a brief summary of calculus

Introduction

Rules:
- A function f, is differentiable at a point x=a, if the derivative f'(x) exists at x=a:
  - ie. must be smooth & continuous;
- If a function is continuous at x=a, then the lim(x→a)f(x) and f(a) both exist and are equal;
- d(x^2-5x+6)/dx = d(x^2)/dx - d(5x)/dx + d(6)/dx = 2x - 5;
- d(uv)/dx = vd(u)/dx + ud(v)/dx;
- dy/dx = (dy/du) * (du/dx);
- dy/dx = 1/(dx/dy);
- if y=u/v, dy/dx = [(vdu/dx)-(udv/dx)]/v^2;
examples
- the cooling rate of a cup of tea
  - the rate of cooling is dependent upon the temperature difference of the tea and the room (ΔT)
  - the difference in temperature changes over time = ΔT(t) = e^-kt
  - rate of change the difference in temperature changes over time = dΔT/dt = -kΔT

eg. x^5 + 2(x^2)(y^3) + y^5 = 0,
d(x^5)/dx + d[2(x^2)(y^3)]/dx + d(y^5)/dx = 0,
and, d(x^5)/dx = 5x^4,
d[2(x^2)(y^3)]/dx = 2x^2d(y^3)/dx + y^3d(2x^2)/dx
- = 2x^2d(y^3)dy/(dy.dx) + 4xy^3
- = (2x^2)(3y^2)dy/dx + 4xy^3,
d(y^5)/dx = d(y^5)dy/(dy.dx)
- = (5y^4)dy/dx,
- ⇒ 5x^4 + 6(x^2)(y^2)dy/dx + 4xy^3 + 5(y^4)dy/dx = 0,
- ⇒ dy/dx =[-x(5x^3 + 4y^3)]/[y^2(6x^2 + 5y^2)], y not=0

these consist of polynomials used to approximate a different function and the higher the level of polynomial you use in the series to approximate, it may either “converge” to a closer approximation or, at some value for x, it may “diverge” from the approximation
for example:
- the approximation for cos(x) near x = 0, is cos(x) approx equals 1 - x²/2
- the approximation for e^x approx equals 1 + x + x²/2! + x³/3! + .. xⁿ/n! and plugging in x = 1 will give you a convergent approximation of the value of e
these Taylor polynomials are created by:
- taking the base Taylor polynomial form P(x) = c₀ + c₁x + c₂x² + …
- set x to be the value you are wanting to most closely approximate (usually 0 but you may use x-a instead of x in the polynomial)
- sequentially differentiate the base form and equate to the same degree of differentiation of the original function and these will give you each of the constant c terms each time you differentiate, note that in the e^x, the differential is e^x each time and for x = 0, the result is 1, hence the Taylor polynomial uses power of x divided by the power factorial (as the sequential differentials of a power of x until the power of x is zero is the factorial of the original power)
  - see Youtube explanation of Taylor polynomials and series

this is used to determine the area bounded by the x axis and a function such that any parts above the x axis are positive areas and those below it are negative areas.
- thus if f(x) = x(2-x) = 2x-x² then the x-axis intercepts (ie. where f(x) = 0) ) are 0 and 2 and f(x) is positive between these intercepts and negative outside these. The integral of f(x) = x² - x³/3 . Thus the area above the x-axis = (0² - 0³/3) - (2² - 2³/3) = 4/3. The area below the x-axis between x = -1 and x = 0 equals (-1² - (-1)³/3) - (0² - 0³/3) = -4/3. This just happens to be equal but opposite sign to the area from x = 0 to x =2 which was first calculated. If you determine the integral from x = -1 to x =2, the answer will then be zero as the two areas will cancel each other out.
it can also be used to determine the area bounded by two functions f(x) and g(x) by first determining the x values of the points of intersection between f(x) and g(x) and then determining the integral from intersect 1 to intersect 2 of f(x) - g(x)