**Calculus**

**Differential and integral
calculus rules:**

**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;

- A function f, is
differentiable at a point x=a, if the derivative
f'(x) exists at x=a:

- 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-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.

- thus if f(x) = x(2-x) = 2x-x
- 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)