Calculus

Differentiation

It is the slope of the line. Given by $\frac{d y}{d x}$ . The more proper definition would be.

\frac{d y}{d x} = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

For example,

f (x) = x^{2}

\begin{array}{rcl} \frac{d y}{d x} & = & lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h} \\ = & lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} - x^{2}}{h} \\ = & 2 x \end{array}

Chain Rule

\frac{d}{d x} f (g (x)) = f^{'} (g (x)) \cdot g^{'} (x)

Product Rule

\frac{d}{d x} f (x) \cdot g (x) = f^{'} (x) g (x) + f (x) g^{'} (x)

Quotient Rule

\frac{d}{d x} \frac{f (x)}{g (x)} = \frac{f^{'} (x) g (x) - f (x) g^{'} (x)}{g (x)^{2}}

$f (x)$	$f^{'} (x)$
$x^{n}$	$n x^{n - 1}$
$e^{f (x)}$	$f^{'} (x) \cdot e^{f (x)}$
$a^{f (x)}$	$\ln a \cdot f^{'} (x) \cdot e^{f (x)}$
$\ln (x)$	$\frac{1}{x}$
$\sin x$	$\cos x$
$\cos x$	$- \sin x$
$\tan x$	$\sec^{2} x$
$\arcsin x$	$\frac{1}{\sqrt{1 - x^{2}}}$
$\arccos x$	$- \frac{1}{\sqrt{1 - x^{2}}}$
$\arctan x$	$\frac{1}{1 + x^{2}}$
$\sec x$	$\sec x \tan x$
$\csc x$	$- \csc x \cot x$
$\cot x$	$- \csc^{2} x$

Integration

Integration is just the opposite of differentiation

Integration by Substitution

Substitute a function as a variable.
Change the limits and dx. Integrate with respect to that said variable

Example

This is a sample question from the Math worksheet
Use substitution of $x = \tan θ$ to find $\int_{0}^{1} \frac{1}{(x^{2} + 1)^{2}} d x$

Step 1. Finding $d x$

\begin{aligned} \frac{d x}{d θ} & = \sec^{2} θ \\ d x & = \sec^{2} θ d θ \end{aligned}

Step 2. Sub in $x = \tan θ$ , and $d x$ and changing the limits

\begin{aligned} \int_{0}^{1} \frac{1}{(x^{2} + 1)^{2}} d x & = \int_{\arctan 0}^{\arctan 1} \frac{1}{(\tan^{2} θ + 1)^{2}} \sec^{2} θ d θ [Using trigo identities] \\ = \int_{0}^{\frac{π}{4}} \frac{1}{\sec^{2} θ} d θ = \int_{0}^{\frac{π}{4}} \cos^{2} θ d θ [Using double angle formulae] \\ = \int_{0}^{\frac{π}{4}} \frac{\cos 2 θ + 1}{2} d θ \\ = {[\frac{\sin 2 θ}{4} + \frac{1}{2} x]}_{0}^{\frac{π}{4}} = \frac{1}{4} + \frac{π}{8} [Q E D] \end{aligned}

Integration by Parts

\begin{aligned} \frac{d}{d x} u v & = u^{'} v + u v^{'} \\ u v & = \int u^{'} v + u v^{'} d x \\ = \int u^{'} v d x + \int u v^{'} d x \end{aligned}

Thus

\int u^{'} v d x = u v - \int u v^{'} d x

Example

This is a sample question from the Math worksheet
Find $\int x \ln x d x$

sub $v$ for the part that is hardest to integrate.
sub $u^{'}$ for the part that is easiest to integrate

Substitute if	$u^{'}$	$v$
Hard to Integrate		$X$
Easy to Integrate	$X$

Since $\ln x$ is hard to integrate and $x$ is easy to integrate, let $v = \ln x$ , and $u^{'} = x$ ,

\begin{aligned} \int x \ln x d x = \int u^{'} v d x & = u v - \int u v^{'} d x \\ = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} \frac{1}{x} d x \\ = \frac{x^{2}}{2} (\ln x - \frac{1}{2}) + C \end{aligned}

A nice application of this formula is in its immense contribution in Proof for Wallis Product.

Trick for inverse trigo functions:

This type of questions typically asks us to solve for $\int \frac{k}{a + (\frac{x}{b})^{2}} d x$ or $\int \frac{k}{\sqrt{a + (\frac{x}{b})^{2}}} d x$
But how do we solve it??

Force the $a$ in the denominator to be $1$ by factoring $a$ out.
Perform integration by substitution on the value $\frac{x}{b}$
Substitute the appropriate inverse trig identity and solve.

Example

This is a sample question from the Math worksheet
Find $\int \frac{3}{\sqrt{4^{2} - (5 x + 3)^{2}}} d x$

Step 1. Force the $a$ in the denominator to be 1 by factoring out 4.

\int \frac{3}{\sqrt{4^{2} - (5 x + 3)^{2}}} d x = \int \frac{3}{4 \cdot \sqrt{1 - (\frac{5 x + 3}{4})^{2}}} d x

Step 2. Perform integration by substitution by letting $u = \frac{5 x + 3}{4}$

\frac{d u}{d x} = \frac{5}{4}

\begin{aligned} \int \frac{3}{4 \cdot \sqrt{1 - (\frac{5 x + 3}{4})^{2}}} d x & = \int \frac{3}{4 \cdot \sqrt{1 - u^{2}}} \frac{4}{5} d u \\ = \frac{3}{5} \int \frac{1}{\sqrt{1 - u^{2}}} d u \end{aligned}

Step 3. Substitute the appropriate inverse trig identity and solve.

\begin{aligned} \frac{3}{5} \int \frac{1}{\sqrt{1 - u^{2}}} d u & = \frac{3}{5} \arcsin u \\ = \frac{3}{5} \arcsin \frac{5 x + 3}{4} \end{aligned}

Solving using Reverse Factor Formulae

Reverse Factor Formulae

$2 \sin A \cos B = \sin (A + B) + \sin (A - B)$
$2 \cos A \sin B = \sin (A + B) - \sin (A - B)$
$2 \cos A \cos B = \cos (A + B) + \cos (A - B)$
$- 2 \sin A \sin B = \cos (A + B) - \cos (A - B)$

Example

Find using the reverse factor formulae

\int \sin 5 x \cos 3 x d x

Step 1. Identify which reverse factor formulae to use.

\sin A \cos B = \frac{1}{2} (\sin (A + B) + \sin (A - B))

Step 2. Substitute the corresponding values into A and B

\sin 5 x \cos 3 x = \frac{1}{2} (\sin 8 x + \sin 2 x)

Step 3. Integrate the expression above

\int \frac{1}{2} (\sin 8 x + \sin 2 x) d x = - \frac{1}{16} \cos 8 x - \frac{1}{4} \cos 2 x + C

Volume of Revolution

Suppose we have $f (x)$ , rotating it around the $x -$ axis would form a volume of

V = π \int_{a}^{b} f (x)^{2} d x

Ordinary Differential Equations (ODE)

Solving a Variable Separable Differential equation

We can sorta treat $\frac{d y}{d x}$ as fractions. Its quite intuitive.

Example

Find the general solution of the following.

\frac{d y}{d x} = \frac{1}{3} \sqrt{5 - y^{2}}

Step 1. Put all the $y$ on one side and the $x$ on the other

\frac{1}{\sqrt{5 - y^{2}}} \frac{d y}{d x} = \frac{1}{3}

Step 2. Multiply $d x$ on both sides

\frac{1}{\sqrt{5 - y^{2}}} d y = \frac{1}{3} d x

Step 3. Put an Integral sign on both sides and solve

\begin{aligned} \int \frac{1}{\sqrt{5 - y^{2}}} d y & = \int \frac{1}{3} d x \end{aligned}

\begin{aligned} \int \frac{1}{\sqrt{5 - y^{2}}} d y & = \int \frac{1}{\sqrt{5} \sqrt{1 - (\frac{y}{\sqrt{5}})^{2}}} d y, sub u = \frac{y}{\sqrt{5}}, \frac{d u}{d y} = \frac{1}{\sqrt{5}} \\ = \int \frac{1}{\sqrt{5}} \frac{1}{\sqrt{1 - u^{2}}} \sqrt{5} d u \\ = \arcsin u + C \\ = \arcsin (\frac{y}{\sqrt{5}}) + C = \frac{1}{3} x \end{aligned}

Thus

\begin{aligned} \arcsin (\frac{y}{\sqrt{5}}) & = \frac{1}{3} x + C \\ y & = \sqrt{5} \sin (\frac{1}{3} x + C) \end{aligned}

Solving homogeneous differential equations by substituting $y = v x$

It argues that this will make your life easier when solving differential functions $\frac{d y}{d x}$ that can be expressed in terms $\frac{y}{x}$ , like $f (\frac{y}{x})$

\frac{d y}{d x} = f (\frac{y}{x})

So it assumes when you divide $y$ by $x$ , you get a function $v$ .

y = v x

thus

\frac{d y}{d x} = v x^{'} + v^{'} x = v + \frac{d v}{d x} x = f (v)

I think its best for an example

Example

Find the general solution of the following.

\frac{d y}{d x} = {(\frac{y}{x})}^{2} + \frac{y}{x} + 1

Step 1. Substitute $\frac{y}{x}$ as $v$

\frac{d y}{d x} = v^{2} + v + 1

Step 2. Since $\frac{d y}{d x} = v + \frac{d v}{d x} x$ , equate that to the above

v^{2} + v + 1 = v + \frac{d v}{d x} x

Step 3. Solve the equation

\begin{aligned} \int \frac{1}{x} d x & = \int \frac{1}{v^{2} + 1} d v \\ \ln | x | + C & = \arctan v \\ v & = \tan (\ln | x | + C) \\ y & = x \cdot \tan (\ln | x | + C) \end{aligned}

Solving a Differential Equation using Integrating Factor

It aims to solve for differential equations in this form

\frac{d y}{d x} + P (x) y = Q (x)

Objectives: I want to separate $y$ on the left and $x$ on the right. We can do this by multiplying throughout by an arbitrary function called Integrating factor $I (x)$

Suppose the Integrating factor $I (x)$ follows this property

I^{'} (x) = P (x) I (x)

Why tho? See that if we multiply throughout by $I (x)$

I (x) \frac{d y}{d x} + I (x) P (x) y = I (x) Q (x)

thus

\begin{aligned} I (x) \frac{d y}{d x} + I^{'} (x) y & = I (x) Q (x) \\ \frac{d}{d x} I (x) y & = I (x) Q (x) [Integrate throughout] \\ I (x) y & = \int I (x) Q (x) d x \\ y & = \frac{\int I (x) Q (x) d x}{I (x)} \end{aligned}

Defining $I (x)$
Since

P (x) = \frac{I^{'} (x)}{I (x)}

Integrating both sides

\begin{aligned} \int P (x) d x & = \int \frac{I^{'} (x)}{I (x)} d x \\ = \ln | I (x) | + C \\ I (x) & = e^{\int P (x) d x + C} \\ = A e^{\int P (x) d x} [where A = e^{C}] \end{aligned}

Thus
For simplicity, let $C = 0$ , thus $A = 1$

y = \frac{\int e^{\int P (x) d x} Q (x) d x}{e^{\int P (x) d x}}

Example

Find the general solution of the following.

\frac{d y}{d x} + y = 3 e^{x}

Note that the coefficient of $\frac{d y}{d x}$ has to be $1$

Step 1. Identify $P (x) = 1$ , find the integrating factor $I (x)$

I (x) = e^{\int 1 d x} = e^{x}

Step 2. Multiply throughout by $I (x)$

e^{x} \frac{d y}{d x} + e^{x} y = 3 e^{2 x}

Step 3. Solve

\begin{aligned} e^{x} \frac{d y}{d x} + e^{x} y & = 3 e^{2 x} \\ \frac{d}{d x} e^{x} y & = 3 e^{2 x} \\ e^{x} y & = \int 3 e^{2 x} d x = \frac{3}{2} e^{2 x} + C \\ y & = \frac{3}{2} e^{x} + C e^{- x} \end{aligned}

Euler's Method

It attempts to approximate the function by using its derivatives. Suppose

\frac{d y}{d x} = f (x, y) \approx \frac{Δ y}{Δ x}

Suppose we know $(x_{0}, y_{0})$ , which is a point lying on $y$ .
For some small increment in $x$ by $Δ x$ .

{\begin{aligned} x_{n + 1} & = x_{n} + Δ x \\ y_{n + 1} & = y_{n} + Δ x \cdot f (x_{n}, y_{n}) \end{aligned}, n = 0, 1, 2, 3 ...

Understand that essentially

y_{n + 1} = y_{n} + Δ y

Limits

$lim_{x \to 0} f (x)$ ask the question, what does the value $f (x)$ approach as $x$ approaches $0$ ?

We are interested in finding the limits of a function in which just subbing in the values of
$x$ would yield an undefined solution like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . Suppose $f (0) = g (0) = 0$ .

lim_{x \to 0} \frac{f (x)}{g (x)} = ? ? ?

L' Hopital rule

It basically says that if the solution is undefined, just differentiate $f (x)$ and $g (x)$ until it yields a valid solution.

For example

lim_{x \to 0} \frac{\sin x}{x} = \frac{\cos x}{1} = 1

Look at Maclaurin Series expansion

The expansion of $\sin x$ yields

lim_{x \to 0} \frac{x + \frac{x^{3}}{3!} + \dots}{x} = lim_{x \to 0} 1 + \frac{x^{2}}{3!} + \dots = 1

We aim to separate a constant that is independent of $x$ from this fraction.

Look that if we differentiate $\sin x$ and $x$ instead. We can also get a constant of $1$ out.

lim_{x \to 0} \frac{x + \frac{x^{3}}{3!} + \dots}{x} = lim_{x \to 0} \frac{1 + \frac{x^{2}}{2!} + \dots}{1} = 1

Differentiation is now a tool used to factor $x$ from $f (x)$ and $g (x)$ . That's why the L'hopital rule works.

Differentiation

Chain Rule

Product Rule

Quotient Rule

Integration

Integration by Substitution

Integration by Parts

Trick for inverse trigo functions:

Solving using Reverse Factor Formulae

Reverse Factor Formulae

Volume of Revolution

Ordinary Differential Equations (ODE)

Solving a Variable Separable Differential equation

Solving homogeneous differential equations by substituting y=vx

Solving a Differential Equation using Integrating Factor

Euler's Method

Limits

L' Hopital rule

Solving homogeneous differential equations by substituting $y = v x$