maths:calculus

see also:

f(x) |
f'(x) |
Integral of f(x) |

x^n | nx^(n-1) | (x^{n+1})/(n+1) |

lnx | x^(-1) | |

ln(ax+b) | a/(ax+b) | |

e^(kx) | ke^(kx) | |

sin(kx) | kcos(kx) | |

cos(kx) | ksin(kx) | |

tan(kx) | k(sec^2)(kx) | |

Arsin(x/a) | 1/SQR(a^2-x^2) | |

Arcos(x/a) | -1/SQR(a^2-x^2) | |

Artan(x/a) | a/(a^2+x^2) | |

g(x)h(x) | g'(x)h(x)+g(x)h'(x) | |

g(x)/h(x) | [g'(x)h(x)-g(x)h'(x)]/[h(x)]^2 | |

goh(x) | g'oh(x).h'(x) |

**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 - x
^{2}/2 - the approximation for e
^{x}approx equals 1 + x + x^{2}/2! + x^{3}/3! + .. x^{n}/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) = c
_{0}+ c_{1}x + c_{2}x^{2}+ … - 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)

- 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-x
^{2}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) = x^{2}- x^{3}/3 . Thus the area above the x-axis = (0^{2}- 0^{3}/3) - (2^{2}- 2^{3}/3) = 4/3. The area below the x-axis between x = -1 and x = 0 equals (-1^{2}- (-1)^{3}/3) - (0^{2}- 0^{3}/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)

- see https://www.youtube.com/watch?v=QUdhxX_Oq3A for an example

maths/calculus.txt · Last modified: 2021/10/03 18:48 by gary1