Differentiation Flashcards
Gradient of lines,
description
formula
note
The gradient of a line is a value that describes how steep the line is
m = ∆y / ∆x
On a line the gradients is constant regardless of where you measure it
Gradient of curves,
description
process of finding
On a curve the gradient is changing all the time
The gradient function is sometimes called the derived function or derivative, and the process of finding the gradient is called differentiation
Notations for the first derivative,
type 1
type 2
type 3
f(x) –> f’(x)
y –> y’
y –> dy/dx
Derivatives of polynomials,
power rule
constant times function
derivative of constant is zero
sum and difference
expanding
simplifying
chain rule (composite functions); 3 types formula
power rule(composite functions);
description
formula
f(x) = x^n f'(x) = n • x^(n-1)
f(x) = ax^n f'(x) = n • ax^(n-1)
f(x) = a f'(x) = 0
f(x) = ax^m + bx^n f'(x) = m • ax^(m-1) + n • bx(n-1)
f(x) = ax^m - bx^n f'(x) = m • ax^(m-1) - n • bx(n-1)
f(x) = x^m(x +a) = x^(m+1) + ax(m)
f'(x) = { (m+1) } • { x^(m+1)-1 } + { m • ax^(m-1) }
f(x) = (ax^m + b) / x^n = (ax^m / x^n) + (b / x^n) =
ax^(m-n) + bx^-n
f’(x) = { (m-n) } • { ax^[(m-n)-1 ] } { - n } x { bx^(-n-1) }
Chain rule:
If y is a function of u , y(u)
and u is a function of x, u(x)
then y is a function of x, y(x)
y’(x) = dy/dx = du/dx • dy/du
Power rule:
This can be solved by power rule, if function notation the general power rule appears as follows
f(x) = [u(x)]^n f'(x) = n • [u(x)]^(n-1) • du/dx
Tangents to curves,
description
process of finding
The gradient at a point on a curve is the same as the gradient of the tangent at that point
Given the coordinates of (x1, y1) of a point on a curve:
1. find the gradient function (derivative)
2. Substitute x1 into the derived function to find m
3. Substitute the gradient m and the point into the equation of the tangent line,
y - y1 = m(x - x1)
Normal to curves,
description
process of finding
The gradient of the normal can be worked out from the gradient of the tangent, m1 x m2 = -1
Given the coordinates of (x1, y1) of a point on a curve:
1. find the gradient function (derivative)
2. Substitute x1 into the derived function to find m
3. Find the negative reciprocal of m
4. Substitute the gradient m and the point into the equation of the tangent line,
y - y1 = m(x - x1)
Notations for the second derivative,
type 1
type 2
type 3
f(x) –> f’(x) –> f”(x)
y –> y’ –> y”
y –> dy/dx –> d^2y / dx^2
Increasing functions,
description
formula
If the gradient is positive over a range of values then the function is said to be increasing, the y value is increasing as the x value is, curve is sloping up from left to right
f’(x) > 0
Decreasing functions,
description
formula
If the gradient is negative over a range of values then the function is said to be decreasing, the y value is decreasing as the x value is increasing, curve is sloping down from right to left
f’(x) < 0
Stationary points,
description
formula
If the gradient of a point is zero then this point is called a stationary point, it can be either a maximum or minimum stationary point, they can also be called turning points
m = 0 hence f’(x) = 0
Point of inflection,
description
formula or finding
Another type of stationary point is called a point of inflection, with this type of point the gradient is zero but the gradient on either side remains positive or negative
find stationary point and then points on either side to decipher whether the point of inflection is negative or positive
Finding stationary points,
steps
second derivative method
- Differentiate the function to find f’(x)
- Put f’(x) = 0 as the gradient at the stationary point
- Solve the equation to find the x-value of the stationary point
- Sub this value into the original equation to find its corresponding y value
- Examine the gradient on either side to determine its nature or use the second derivative method
Second derivative:
- f”(x) < 0 means max point
- f”(x) > 0 means min point
- f”(x) = 0 means point of inflection
Applications of calculus,
finding max and min points
distance, velocity and acceleration problems;
velocity formula
acceleration formula
kinematic problem notes;
distance
velocity
acceleration
rates of change
connected rates of change
The same techniques for finding max and min points of graphs can be used to find the max and min of problems
v = s’ = ds/dt
a = v’ = dv/dt
Distance:
s = 0, object at starting point
s > 0, object above or to the right of starting point
s < 0, object below or the the left of starting point
Velocity:
v = 0, object is stationary
v > 0, object moving right or upwards
v < 0, object moving left or backwards
Acceleration:
a = 0, object moving at constant velocity
a > 0, object speeding up
a < 0, object slowing down
Differentiation gives a measure of a rate of change at a particular instant, giving an instantaneous rate of change
dy/dx = du/dx • dy/du
