Mathematics: Introduction to Calculus

• Calculus is the study of continous change
• Calculus is used to find the gradient of a graph, the maximum and minimum values, the area under or over a graph, inflection points, etc.

Derivation

• A derivation is a function which generalises some property of another function
• the derivative of $f(x)$ (say “f of x”) is $f\prime (x)$ (say “f prime of x”)

Rules (Copy into notes)

• $f\prime (x)=\frac{dy}{dx}x=y\prime$
• $f\prime (a)=0$ if $a$ is a number/constant
• $f\prime (ax)=a$ if $a$ is a number/constant
• $f\prime (a{x}^n)=an{x}^{n-1}$ if $a$ and $n$ are a numbers/constants
• $f\prime (e^x)=e^x$
• $f\prime (a^x)=\ln (a)a^x$
• $ln(x)=\frac{1}{\ln (x)}\cdot \frac{1}{x}$

Common Graphs

Quadratic

Cubic

Quartic

Hyperbola

Exponentials other than $e^x$

$ln$

First Principle

• The first principle is used to calculate the derivative of a graph

$f\prime x={\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$

Example of First Principle

Find the first derivative of $f(x)=x^4+5x+3$

1. Define f(x)

$f(x)=x^4+5x+3$

2. define f(x+h)

$f(x+h)=(x+h{)}^4+5(x+h)+3$

3. State the First Principle

$f\prime x={\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$

5. Substitute f(x) and f(x+h) into First Principle

$={\displaystyle \underset{h\to 0}{lim}\frac{(x+h{)}^4+5(x+h)+3-(x^4+5x+3)}{h}}$

6. Expand f(x) and f(x+h) where possible

$=\frac{x^4+4x^{3}h+6x^2{h}^2+4x{h}^3+h^4+5x+5h+3-x^4-5x-3}{h}$

7. Eliminate as many variables as possible from the numerator

$=\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4x{h}^3+h^4+\cancel{5x}+5h\cancel{+3}\cancel{-x^4}\cancel{-5x}\cancel{-3}}{h}$ $=\frac{4x^{3}h+6x^2{h}^2+4x{h}^3+h^4+5h}{h}$

8. factorise numerator to eliminate h

$=\frac{\cancel{h}(4x^3+6x^{2}h+4x{h}^2+h^3+5)}{\cancel{h}}$ \(=4x^3 +6x^2 h +4xh^2 +h^3 +5)

9. Substitute $h=0$

$(4x^3)\cancel{+(6x^2\cdot 0)}+\cancel{+(4x\cdot (0{)}^2)}\cancel{+(0{)}^3}+5$ $\boxed{=4x^3+5}$

10. Answer the question

$\boxed{\therefore \frac{d}{dx}(x^4+5x+3)=4x^3+5}$

Shortcuts for the First Princple

• Here’s some convenient shortcuts for people who don’t want to waste time doing all the steps of the First Principle
• Use these unless a question SPECIFICALLY asks for First Principle
• Note that anything except $x$, $y$ and $t$ represent NUMBERS
• Remember, differentiation is just finding the gradient of a curve!!!

1. $\frac{d}{dx}(a)=0$
2. $\frac{d}{dx}ax=a$
3. $\frac{d}{dx}a{x}^n=(a\cdot n)x^{n-1}$
4. $\frac{d}{dx}(A+B)=\frac{d}{dx}(A)+\frac{d}{dx}(B)$
5. $\frac{d}{dx}(A-B)=\frac{d}{dx}(A)-\frac{d}{dx}(B)$
6. $(AB)\prime =A\prime B+B\prime A$
7. $\frac{d}{dx}\frac{A}{B}=\frac{A\prime B-B\prime A}{B^2}$
8. $\frac{d}{dx}((f(x){)}^n)=n(f\prime (x){)}^{n-1}$
9. $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
10. $\frac{d}{dx}af(x)=a(f\prime (x))$

Graphing Derivatives

Chain Rule

• The chain rule is used when one function is acting on another (e.g. $f(g(x))$), and you need to find the derivative of the whole thing

$f(g(x))\prime =f\prime (g(x))g\cdot \prime (x)$

Steps

$\text{Example question: find }f\prime (x)\text{ where }f(x)=(3x^2+5{)}^{10}$

1. Bring the exponent to the front (Parentheses stay the same)

$10(3x^2+5{)}^{10}$

2. Subtract 1 from exponent

$10(3x^2+5{)}^9$

3. Multiply by derivative of inside

$10(3x^2+5{)}^9\cdot (6x)$

4. Solve and answer the question

$\therefore f\prime (x)=60x{(3x^2+5)}^9$

Product Rule (UWU Rule)

• The product rule is used to find the product of two differentials when given the original multipliers, for example:

$\text{Find }f\prime (x)\text{ where }f(x)=(2x+5)(4x^2+5)$

• The product rule is as follows:

$\text{If }f(x)=u(x)\cdot v(x)\text{, then }f\prime (x)=u(x)\prime v(x)+v(x)\prime u(x)$

Steps

1. Find $u\prime (x)$ and $v\prime (x)$

$u\prime (x)=2$ $v\prime (x)=8x$

2. Find $u\prime v$ and $v\prime u$

$u\prime v=2\cdot (4x^2+5)=8x^2+10$

$v\prime u=8x\cdot (2x+5)=16x^2+40x$

3. Solve and answer the question

$f\prime (x)=8x^2+10+16x^2+40x$

$\therefore f\prime (x)=24x^2+40x+10$

Quotient Rule

• The Quotient rule is used to find the quotient of two differentials when given the divisor and dividend, for example:

$\text{Find }f\prime (x)\text{ where }f(x)=\frac{2x+5}{4x^2+5}$

Yes I used the same functions. Deal with it 😎

- Pranav

Example Question

$\text{Find }f\prime (x)\text{ where }f(x)=\frac{2x+5}{4x^2+5}$

1. Find $u(x),u\prime (x),v(x),andv\prime (x)$

   $u(x)=2x+5$

   $u\prime (x)=2$

   $v(x)=4x^2+5$

   $v\prime (x)=8x$

2. Substitute values into the Quotient formula

   $=\frac{u\prime v-v\prime u}{v^2}$

   $f\prime (x)=\frac{2(4x^2+5)-8(2x+5)}{(4x^2+5{)}^2}$

3. Solve/Simplify

$2(4x^2+5)-8(2x+5)=8x^2-16x-30$

4. Answer the question

$\therefore f\prime (x)=\frac{8x^2-16x-30}{(4x^2+5{)}^2}$

Differentiating Sine, Cosine, and Tangent equations

TL;DR

• $\frac{dsin(x)}{dx}=cos(x)$
• $\frac{dcos(x)}{dx}=-sin(x)$
• $\frac{dtan(x)}{dx}=se{c}^2(x)$

(WHAT A PLOT TWIST) - Pranav Sharma

The Long Version

• When differentiating Sin, Cos and Tan, it’s important to remember that $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$
• The First Principles (I hate it too, but it’s occasionally useful) can be used to find the derivatives of Sin, Cos, and Tan.

First derivative of Sine ($f(x)=sin(x)$)

$f\prime x={\displaystyle \underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}}$

• So in the case of $f(x)=\sin (x)$, we have:

$f\prime x={\displaystyle \underset{h\to 0}{lim}\frac{\sin (x+h)-\sin (x)}{h}}$

• From this, we can use the Sum-Difference identity $\sin (A+B)=\sin A\cos B+\sin B\cos A$ to get

$f\prime (x)={\displaystyle \underset{h\to 0}{lim}\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}}$

$={\displaystyle \underset{h\to 0}{lim}\frac{\sin x(\cos h-1)+\sin h\cos x}{h}}$ $={\displaystyle \underset{h\to 0}{lim}(\frac{\sin x(\cos h-1)}{h}+\frac{\sin h\cos x}{h})}$

$=(\sin x){\displaystyle \underset{h\to 0}{lim}\frac{\cos h-1}{h}+\cos (x){\displaystyle \underset{h\to 0}{lim}\frac{\sin h}{h}}}$

• Based on our limits, and some lad named L’Hopital, we know that ${\displaystyle \underset{h\to 0}{lim}\frac{\sin h}{h}=1}$, and ${\displaystyle \underset{h\to 0}{lim}\frac{\cos h-1}{h}=0}$
• Therefore:

$f\prime (x)=(\sin x){\displaystyle \underset{h\to 0}{lim}\frac{\cos h-1}{h}+\cos (x){\displaystyle \underset{h\to 0}{lim}\frac{\sin h}{h}=1(\cos x)+0}}$ $\therefore f\prime (x)=\cos (x)$