Calculus

• differential and integral calculus rules
• implicit differentiation
• see also:
  • Algebra
  • Mensuration
  • Circular functions
  • Coordinate geometry, parabolae, etc
  • Finance equations
  • Logarithmic, exponential and hyperbolic functions
  • Probability theory

Differential and integral calculus rules:

• f(x)	f'(x)	Integral of f(x)
  x^n	nx^(n-1)	(xn+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;

Implicit Differentiation:

• 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

Integral calculus:

• 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)