- 1 Differentiation
- 2 Integration
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.
The following are various notations for derivatives.
Note that: y = ƒ(x) = ƒ
dy / dx
dƒ / dx
dƒ(x) / dx
Notation for second derivative:
d2y / dx2
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)
Differentiation can be used to find the x-intercepts of a line.
The following formulae is used to find this:
xn+1 = xn - ƒ(xn) / ƒ′(xn)
Applying this function until ƒ(x) ≈ 0
The initial x is just a guess.
Note: if this results in divergence (x does not converge at 0 or x → ∞) then x1 was a bad guess.
Approximating Roots and Powers
Differentiation can be used to approximate roots.
For example approximating the answer to:
First round to the nearest known value:
3√(125) = 5
Differentiate the original equation:
ƒ(x) = 3√x = x1/3
ƒ′(x) = x-2/3 / 3
Substitute the approximation:
ƒ′(120) = 120-2/3 / 3
120-2/3 / 3 = 0.014
Place this into the equation of gradient:
(y - 5) / (x - 125) = 0.014
Prearrange to the formulae of a line:
y = 0.014x + 3.29
Substitute for the number to approximate with x:
y = 0.014(120) + 3.29
y = 4.934
3√(120) ≈ 4.934
3√(120) = 4.932 (calculated)
A turning point is the point where the gradient is 0, an equation may have multiple, quadratics usually have one, cubic functions usually have 2.
The turning point can be found by obtaining the first derivative (ƒ′(x)).
Then solve for when ƒ′(x) = 0.
The results of this are the x values of the turning points, the y values can be found by substituting the x values into the original equation.
Minimum and Maximum
A Minimum is the the turning point that is Concave Up.
A Maximum is the the turning point that is Concave Down.
The Minimum can be found where the first derivative (ƒ′(x)) is 0 and the second derivative (ƒ′′(x)) is positive.
ƒ′(x) = 0 and ƒ′′(x) > 0
The Maximum can be found where the first derivative (ƒ′(x)) is 0 and the second derivative (ƒ′′(x)) is negative.
ƒ′(x) = 0 and ƒ′′(x) < 0
Concave Up and Down
Concave Up is the part of the graph that is U shape.
Concave Down is the part of the graph that is n shape.
Cubic functions have one of each, quadratics have one or the other.
To find out if a point is Concave Up or Down, find the second derivative (ƒ′′(x)).
If ƒ′′(x) > 0, then the point is Concave Up.
If ƒ′′(x) < 0, then the point is Concave Down.
Point of Inflection
The Point of Inflection (P.O.I.) is the point on the graph where Concave Up changes to Concave Down.
The P.O.I. is when the second derivative (ƒ′′(x)) is equal to 0 and it changes from a negative value to a positive one immediately beforehand (or vice versa).
ƒ′′(x) = 0
ƒ′′(x - 0.1) < 0
ƒ′′(x + 0.1) > 0
ƒ′′(x) = 0
ƒ′′(x - 0.1) > 0
ƒ′′(x + 0.1) < 0
x = P.O.I.
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.
k represents a constant.
∫k dx = k + c
∫kx dx = k∫x dx
∫xn dx = xn + 1 / (n + 1)
∫(xn + xm) dx = ∫xn dx + ∫xm dx
∫( ƒ(x) ± g(x) )dx = ∫ƒ(x)dx ± ∫g(x)dx
∫(1 / x) dx = Ln(x) + c
Note: Ln(x) = Loge(x)
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.
Solid of Revolution
A cool use for integration is to calculate a solid of revolution - the volume of a shape that is made from the area of two points spun around a axis, for example a triangle could create a cone.
The formulae for this is as follows:
V = Π a∫b ƒ2(x) dx
Note: ƒ2(x) = ƒ(x) * ƒ(x)