Example 1 Find the derivative of Function is

Example 1 Find the derivative of Function is written in terms of f(x), answer should be written in terms of f ′ (x)

Example 2 Find the derivative of Function is written in terms of y = answer should be written in terms of dy/dx =

Example 3 Differentiate s with respect to t for Function is written in terms of s and t so derivative will be ds/dt =

Example 4 Harder examples Find the derivative of

Example 5 Harder examples Find the derivative of

Example 6 Harder examples Find the derivative of

Example 7 Harder examples Find the derivative of

Example 8 Harder examples Find the derivative of

Example 9 Harder examples Find the derivative of

Stationary Points Example A function is given by f (x) = 4 x 3 – x 4. Find: (i) the coordinates and nature of the stationary points on the curve; (ii) the intervals on which f (x) is increasing or decreasing.

Stationary Points Step 1 Find the stationary points Stationary points occur when f ′ (x) = 0 Finding y coordinates of function f (x) When x = 0, When x = 3, Stationary points occur at ( 0 , 0 ) and ( 3 , 27 )

Step 2 points Determine the nature of the stationary Consider the gradient f ′ (x) around each stationary point: x f ′(x) ® 0 + ve ® 0 3 + ve ® 0 − ve slope pick a value At the point (0, 0) the shape of the graph is + ve 0 pick + ve a value for x > 3 ® at means approaching 0, so test between 0 and 3 ∴ the xpoint (0, 0) there is athe rising point of inflexion gradient by picking a suitable value for x < 0 and substituting in to f ′(x) At the point (3, 27) the shape of the graph is + ve 0 - ve ∴ at the point (3, 27) there is a maximum turning point

(ii) From the table we can see the shape of the whole graph: Stationary Points x ® 0 ® 3 ® f ′(x) + ve 0 − ve slope ↗ → ↘ This tells us that f is increasing for values of x up to 3, except at 0. f is increasing for and that f is decreasing for values of x greater than 3. f is decreasing for Back

Curve Sketching Sketch the curve y = x 3 – 3 x 2. Step 1 The y-intercept The curve cuts the y-axis when x = 0 the y-intercept is at ( 0 , 0 ). Step 2 The x-intercept The curve cuts the x-axis when y = 0 the x-intercepts are at ( 0 , 0 ) and (3, 0).

Curve Sketching Step 3 Stationary points So For stationary values, To find y coordinates when x = 0 To find x coordinates when x = 2 Stationary values occur at ( 0 , 0 ) and ( 2 , − 4 )

Curve Sketching Step 3 Stationary points cont’d. Stationary values occur at ( 0 , 0 ) and ( 2 , − 4 ) The nature of the stationary values is best described through the gradient table as before x ® + ve 0 ® 0 2 − ve ® 0 + ve slope Remember to use dy/dx and select suitable values to test the gradient. (− 1) (1) == 9 (3) -3 9 Maximum Turning Point at ( 0 , 0) and Minimum turning point at ( 2 , − 4 )

Curve Sketching Step 4 Large positive and negative values of x The function y = x 3 – 3 x 2 will behave like x 3 for very large values of x. If x is very large and positive x 3 is large and positive so y is large and positive If x is very large and negative x 3 is large and negative so y is large and negative

Step 5 Curve Sketching Sketch the graph y = x 3 – 3 x 2 (0, 0) (3, 0) (2, -4)

Closed Intervals Example 1 Max 2 6 (2, − 2) Min (4, − 5) For the interval 2 ≤ x ≤ 6 sometimes written as [2, 6] The minimum value is − 5 at x = 4 The maximum value is 0 at x = 6

Closed Intervals Example 2 Find the maximum and minimum values of f(x) = 4 x 3 −x 2 − 4 x + 1 in the closed intervals (a) − 1 ≤ x ≤ 2 (b) [− 1, 1] Maximum and minimum values are either at a stationary point or at an end point, so start by finding these points. Step 1 End Points

Closed Intervals Step 2 Stationary Points For stationary points To find y-coordinates Stationary points occur at and

Closed Intervals Step 2 Stationary Points (Nature Table) x ® ® f ′ (x) + ve 0 − ve 0 + ve slope ↗ → ↘ → ↗ Maximum turning point at And, minimum turning point at ®

Closed Intervals Step 3 Back Sketch the graph 21 9/4 1 -1 -1 2 -25/27 (a) For -1 ≤ x ≤ 2 (b) For [-1, 1] or -1 ≤ x ≤ 1 The maximum value of f is 21 The maximum value of f is 9/4 The minimum value of f is -25/27

Problem Solving The volume of the square based display case shown is 500 cm 3. The length of the base is x cm. The base of the case is not made of glass. h x cm (a) Show that the area of glass in the case is (b) Find the dimensions of the case that minimise the use of glass and calculate this minimum area. This problem can best be solved using differentiation. Find a formula for the volume in terms of x and h (height). Rearrange to get an equation for h in terms of x Use that to find a formula for the total area of glass in terms of x.

Problem Solving (a) The total area of glass is given by, A(x) = area of top + area of 4 sides h x cm To find the area of glass we need to find an expression for the height of the case. Volume = length x breadth x height Total area of glass, A(x) = area of top + 4 x area of a side x cm

Problem Solving (b) By finding the minimum stationary point for A(x) we can obtain the value of x for which the area is a minimum. For a stationary point Check that the stationary point at 10 is a minimum x ® 10 ® f ′ (x) - ve 0 + ve slope ↘ → ↗ A minimum occurs when x = 10. When x = 10, length =10, breadth = 10 and height = 5 The minimum area of glass is

Rate of Change The number of bacteria, N(t), in a certain culture is calculated using the formula N(t) = 4 t 3 + 400 t + 1000 where t is the time in hours from the start of the growth. Calculate the growth rate of this culture when t=5. The rate of change is a derative so we need to find N ′(t). t=5, so substitute 5 into N’(t) for t. The growth rate of the culture is 700 bacteria per hour after 5 hours. Back

Velocity, Speed and Acceleration The speed of a particle, relative to its starting position, is given by where t is the time in seconds and d(t) is the distance in cm. (a) Calculate the speed of the particle when t = 5 (b) (i) Find a function to describe the acceleration of (ii) What is the acceleration when t = 5? the particle

(a) Velocity, Speed and Acceleration The speed of the particle is its rate of change of distance with time, given by d’(t). So we need to differentiate : The speed of the particle after 5 seconds is 143 cm/s

(b) (i) Velocity, Speed and Acceleration To find the acceleration of the particle we need to find how quickly it is changing speed i. e. the rate of change of speed with time. We find this by differentiating the speed. The speed was found in (a) as d′(t). We can call this s(t) where s is speed. Acceleration will then be s′(t) (b) (ii) So the acceleration of the particle at t = 5 is 42 cm/s per second