Differentiation

Question 1

Angle of slope at a point inflection

Answer 1

tan^-1 (f’(x))

Question 2

Limits

Answer 2

The limits of a function are y-values that are found by approaching an x-value from the left and right.

Question 3

Cases where there is no limit

Answer 3

• 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)

Question 4

Calculating limits using a calculator

Answer 4

Calculate the value of the function for values of x close to a (above and below)

Question 5

Calculating limits using direct substitution

Answer 5

Evaluate the limit of f(x) as x approaches a = f(a)

Question 6

Calculating limits using algebraic cancellation

Answer 6

Factorise the numerator and/or denominator, ‘canceling’ common factors and then substituting the x value given to solve

Question 7

Summary of calculating limits

Answer 7

1. ‘Sensible’ answer - This is the limit
2. number ≠ 0 / 0 - No limit
3. 0 / number ≠ 0 - Limit is 0
4. 0 / 0 - Factorise, cancel, then repeat process

Question 8

Limits as x tends to infinity

Answer 8

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

Question 9

Differentiating exponential functions with base e

Answer 9

If f(x) = e^(g(x)), then f’(x) = g’(x) x e^(g(x))

Question 10

General advice for differentiating

Answer 10

• 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

Question 11

Continuity

Answer 11

A function is continuous if the value of the limit at a point is equal to the value of the function at the point.

Question 12

Cases where there is a discontinuity

Answer 12

• 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)

Question 13

Differentiable

Answer 13

A function is differentiable at a point if the derived function is defined at that point.

Question 14

What does discontinuity imply?

Answer 14

Discontinuity implies that it is not differentiable at a point.

Question 15

What does differentiability imply?

Answer 15

Differentiability implies that it is continuous at a point.

Question 16

When is a function considered continuous?

Answer 16

A function is only considered to be continuous if it is continuous at every point in its domain

Question 17

When is a function considered differentiable?

Answer 17

A function is only considered to be differentiable if it is differentiable at every point in its domain

Question 18

Differentiating a log function with base e

Answer 18

If f(x) = ln [g(x)], then f’(x) = g’(x) / g(x)

• Rearranging can be done using k • logb (m) = logb (m^k)

Question 19

When differentiating fractions

Answer 19

• 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

Question 20

When differentiating roots

Answer 20

• 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

Question 21

General advice for differentiating roots

Answer 21

• 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)

Question 22

Differentiating exponential functions with base a

Answer 22

If f(x) = a^(g(x)), then f’(x) = ln(a) x g’(x) x a^(g(x))

Question 23

Simplifying complex expressions

Answer 23

1. Simplify to a form where brackets in each term are equal
2. Take out common factors using the lowest powers to put in front of the brackets. Write the remaining terms inside the [ ] brackets
3. Simplify the [ ] brackets

Question 24

Fraction rules

Answer 24

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).

Question 25

Perfect square formulas

Answer 25

(a + b)^2 = a^2 + 2ab + b^2 | (a - b)^2 = a^2 - 2ab + b^2

Question 26

Differentiating trigonometry functions

Answer 26

- Simplify using trigonometric rules | - Differentiate using chain rule

Question 27

Stationary points

Answer 27

At a stationary point, f'(x) = 0 | max, min or point of inflection

Question 28

Calculating the co-ordinates of a turning point on the graph of a function

Answer 28

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

Question 29

Positive and negative functions

Answer 29

A function is positive when f(x) > 0 (above x - axis) | A function is negative when f(x) < 0 (below x - axis)

Question 30

Increasing, non-decreasing and decreasing functions

Answer 30

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

Question 31

Perfect cubes formulas

Answer 31

a^3 + b^3 = (a + b)(a^2 - ab + b^2) | a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Question 32

Finding stationary points

Answer 32

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

Question 33

General solutions

Answer 33

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

Question 34

When is a function undefined?

Answer 34

A function is undefined when the denominator is equal to 0

Question 35

Calculating the gradient at a point on a curve

Answer 35

1. Find f'(x) | 2. Substitue the x-value of the point into f'(x) and solve to find the m-value (gradient)

Question 36

Calculating the equation of a tangent to a curve

Answer 36

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

Question 37

Converting between the normal gradient and the tangent gradient

Answer 37

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)

Question 38

Calculating the equation of a normal to a curve

Answer 38

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)

Question 39

Calculating points on a curve which have a particular gradient

Answer 39

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

Question 40

Calculating the inital displacement, velocity and acceleration

Answer 40

s(0) gives inital displacement v(0) gives inital velocity a(0) gives inital acceleration

Question 41

Meaning when distance/displacement, velocity and acceleration = 0

Answer 41

``` 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 ```

Question 42

Calculating the distance travelled between t1 and t2 seconds

Answer 42

t2∫t1 v(t) gives the distance travelled between t1 and t2 seconds

Question 43

Meaning when distance/displacement, velocity and acceleration are < or > 0

Answer 43

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)

Question 44

Related rates of change

Answer 44

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

Question 45

Obtaining the gradient dy/dx at any point on the curve in terms of a parameter t

Answer 45

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)

Question 46

Calculating the equation of a tangent to a parametrically defined curve

Answer 46

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

Question 47

Calculating the equation of a normal to a parametrically defined curve

Answer 47

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

Question 48

Tangents and normals of vertical lines

Answer 48

- Vertical lines have an undefined gradient (denominator of gradient fraction is 0)

Question 49

Implicit differentation for when curves have more than one x value associated with the y value

Answer 49

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

Question 50

Differentiating composite functions

Answer 50

d/dx [u^n] = n [u]^n-1 • u'

Question 51

Differentiating trig functions

Answer 51

- 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)