that

Example 3 :
- WRONG.
- $\begin{align} y&={\color{Red}{x}^{2}+\sqrt{x}}\\ &= {\color{Red}{x}^{2}+{x}^{\frac{1}{2}}}\\ \frac{dy}{dx}&={\color{Blue}2x-\frac{1}{2}{x}^{-\frac{1}{2}}}\\ &={\color{Blue}2x-\frac{1}{2\sqrt{x}}} \end{align}$

e^x, ln x, sin x, cos x, tan x

Note : We don't really need to know the why's of each formula below, just need to memorize it.

E) $\frac{d}{dx}{e}^{x}$ $={e}^{x}\,$

Remember : differentiate ${e}^{x}\,$ we get back ${e}^{x}\,$ itself.

F) $\frac{d}{dx}{a}^{x}$ $={a}^{x}\ln a \,$

Remember :

G) $\frac{d}{dx}\ln x$ $=\frac{1}{x}$

Remember : differentiate $\ln x\,$ we get $\frac{1}{x}$

H) $\frac{d}{dx}\sin x$ $=\cos x \,$

Remember : differentiate $\sin x\,$ we get $\cos x\,$

I) $\frac{d}{dx}\cos x$ $=-\sin x \,$

Remember : differentiate $\cos x\,$ we get ${\color{Red}-}\sin x\,$

J) $\frac{d}{dx}\tan x$ $={\sec}^{2}x \,$

Remember : differentiate $\tan x\,$ we get $={\sec}^{{\color{Red}2}}x$

Examples

Find $f'\left(x\right)$ for the following

- $f'\left(x\right)={e}^{x}+\frac{1}{x}$
- $f'\left(x\right)=2\cos x+\sin x + \frac{1}{3}{\sec}^{2}x$
- $f'\left(x\right)=2\left(12{x}^{3}-2\cos x \right)$

Exercise 1

Differentiate the following w.r.t. $x\,$

a) $f\left(x\right)={x}^{7}+5{x}^{2}-3{x}^{-2}+3{x}^{\frac{1}{2}}$

$f'\left(x\right)=7{x}^{6}+10x+6{x}^{-3}+\frac{3}{2}{x}^{-\frac{1}{2}}$

b) $f\left(x\right)=2{x}^{5}-4x+\frac{3}{x}+5\sqrt{x}-\frac{2}{\sqrt{x}}$

$f'\left(x\right)=10{x}^{4}-4-\frac{3}{{x}^{2}}+\frac{5}{2\sqrt{x}}+\frac{1}{{x}^{\frac{3}{2}}}$

c) $f\left(x\right)=2\ln x-{e}^{x}+{2}^{x}-{3}^{x}$

$f'\left(x\right)=\frac{2}{x}-{e}^{x}+{2}^{x}\ln 2-{3}^{x}\ln 3$

d) $f\left(x\right)=4\tan x+7\cos x-3\sin x$

$f'\left(x\right)=4{\sec}^{2}x- 7\sin x-3\cos x$

e) $f\left(x\right)=\frac{2}{{x}^{4}}-\frac{\ln x}{4}+\frac{2\tan x}{3}$

$f'\left(x\right)=-\frac{8}{{x}^{5}}-\frac{1}{4x}+\frac{2}{3}{\sec}^{2}x$

Products and Quotients

K) if $y=uv\,$ , where $u,v\,$ are functions of $x\,$

$=u\frac{dv}{dx}+v\frac{du}{dx}$
- Remember :

L) if $y=\frac{u}{v}$ , where $u,v\,$ are functions of $x\,$

$=\cfrac{v\cfrac{du}{dx}-u\cfrac{dv}{dx}}{{v}^{2}}$
- Remember :

Important Note:
are functions of , if constants, the formulas are NOT needed (just use formula D)
- Example : $\left(x+2\right)\left(x-2\right), x{e}^{x}, {x}^{2}{\sin x}, \frac{\tan x}{x}$
- Example : $3{e}^{x},\pi \sin x, \frac{\tan x}{3}$

Differentiating Complicated Functions

When $u,v\,$ are complicated functions, we can differentiate it separately away from the main workings.

Say, for example, we have ${e}^{\sin x}\cos \left(ax+b\right)$ , we can see it's something times something ${\color{Red}{e}^{\sin x}}{\color{Blue}\cos \left(ax+b\right)}$ . Differentiating those parts aren't really that difficult once we get the hang of it, but it's very easy to make careless mistakes if we do it too quickly. My suggestion is to do some side working.

One way we could do is $\frac{dy}{dx}$
Instead, we do it as side working
- First we write $\frac{dy}{dx}={e}^{\sin x}$
- $\begin{array}{ll} y={e}^{\sin x}\qquad\qquad\qquad\qquad\qquad\quad & \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}$
- $\begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)\qquad\quad\qquad\qquad\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}$
- $\begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit} \end{array}$
- $\begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]\qquad\;& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit}\quad \cfrac{d}{dx}{e}^{\sin x}={\color{Blue}\bigcirc} \end{array}$
- $\begin{array}{ll} y={e}^{\sin x}\left({\color{Red}\heartsuit}\right)+\left[\cos\left(ax+b\right)\right]{\color{Blue}\bigcirc}\quad& \cfrac{d}{dx}\cos \left(ax+b\right)={\color{Red}\heartsuit}\quad \cfrac{d}{dx}{e}^{\sin x}={\color{Blue}\bigcirc} \end{array}$

Examples

Differentiate the following w.r.t $x\,$

$\frac{dy}{dx}=x\cos x+\sin x$
- Analyze :
  - Copy first thing : $\frac{dy}{dx}={\color{Red}x}$
  - Multiply differentiate second thing: $\frac{dy}{dx}={\color{Red}x}{\color{Blue}\cos x}$
  - Plus copy second thing: $\frac{dy}{dx}={\color{Red}x}{\color{Blue}\cos x}+{\color{Blue}\sin x}$

$\frac{dy}{dx}=\frac{{x}^{2}+1}{x}+2x\ln x$
- Analyze : Something times something. ${\color{Red}\left({x}^{2}+1\right)}{\color{Blue}\ln x}$

$\frac{dy}{dx}=\frac{{e}^{x}\left(x-1\right)}{{x}^{2}}$
- Analyze :
$\frac{dy}{dx}=\frac{2x\ln x+2x+1}{x{\left(\ln x\right)}^{2}}$
- Analyze :
  - Differentiate as usual : $\frac{dy}{dx}=\frac{\left(\ln x\right)2+\cfrac{\left(2x+1\right)}{x}}{{\left(\ln x\right)}^{2}}$
  - Using the same method, $\frac{dy}{dx}=\frac{2{\color{Red}x}\ln x+2x+1}{{\color{Red}x}{\left(\ln x\right)}^{2}}$

$\frac{dy}{dx}=\frac{{e}^{x}\left(2x-1\right)}{2{x}^{\frac{3}{2}}}$
- Analyze : Something over something.
  - Thus, differentiating, we get $\frac{dy}{dx}=\frac{\sqrt{x}\left({e}^{x}\right)-{e}^{x}\left(\cfrac{1}{2\sqrt{x}}\right)}{{\left(\sqrt{x}\right)}^{2}}$
  - $\frac{dy}{dx}=\frac{2x\left({e}^{x}\right)-{e}^{x}}{2\sqrt{x}\left(x\right)}$

Exercise 2

Differentiate the following w.r.t. $x\,$

a) $y=\left(3x+1\right)\ln x$

$\frac{dy}{dx}=\frac{\left(3x+1\right)}{x}+3\ln x$

b) $y={x}^{2}\cos x\,$

$\frac{dy}{dx}=x\left(2\cos x-x\sin x \right)$
- $\begin{align} y &={x}^{2}\cos x \\ \frac{dy}{dx} & =-{x}^{2}\sin x+2x\cos x\\ & =x\left(2\cos x-x\sin x \right) \end{align}$

c) $y ={e}^{x}\sin x\,$

$\frac{dy}{dx}={e}^{x}\left(\cos x+\sin x\right)$
- $\begin{align} y &={e}^{x}\sin x \\ \frac{dy}{dx} & ={e}^{x}\cos x +\left( \sin x\right){e}^{x} \\ &= {e}^{x}\left(\cos x+\sin x\right) \end{align}$

d) $y =x{e}^{x}\cos x \,$

$\frac{dy}{dx}={e}^{x}\left(x\cos x-x\sin x+\cos x \right)$
- $\begin{align} y &=x{e}^{x}\cos x \\ &=\left(x{e}^{x}\right)\cos x \\ \frac{dy}{dx} & =-\left(x{e}^{x}\right)\sin x +\left(\cos x\right)\left(x{e}^{x}+{e}^{x} \right) \\ & ={e}^{x}\left(x\cos x-x\sin x+\cos x \right) \end{align}$ $\begin{align} & \mbox{Let } y= x{e}^{x} \\ & \frac{dy}{dx} = xe^{x}+e^{x} \end{align}$

e) $y =\left({e}^{x}-3\right)\left(\sin x+2x \right)$

$\frac{dy}{dx} =\left({e}^{x}-3\right)\left(\cos x+2\right)+{e}^{x}\left(\sin x+2x \right)$

f) $y ={e}^{x}\left(2\cos x-3{x}^{2} \right)$

$\frac{dy}{dx}={e}^{x}\left(-2\sin x+2\cos x-3{x}^{2}-6x\right)$
- $\begin{align} y &={e}^{x}\left(2\cos x-3{x}^{2} \right) \\ \frac{dy}{dx} & ={e}^{x}\left(-2\sin x-6x\right)+\left(2\cos x-3{x}^{2} \right){e}^{x}\\ &={e}^{x}\left(-2\sin x+2\cos x-3{x}^{2}-6x\right) \end{align}$

g) $y =\frac{x+1}{x+2}$

$\frac{dy}{dx}=\frac{1}{{\left(x+2\right)}^{2}}$
- $\begin{align} y &=\frac{x+1}{x+2} \\ \frac{dy}{dx} & = \frac{\left(x+2\right)-\left(x+1\right)}{{\left(x+2\right)}^{2}}\\ &= \frac{1}{{\left(x+2\right)}^{2}} \end{align}$

h) $y =\frac{6{x}^{3}}{{x}^{2}-1}$

$\frac{dy}{dx} = \frac{6{x}^{2}\left({x}^{2}-3\right)}{{\left({x}^{2}-1\right)}^{2}}$
- $\begin{align} y &=\frac{6{x}^{3}}{{x}^{2}-1} \\ \frac{dy}{dx} & = \frac{\left({x}^{2}-1\right)\cdot 18{x}^{2} -6{x}^{3}\cdot 2x}{{\left({x}^{2}-1\right)}^{2}}\\ &= \frac{6{x}^{2}\left(3{x}^{2}-3-2{x}^{2}\right)}{{\left({x}^{2}-1\right)}^{2}}\\ &= \frac{6{x}^{2}\left({x}^{2}-3\right)}{{\left({x}^{2}-1\right)}^{2}} \end{align}$

i) $y =\frac{1-x}{\sqrt{x+1}}$

$\frac{dy}{dx}=\frac{-x-3}{2{\left(x+1\right)}^{\frac{3}{2}}}$
- $\begin{align} y &=\frac{1-x}{\sqrt{x+1}} \\ \frac{dy}{dx} & = \frac{\sqrt{x+1}\left(-1\right)-\left(1-x\right)\cfrac{1}{2\sqrt{x+1}}}{x+1}\\ & = \frac{-2\left(x+1\right)-\left(1-x\right)}{2\sqrt{x+1}\left(x+1\right)}\\ & = \frac{-x-3}{2{\left(x+1\right)}^{\frac{3}{2}}}\\ \end{align}$

j) $y =\frac{\ln x}{x}$

$\frac{dy}{dx}=\frac{1-\ln x}{{x}^{2}}$
- $\begin{align} y &=\frac{\ln x}{x} \\ \frac{dy}{dx} & = \frac{x\left(\cfrac{1}{x}\right)-\ln x}{{x}^{2}}\\ & = \frac{1-\ln x}{{x}^{2}}\\ \end{align}$

k) $y=\frac{{e}^{x}}{{x}^{3}}$

$\frac{dy}{dx} =\frac{{e}^{x}\left(x-3 \right)}{{x}^{4}}$
- $\begin{align} y &=\frac{{e}^{x}}{{x}^{3}} \\ \frac{dy}{dx} & = \frac{{x}^{3}{e}^{x}-{e}^{x}\cdot 3{x}^{2}}{{x}^{6}}\\ & = \frac{{x}^{2}{e}^{x}\left(x-3 \right)}{{x}^{6}}\\ & = \frac{{e}^{x}\left(x-3 \right)}{{x}^{4}}\\ \end{align}$

l) $y=\frac{x+1}{\sin x}$

$\frac{dy}{dx} =\frac{\sin x-\left(x+1\right)\cos x}{{\left(\sin x\right)}^{2}}$

m) $y=\frac{\sin x}{\sqrt{x}}$

$\frac{dy}{dx} =\frac{2x\cos x-\sin x}{2{x}^{\frac{3}{2}}}$
- $\begin{align} y &=\frac{\sin x}{\sqrt{x}} \\ \frac{dy}{dx} &=\frac{\sqrt{x}\cos x-\left(\sin x \right)\left(\cfrac{1}{2\sqrt{x}} \right)}{x}\\ & = \frac{2x\cos x-\sin x}{2{x}^{\frac{3}{2}}} \end{align}$

n) $y=\frac{x}{2\ln x}$

$\frac{dy}{dx} =\frac{\ln x -1}{2{\left(\ln x \right)}^{2}}$
- $\begin{align} y &=\frac{x}{2\ln x}\\ \frac{dy}{dx} &=\frac{1}{2}\left[\frac{\ln x-x\left(\cfrac{1}{x}\right)}{{\left(\ln x \right)}^{2}}\right]\\ &=\frac{\ln x -1}{2{\left(\ln x \right)}^{2}} \end{align}$

Differentiation Part1

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Notation

Basic Formulas

Constants

X power n

Examples

Useful Formulas

Plus/Minus/Multiply with Constant

Example

Writing the Working

e^x, ln x, sin x, cos x, tan x

Examples

Exercise 1

Products and Quotients

Differentiating Complicated Functions

Examples

Exercise 2

Personal tools

Namespaces

Variants

Views

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