Calculus

Functions and models

• even functions: \(f(x) = f(-x)\)

• odd functions: \(f(-x) = -f(x)\)

• polynomials: \(f(x) = x^2 + 2x + 1\)

• power functions: \(f(x) = x^a\)

• rational functions: \(f(x) = \frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials

• algebraic functions

• transcendental functions: \(\sin x \cos x \, e^x \, \ln x\)

• exponential functions: \(f(x) = b^x\)

• logarithmic functions: \(f(x) = \log_b x\)

Translation:

• up/down: \(y = f(x) \pm c\)

• right/left: \(y = f(x \mp c)\)

Stretching:

• vertical: \(y = cf(x)\)

• horizontal \(y = f(c^{-1}x)\)

Reflection:

• about the x-axis \(y = -f(x)\)

• about the y-axis \(y = f(-x)\)

The composition of two functions:

\[ (f \circ g)(x) = f(g(x)) \]

Note that

\[ f \circ g \ne g \circ f \]

Inverse functions: reflect about the line \(y = x\)

Laws of logarithms;

\[ \log_b x^r = r\log_b x \]

\[\begin{split} \log_b x + \log_b y = \log_b xy \\ \log_b x - \log_b y = \log_b \frac x y \end{split}\]

\[ \log_b x = \frac{\ln x}{\ln b} \]

Limits and derivatives

Limits

\[\begin{split} \lim_{x \to a} f(x) = L \\ f(x) \to L \text{ as } x \to a \end{split}\]

One-sided limits:

\[\begin{split} \begin{array}{lr} \lim_{x \to a^+} f(x) = L & \text{right-handed} \\ \lim_{x \to a^-} f(x) = L & \text{left-handed} \\ \end{array} \end{split}\]

The Heaviside function:

\[\begin{split} H(t) = \begin{cases} 0 & \text{if} & t < 0 \\ 1 & \text{if} & t \ge 0 \end{cases} \end{split}\]

Strategy: Calculating limits

• try to reduce the number of occurrences of \(x\) in \(f(x)\)

• try to make the expression inside the limit as simple as possible by applying limit laws

• consider using two one-sided limits to determine a limit

The precise definition of a limit:

\[ \lim_{x \to a} f(x) = L \]

if for every number \(\varepsilon > 0\) there is a number \(\delta > 0\) such that

\[ \text{if} \; 0 < |x - a| < \delta \;\text{then}\; |f(x) - L| < \varepsilon \]

Continuity:

\(f(x)\) is continuous at a number \(a\) if

\[ \lim_{x \to a} f(x) = f(a) \]

Tip

\[ \lim_{x \to a} f(g(x)) = \lim_{t \to \lim_{x \to a} g(a)} f(t) \]

Warning

Limit laws don’t apply to infinite limits!

The following is wrong!!!

Error

\[ \lim_{x \to \infty}(x^2 - x) = \lim_{x \to \infty} x^2 - \lim_{x \to \infty} x = \infty - \infty \]

Derivatives

For function

\[ y = f(x) \]

Its derivative is

\[\begin{split} f'(x) = y' = \frac{dy}{dx} = \frac{df}{dx} = \frac{d}{dx}f(x) = Df(x) = D_xf(x) \\ = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \end{split}\]

Differentiation rules:

\[\begin{split} \begin{align*} \frac{d}{dx}c & = 0 \\ \frac{d}{dx}x^n & = nx^{n-1} \\ \frac{d}{dx}e^x & = e^x \\ \frac{d}{dx}\sin x & = \cos x \\ \frac{d}{dx}\cos x & = -\sin x \\ \frac{d}{dx}\tan x & = \sec^2 x \\ \frac{d}{dx}\cot x & = -\csc^2 x \\ \frac{d}{dx}\sec x & = \sec x\tan x \\ \frac{d}{dx}\csc x & = -\csc x\cot x \\ \frac{d}{dx}\sin^{-1} x & = \frac{1}{\sqrt{1-x^2}} \\ \frac{d}{dx}\cos^{-1} x & = -\frac{1}{\sqrt{1-x^2}} \\ \frac{d}{dx}\tan^{-1} x & = \frac{1}{1+x^2} \\ \frac{d}{dx}\cot^{-1} x & = -\frac{1}{1+x^2} \\ \frac{d}{dx}\sinh x & = \cosh x \\ \frac{d}{dx}\cosh x & = \sinh x \\ \end{align*} \end{split}\]

\[\begin{split} (f \pm g)' = f' \pm g' \\ (fg)' = f'g + fg' \\ \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \end{split}\]

Important

The Chain Rule:

\[ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \]

Or alternatively:

\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} \]

Tip

The Power Rule combined with the Chain Rule:

\[ y = [g(x)]^n = u^n \]

\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = nu^{n-1}\frac{du}{dx} = n[g(x)]^{n-1}g'(x) \]

The derivative of any exponential function:

\[ \frac{d}{dx}b^x = \frac{d}{dx}(e^{\ln b})^x = \frac{d}{dx}e^{(\ln b)x} = \frac{d}{du}e^u\frac{d}{dx}(\ln b)x = e^{(\ln b) x}\ln b = b^x \ln b \]

Tip

Using the Chain Rule repeatedly:

\[ \frac{dy}{dt} = \frac{dy}{du}\frac{du}{dx}\frac{dx}{dt} \]

Recursively “peel” the function by differentiating the outmost function evaluated at the inner function and then multiplying the result by the derivative of the inner function.

Implicit differentiation

Important

Remember \(y\) is a function of \(x\)

Warning

Failing to differentiate a constant \(c\) and leaving it non-zero is a common error!

Error

\[\begin{split} \begin{align*} x^2 - 4xy + y^2 & = 4 \\ 2x - 4y - 4xy' + 2yy' & = 4 \gets \text{this should be 0!}\\ \end{align*} \end{split}\]

Logarithmic differentiation

\[\begin{split} \begin{align*} \frac{d}{dx}\log_b x & = \frac{1}{x\ln b} \\ \frac{d}{dx}\ln x & = \frac{1}{x} \\ \frac{d}{dx}\ln |x| & = \frac{1}{x} \\ \end{align*} \end{split}\]

Example

Differentiate

\[ y = \frac{e^{-x}\cos^2x}{x^2 + x + 1} \]

1. analyze: \(e^{-x} > 0\); \(\cos^2 x \ge 0\); \(x^2 + x + 1 > 0\), therefore \(y \ge 0\) and we can take the logarithm without using absolute values

2. take the natural logarithm of both sides:

   \[ \ln y = -x + \ln \cos^2 x - \ln(x^2 + x + 1) \]

3. differentiate implicitly with respect to x:

   \[ \frac{1}{y}\frac{dy}{dx} = -1 + \frac{1}{\cos^2 x}[2\cos x\cdot(-\sin x)] - \frac{1}{x^2 + x + 1}(2x + 1) \]

4. substitute \(y\) with the original function and simplify

   \[ \frac{dy}{dx} = -\frac{e^{-x}\cos^2x}{x^2 + x + 1}(1 + \frac{\sin 2x}{\cos^2 x} + \frac{2x + 1}{x^2 + x + 1}) \]

Rates of change in physics

Position, velocity and acceleration

Instantaneous velocity is the derivative of the position function with respect to time:

\[ v = \frac{ds}{dt} \]

Acceleration is the derivative of the velocity function with respect to time:

\[ a = \frac{dv}{dt} = \frac{d^2s}{dt^2} \]

---

Charge and current

\[ I = \frac{dQ}{dt} \]

Exponential growth and decay

Consider the differential equation

\[ \frac{dy}{dt} = ky \]

It has only one solution:

\[ y(t) = y(0)e^{kt} \]

This function is extremely common in various sciences.

Related rates

The essence of relating derivatives with each other is the Chain Rule:

\[ \frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} \]

Therfore, to relate \(dy/dt\) with \(dx/dt\), first find an equation that describes the relationship between \(y\) and \(x\). Then differentiate implicitly with respect to \(t\).

Warning

A common error is to forget the minus sign before \(dy/dx\) when \(x\) is increasing while \(y\) is decreasing.

Tip

When circles are involved, try to use the parametric form (with \(r\) and \(\theta\)) instead of using the classic implcit formula.

Applications of differentiation

Maximum and minimum values

Fermat’s theorem

If \(f\) has a local extremum at \(c\), and if \(f'(c)\) exists, then \(f'(c) = 0\).

Critical number

a number \(c\) such that \(f'(c) = 0\) or \(f'(c)\) doesn’t exist

Strategy: The closed interval method

To find the absolute maximum/minimum of a continuous function \(f\) on a closed interval \([a, b]\),

1. calculate \(f(c)\) for every critical number \(c\) on the open interval \((a, b)\)

2. calculate the endpoints

• The mean value theorem

• Increasing/decreasing test (first derivative)

• The first derivative test

• Concavity test

• Inflection point

• The second derivative test

Important

L’Hospital’s Rule:

If limit

\[ \lim_{x \to a}\frac{f(x)}{g(x)} \]

is of the indeterminate form (\(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)), then

\[ \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \]

provided that \(f'(x)\) and \(g'(x)\) exist.

Curve sketching:

1. Domain

2. Intercepts (if the equation is easy to solve)

3. Symmetry (even/odd) / periodic

4. Asymptotes

   • horizontal: \(\lim_{x \to \pm\infty}\)

   • vertical: \(\lim = \pm\infty\)

   • slant: \(\lim_{x \to \pm\infty}[f(x)-(mx+b)] = 0\)

5. Intervals of increase/decrease

6. Local max/min values

7. Concavity and points of inflection

Newton’s method:

\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]

Integrals

The definite and indefinite integrals

\[ \int_a^bf(x)\,dx = \lim_{n \to \infty}\sum_{i=1}^n f(x_i^*)\Delta x \]

\[ \int f(x)\,dx = F(x) \]

Table of indefinite integrals:

\[\begin{split} \begin{align*} \int k\,dx & = kx + C \\ \int x^n\,dx & = \frac{x^{n+1}}{n+1} + C\;(n \ne -1) \\ \int \frac{1}{x}\,dx & = \ln|x| + C \\ \int e^x\,dx & = e^x + C \\ \int b^x\,dx & = \frac{b^x}{\ln b} + C \\ \int \sin x\,dx & = -\cos x + C \\ \int \cos x\,dx & = \sin x + C \\ \int \sec^2 x\,dx & = \tan x + C \\ \int \csc^2 x\,dx & = -\cot x + C \\ \int \sec x\tan x\,dx & = \sec x + C \\ \int \csc x\cot x\,dx & = -\csc x + C \\ \int \frac{1}{1 + x^2}\,dx & = \tan^{-1} x + C \\ \int \frac{1}{\sqrt{1-x^2}}\,dx & = \sin^{-1} x + C \\ \end{align*} \end{split}\]

Some more exotic ones:

\[\begin{split} \begin{align*} \int \tan x\,dx & = \ln |\sec x| + C \\ \int \cot x\,dx & = \ln |\sin x| + C \\ \int \sec x\,dx & = \ln |\sec x + \tan x| + C \\ \int \csc x\,dx & = \ln |\csc x - \cot x| + C \\ \int \sinh x\,dx & = \cosh x + C \\ \int \cosh x\,dx & = \sinh x + C \\ \int \frac{dx}{x^2 + a^2} & = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) \\ \int \frac{dx}{\sqrt{a^2 - x^2}} & = \sin^{-1}\left(\frac{x}{a}\right), \;a > 0 \\ \int \frac{dx}{x^2 - a^2} & = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| \\ \int \frac{dx}{\sqrt{x^2 \pm a^2}} & = \ln\left|x + \sqrt{x^2 \pm a^2}\right| \\ \int \frac{1}{1 + k\cos\theta}\,d\theta & = 2\sqrt{\frac{1}{1 - k^2}}\tan^{-1}\left(\sqrt{\frac{1+k}{1-k}}\tan\frac{\theta}{2}\right) \\ \end{align*} \end{split}\]

The FUNDAMENTAL theorem of calculus

Part 1

If

\[ g(x)= \int_a^x f(t)\,dt \]

then

\[ g'(x) = f(x) \]

Part 2

\[ \int_a^b f(x)\,dx = F(b) - F(a) \]

where \(F\) is any antiderivative of \(f\) (\(F' = f\)).

Important

A few VERY useful trigonometric identities:

Let \(u = \tan\frac{x}{2}\), then

\[ \sin x = \frac{2u}{1 + u^2} \; \csc x = \frac{1 + u^2}{2u} \]

\[ \cos x = \frac{1 - u^2}{1 + u^2} \; \sec x = \frac{1 + u^2}{1 - u^2} \]

\[ \tan x = \frac{2u}{1 - u^2} \; \cot x = \frac{1 - u^2}{2u} \]

Integration techniques

This section is particularly important because integration is a tricky process and requires a significant amount of skill as well as ingenuity.

1. try the formulas given above

2. simplify the integrand if possible, e.g.

   \[ \int \frac{\tan x}{\sec^2 x}\,dx = \int \sin x\cos x\,dx = \frac{1}{2}\int\sin 2x\,dx \]

3. try the substitution rule:

   1. look for a part in the expression whose derivative also exists in the integrand

   2. let the part be \(u\) where \(u = g(x)\)

   3. evaluate \(du = g'(x)\,dx\)

   4. substitue every \(x\) in the integrand with appropriate forms of \(u\)

   5. evaluate the integral \(\int f(u)\,du\)

   6. If it is an indefinite integral, substitute \(x\) back in. Otherwise, replace the limits of integration for \(x\) with the corresponding values for \(u\).

4. classify the integrand:

   • trigonometric functions:

     • powers of \(\sin x\) and \(\cos x\), or powers of \(\tan x\) and \(\sec x\), or powers of \(\cot x\) and \(\csc x\):

       use trigonometric identities to transform the expression and then use the substitution rule

     • others:

       apply the VERY useful identity above to eliminate everything but \(\tan\frac{x}{2}\)

   • rational functions:

     use partial fractions:

     1. long division

        \[ f(x) = \frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)} \]

     2. write \(f(x)\) in this way

        \[ f(x) = \sum_{i=1}^m\sum_{u=1}^{u_i}\frac{A_{iu}}{(a_ix + b_i)^u} + \sum_{j=1}^n\sum_{v=1}^{v_j}\frac{A_{jv}x + B_{jv}}{(a_jx^2 + b_jx + c_j)^v} \]

     3. evaluate the transformed integral section by section, completing squares and substituting with \(u\) when necessary

   • radicals:

     • \(\sqrt{\pm x^2 \pm a^2}\):

       Expression	Substitution	Identity
       \(\sqrt{a^2-x^2}\)	\(x = a\sin\theta, \;-\pi/2 \le \theta \le \pi/2\)	\(1 - \sin^2\theta = \cos^2\theta\)
       \(\sqrt{a^2+x^2}\)	\(x = a\tan\theta, \;-\pi/2 < \theta < \pi/2\)	\(1 + \tan^2\theta = \sec^2\theta\)
       \(\sqrt{x^2-a^2}\)	\(x = a\sec\theta, \;0 \le \theta < \pi/2\;\text{or}\;\pi \le \theta < 3\pi/2\)	\(\sec^2\theta - 1 = \tan^2\theta\)

     • \(\sqrt[n]{ax + b}\):

       use substitution \(u = \sqrt[n]{ax + b}\)

   • other:

     try integration by parts:

     1. separate the integrand into two parts \(u\) and \(dv\) keeping in mind that \(u\) should be as simple as possible while \(dv\) should be easy to integrate

     2. plug everything in:

        \[ \int u\,dv = uv - \int v\,du \]

Differential equations

First-order ODEs

Separable equations

Differential equations that can be written in the form

\[ \frac{dy}{dx} = \frac{g(x)}{h(y)} \]

To solve separable equations,

1. put all of \(y\) on one side and all of \(x\) on the other side

   \[ h(y)\,dy = g(x)\,dx \]

2. integrate both sides

   \[ \int h(y)\,dy = \int g(x)\,dx \]

Linear equations

Differential equations that can be put into the form

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

To solve linear equations,

1. let

   \[ I(x) = e^{\int P(x)\,dx} \]

2. multiply both sides by \(I(x)\)

3. transform the left side into a derivative with the product rule

4. integrate both sides

Second-order ODEs

Second-order linear homogeneous equations

equations in the form

\[ P(x)\frac{d^2y}{dx^2} + Q(x)\frac{dy}{dx} + R(x)y = 0 \]

If \(P(x)\), \(Q(x)\) and \(R(x)\) are constant-valued functions (\(a\), \(b\) and \(c\)), the second-order differential equation can be easily solved by using auxiliary equations.

Auxiliary equations

the quadratic equations

\[ ar^2 + br + c = 0 \]

After solving the auxiliary equation, we can get the solution for the original second-order homogeneous linear differential equation (OMG, that’s really a mouthful):

Roots of the auxiliary equation	General solution
\(r_1\), \(r_2\), real and distinct	\(y = c_1e^{r_1x} + c_2e^{r_2x}\)
\(r_1 = r_2 = r\)	\(y = c_1e^{rx} + c_2xe^{rx}\)
\(r_1\), \(r_2\), complex: \(\alpha + i\beta\)	\(y = e^{\alpha x}(c_1\cos\beta x + c_2\sin\beta x)\)

For non-homogeneous linear equations such as

\[ a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = G(x) \]

it can be proven that the general solution is

\[ y = y_p + y_c \]

where \(y_c\) is the general solution of the complementary homogeneous equation

\[ a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \]

and \(y_p\) is a particular solution of the original equation.

To find a particular solution, use either the method of undetermined coefficients or the method of variation of parameters.

Method: Undetermined coefficients

1. guess the form of the solution by classifying the function \(G(x)\)

   • \(Ce^{kx}\;\to\;Ae^{kx}\)

   • a polynomial \(P(x)\;\to\;Q(x)\), where \(Q(x)\) is a polynomial of the same degree as \(P(x)\)

   • \(C\sin kx \;\text{or}\; C\cos kx\;\to\;A\cos kx + B\sin kx\)

   • Everything above can be freely combined or superimposed together.

   • If the guessed form is already a solution of the complementary equation, multiply it by \(x\) or \(x^2\)

2. differentiate the function and substitute it into the differential equation

3. solve for the coefficients

Method: Variation of parameters

1. replace the constants \(c_1\) and \(c_2\) in \(y_c\) with functions \(u_1(x)\) and \(u_2(x)\)

2. differentiate the altered function \(y_p\) and simplify it by imposing conditions

   Tip

   The condition

   \[ u_1'y_1 + u_2'y_2 = 0 \]

   often works well.

3. plug everything into the original equation and let

   \[ ay'' + by' + cy \]

   be zero, because \(y_1\) and \(y_2\) are particular solutions of the complementary equation.

4. solve for \(u_1\) and \(u_2\)

If, however, the coefficients of \(y\) are not constant-valued functions, some of the second-order ODEs can still be solved by downgrading them to first-order ODEs.

If the equation is in the form

\[ y'' = f(y, y') \]

then let \(p = y'\). Therefore

\[ y'' = \frac{dp}{dx} = \frac{dp}{dy}\frac{dy}{dx} = p\frac{dp}{dy} \]

and the original equation turns into a first-order ODE:

\[ p\frac{dp}{dy} = f(y, p) \]

If the equation is in the form

\[ y'' = f(x, y') \]

then simply substitute \(y'\) with \(p\) and integrate \(p\) after solving the transformed equation.

Conic sections

Parabolas

Parabola, focus, directrix

The set of points in a plane that are equidistant from a fixed point \(F\) (called the focus) and a fixed line (called the directrix)

Vertex

The point halfway between the focus and the directrix

Axis

The line through the focus perpendicular to the directrix

An equation of the parabola with focus \((0, p)\) and directrix \(y = -p\) is

\[ x^2 = 4py \]

Or equivalently:

\[ y = \frac{x^2}{4p} \]

Ellipses

Ellpise, foci

An ellipse is the set of points in a plane the sum of whose distances from two fixed points \(F_1\) and \(F_2\) is a constant. The two fixed points are the foci.

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \; (a \ge b) \]

Vertices

The points \((a, 0)\) and \((-a, 0)\) in the equation above

Foci:

\[ (\pm c, 0), \text{where}\; c^2 = a^2 - b^2 \]

Hyperbolas

Hyperbola

The set of points in a plane the difference of whose distances from two fixed points \(F_1\) and \(F_2\) is a constant.

The hyperbola

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

has foci \((\pm c, 0)\) where \(c^2 = a^2 + b^ 2\), vertices \((\pm a, 0)\) and asymptotes \(y = \pm(b/a)x\).

Conic sections in polar forms

Eccentricity \(e\), directrix \(x = \pm d\)

\[ r = \frac{ed}{1 \pm e\cos\theta} \]

Eccentricity \(e\), directrix \(y = \pm d\)

\[ r = \frac{ed}{1 \mp e\cos\theta} \]

Specifically, the polar equation of an ellipse with focus at the origin, semimajor axis \(a\), eccentricity \(e\), and directrix \(x = d\) can be written as:

\[ r = \frac{a(1-e^2)}{1 + e\cos\theta} \]

In the equation above, the perihelion distance is \(a(1 - e)\) and the aphelion distance is \(a(1 + e)\).

Multivariable calculus

TODO

Vector calculus

TODO

Complex numbers

A complex number

\[ z = a + bi \]

has a real part and an imaginary part:

\[\begin{split} \Re(z) = a \\ \Im(z) = b \end{split}\]

Its complex conjugate is

\[ \overline{z} = a - bi \]

The modulus of \(z\) is

\[ |z| = \sqrt{a^2 + b^2} \]

\[ z\overline{z} = |z|^2 \]

The polar form of a complex number \(z\):

\[ z = r(\cos\theta + i\sin\theta) \]

To multiply (or divide) complex numbers in polar form, simply multiply (or divide) the moduli and add (or subtract) the arguments.

Important

Euler’s formula:

\[ e^{ix} = \cos x + i\sin x \]

Therefore

\[\begin{split} \begin{align*} e^{-ix} & = \cos(-x) + i\sin(-x) \\ & = \cos x - i\sin x \\ \cos x & = \frac{e^{ix} + e^{-ix}}{2} \\ \sin x & = \frac{e^{ix} - e^{-ix}}{2i} \end{align*} \end{split}\]