1
Q
Angle of slope at a point inflection
A
tan^-1 (f’(x))
2
Q
Limits
A
The limits of a function are y-values that are found by approaching an x-value from the left and right.
3
Q
Cases where there is no limit
A
- When a function has two limits at a point (gap)
- When a function is undefined at a point (exponential)
- When a function tends to infinity at a point (hyperbola asymptote)
4
Q
Calculating limits using a calculator
A
Calculate the value of the function for values of x close to a (above and below)
5
Q
Calculating limits using direct substitution
A
Evaluate the limit of f(x) as x approaches a = f(a)
6
Q
Calculating limits using algebraic cancellation
A
Factorise the numerator and/or denominator, ‘canceling’ common factors and then substituting the x value given to solve
7
Q
Summary of calculating limits
A
- ‘Sensible’ answer - This is the limit
- number ≠ 0 / 0 - No limit
- 0 / number ≠ 0 - Limit is 0
- 0 / 0 - Factorise, cancel, then repeat process
8
Q
Limits as x tends to infinity
A
x → ∞ indicates that the limit of f(x) tends to infinity. To solve, divide each term by the highest power of x in the denominator, letting (a/x) = 0
9
Q
Differentiating exponential functions with base e
A
If f(x) = e^(g(x)), then f’(x) = g’(x) x e^(g(x))
10
Q
General advice for differentiating
A
- The constant stays when differentiating
- Simplify before differentiating if needed (eg. move terms to top of the fraction by inverting the sign of the power)
- Differentiate sums separately
11
Q
Continuity
A
A function is continuous if the value of the limit at a point is equal to the value of the function at the point.
12
Q
Cases where there is a discontinuity
A
- When a function has a limit at a point
- When a function has different rules for different parts of domain (gap between limit and defined point)
- When a function tends to infinity at a point (hyperbola asymptote)
13
Q
Differentiable
A
A function is differentiable at a point if the derived function is defined at that point.
14
Q
What does discontinuity imply?
A
Discontinuity implies that it is not differentiable at a point.
15
Q
What does differentiability imply?
A
Differentiability implies that it is continuous at a point.
16
Q
When is a function considered continuous?
A
A function is only considered to be continuous if it is continuous at every point in its domain
17
Q
When is a function considered differentiable?
A
A function is only considered to be differentiable if it is differentiable at every point in its domain
18
Q
Differentiating a log function with base e
A
If f(x) = ln [g(x)], then f’(x) = g’(x) / g(x)
- Rearranging can be done using k • logb (m) = logb (m^k)
19
Q
When differentiating fractions
A
- Move any powers of x in the bottom half of the fraction to the top half by inverting the sign on the power (and multiplying with the existing terms)
- Differentiate the top half of the fraction
- Simplify by moving negative powers of x in the top half of the fraction to the bottom half by inverting the sign on the power again
20
Q
When differentiating roots
A
- Remove the root sign by using powers
- Simplify positive fractional powers
- Differentiate
- Remember squares under the root sign can be cancelled out by moving the x value in front of the root sign
21
Q
General advice for differentiating roots
A
- Use radical rule to simplify if needed √a√a = a
- Separate square roots in numerator and denominator √(x + a / x + b) = √(x+a) / √(x+b)
22
Q
Differentiating exponential functions with base a
A
If f(x) = a^(g(x)), then f’(x) = ln(a) x g’(x) x a^(g(x))
23
Q
Simplifying complex expressions
A
- Simplify to a form where brackets in each term are equal
- Take out common factors using the lowest powers to put in front of the brackets. Write the remaining terms inside the [ ] brackets
- Simplify the [ ] brackets
24
Q
Fraction rules
A
b / (a/c) = bc / a
(b/a) / c = b / ac
- The bottom number of the ‘small’ fraction always moves, but if the ‘small’ fraction is on the top it moves to the bottom, and if it is on the bottom it moves to the top (to be multiplied with the other number).
25
Perfect square formulas
(a + b)^2 = a^2 + 2ab + b^2
| (a - b)^2 = a^2 - 2ab + b^2
26
Differentiating trigonometry functions
- Simplify using trigonometric rules
| - Differentiate using chain rule
27
Stationary points
At a stationary point, f'(x) = 0
| max, min or point of inflection
28
Calculating the co-ordinates of a turning point on the graph of a function
1. Differentiate the function
2. Make f'(x) = 0
3. Solve to find x
4. Substitute x back into the function
5. Solve to find y
6. Turning point is at (x, y)
- Graph can then be drawn
29
Positive and negative functions
A function is positive when f(x) > 0 (above x - axis)
| A function is negative when f(x) < 0 (below x - axis)
30
Increasing, non-decreasing and decreasing functions
A function is increasing when f'(x) > 0 for all values of x
A function is non-decreasing when f'(x) ≥ 0 for all values of x (thus includes stationary points)
A function is decreasing when f'(x) < 0 for all values of x
31
Perfect cubes formulas
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
| a^3 - b^3 = (a - b)(a^2 + ab + b^2)
32
Finding stationary points
1. Solve f'(x) = 0
2. Solve f''(x)
- If f''(x) < 0, it is a maximum point (concave down)
- If f''(x) > 0, it is a minimum point (concave up)
- If f''(x) = 0, the point is indeterminate and further investigation is necessary
3. In indeterminate case, calculate the value of the function at points either side of the point. Check the signs of each value to determine if it is a minimum, maximum or point of inflection.
- If values are positive & negative, it is a point of inflection
- If values are both positive, it is a minimum point
- If values are both negative, it is a maximum point
33
General solutions
1. Solve equation to find the value of the trig function
2. Solve trig function to find the value of x. This is possible solution (1)
3. Solve π - (1) to find the other value of x. This is possible solution (2)
4. Use the given range to determine the appropriate value of x (positive or negative)
5. Substitute the value of x into the equation to solve for the value of y
34
When is a function undefined?
A function is undefined when the denominator is equal to 0
35
Calculating the gradient at a point on a curve
1. Find f'(x)
| 2. Substitue the x-value of the point into f'(x) and solve to find the m-value (gradient)
36
Calculating the equation of a tangent to a curve
1. Find f'(x)
2. Substitude x-value of point into f'(x) and solve to find the m-value (tangent gradient)
3. Substitute values into equation y - y1 = m(x - x1), where y1 is the y value of the point, x1 is the x value of the point and m is the tangent gradient
4. Rearrange equation into y = mx + c format
37
Converting between the normal gradient and the tangent gradient
Normal gradient x tangent gradient = -1
(The normal gradient is the negative reciprocal or the tangent gradient, except when tangent is a horizontal or vertical line)
38
Calculating the equation of a normal to a curve
1. Find f'(x)
2. Substitute x-value of point into f'(x) and solve to find m-value (gradient of tangent)
3. Calculate the normal gradient from the tangent gradient
4. Substitue values into equation y - y1 = m(x - x1), where y1 is the y value of the point, x1 is the x value of the point and m is the normal gradient
5. Rearrange equation into y = mx + c format
(If not initially given the y-value, substitute value of x given into the equation and solve to find y)
39
Calculating points on a curve which have a particular gradient
1. Find f'(x)
2. Make f'(x) equal to the given gradient
3. Rearrange the equation to equal 0
4. Solve to find values of x
5. Substitue x values back into the original equation and solve to find the corresponding y values
40
Calculating the inital displacement, velocity and acceleration
s(0) gives inital displacement
v(0) gives inital velocity
a(0) gives inital acceleration
41
Meaning when distance/displacement, velocity and acceleration = 0
```
s = 0 means the object is at the origin
v = 0 means the object is momentarily at rest or at maximum/minimum distance from origin
a = 0 means the velocity is constant and that the object is not accelerating
```
42
Calculating the distance travelled between t1 and t2 seconds
t2∫t1 v(t) gives the distance travelled between t1 and t2 seconds
43
Meaning when distance/displacement, velocity and acceleration are < or > 0
s < 0 means the object is below or to the left of the origin
s > 0 means the object is above or to the right of the origin
v < 0 means the object is travelling backwards
v > 0 means the object is travelling forwards
a < 0 means the object is slowing down (deceleration)
a > 0 means the object is speeding up (acceleration)
44
Related rates of change
1. Define variables from the question
2. Determine the relationship between two variables given in the question to find a derivative (figured out)
3. Determine another derivative given from the question that is equal to a value (given)
4. Determine the last derivative (calculated) by writing an equation to link the three derivatives together using the chain rule
- Two of the derivatives usually involve a rate of change with respect to time
6. Substitute the values for the two known derivatives
7. Use the value for a variable given in the question to solve for the final derivative
45
Obtaining the gradient dy/dx at any point on the curve in terms of a parameter t
First derivative -
The gradient = dy/dx = y'/x'
Second derivative -
The gradient = d^2y/dx^2 = d(dy/dx) / dx
(d(dy/dx) is found by first calculating the first derivative)
46
Calculating the equation of a tangent to a parametrically defined curve
1. A point, gradient and parameter is needed to determine the equation
- To calculate the parameter, make each equation for x and y respectively equal to the x and values given from the point and solve to calculate the parameter value. This will give two values, the cross over of which is the correct value.
- To calculate the point, solve each equation for x and y respectively by substituting in the parameter value given
- To calculate the gradient, calculate the first derivative (dy/dx = y'/x')
2. Substitue values into equation y - y1 = m(x - x1), where y1 is the y value of the point, x1 is the x value of the point and m is the tangent gradient
5. Rearrange equation into y = mx + c format
47
Calculating the equation of a normal to a parametrically defined curve
1. A point, gradient and parameter is needed to determine the equation
- To calculate the parameter, make each equation for x and y respectively equal to the x and values given from the point and solve to calculate the parameter value. This will give two values, the cross over of which is the correct value.
- To calculate the point, solve each equation for x and y respectively by substituting in the parameter value given
- To calculate the gradient, calculate the first derivative (dy/dx = y'/x'). This gives the tangent gradient, which the negative reciprocal will then need to be taken of to determine the normal gradient.
2. Substitue values into equation y - y1 = m(x - x1), where y1 is the y value of the point, x1 is the x value of the point and m is the normal gradient
5. Rearrange equation into y = mx + c format
48
Tangents and normals of vertical lines
- Vertical lines have an undefined gradient (denominator of gradient fraction is 0)
49
Implicit differentation for when curves have more than one x value associated with the y value
1. Differentiate both sides of the equation with respect to x
- To differentiate powers of y with respect to x, treat it as a composite function and use the chain rule
- Remember to account for y by adding dy/dx after it
2. Solve the resulting equation for dy/dx
50
Differentiating composite functions
d/dx [u^n] = n [u]^n-1 • u'
51
Differentiating trig functions
- If trig function is to a power, consider it as a composite function (rewrite with power outside the brackets)
- Calculate the derivative of each term starting from the outside of the brackets and working inwards
- Multiply each derivative together to get the overall answer
eg.
tan^2(2x) = 2 tan(2x) x sec^2(2x) x 2 = 4 tan(2x)sec^2(2x)
tan(2x) = sec^2(2x) x 2 = 2 sec^2(2x)
tan(x) = sec^2(x) x 1 = sec^2(2x)
