User:Saul/calculus
Differentiation
We will use the function notation ƒ(x) which just applies some action to x like this function will double x: ƒ(x) = 2x
Differentiation of a function is taking a function (usually a curve) and finding the gradient at the single instant of x.
For example the function ƒ(x) = x2 will represent a 'U' shaped curve, the gradient at point x will be: ƒ′(x) = 2x
A derivative of a function will be notated with ƒ′, second derivatives are marked ƒ′′ and so on.
Various Rules
Note: n represents a real number, and a represents a constant.
ƒ(x) = xn
ƒ′(x) = nxn-1
ƒ(x) = axn
ƒ′(x) = anxn-1
ƒ(x) = a
ƒ′(x) = 0
ƒ(x) = x
ƒ′(x) = 1
ƒ(x) = fg
ƒ′(x) = f′g + g′f
ƒ(x) = f/g
ƒ′(x) = (f′g - g′f) / g2
ƒ(x) = sin(x)
ƒ′(x) = cos(x)
ƒ(x) = cos(x)
ƒ′(x) = -sin(x)
ƒ(x) = tan(x)
ƒ′(x) = 1 / cos2(x) = sec2(x)
Integration
Integration is the reverse of differentiation, sometimes known as anti-derivative.
Integration is usually noted like so:
∫ƒ′(x)dx = ƒ(x) + c
Where dx shows that it will be the integral relative to x (there may be other variables in the equation).
c represents a unknown constant - this must exist because when a function is derived (differentiation) some information is lost so c is to make up for that loss.
Various Rules
k represents a constant.
∫k dx = k + c
∫kx dx = k∫x dx
∫xn dx = xn + 1 / (n + 1)
∫( ƒ(x) ± g(x) )dx = ∫ƒ(x)dx ± ∫g(x)dx
∫(1 / x) dx = Ln(x) + c
Note: Ln(x) = Loge(x)
Uses
Area
A use of integration is to find the area between a line and the x axis between two points on a graph.
The area between the two points a and b (where x = 0) can be found like so:
First find the derivative:
a∫bƒ′(x) dx = ƒ(x)
Say that the derivative is F, then find the difference of the values for F for each limit like so:
F(b) - F(a) = Area
Note that the area will be negative if it is under the x axis and positive if above.
Also if the line crosses the x axis more than twice then there will be more than two values where x = 0 and at least one positive area and a negative area, these will at least partially cancel each other out if only the left most and right most x values are taken, so to effectively calculate the area the absolute areas between every set of points (where x = 0) muse be calculated.