Higher Unit 1 Applications Finding the gradient for a polynomial Differentiating Easy Functions Differentiating Harder Functions Differentiating with Leibniz Notation Equation of a Tangent Line Increasing / Decreasing functions Max / Min and inflexion Points Curve Sketching Max & Min Values on closed Intervals Optimization Higher Mind Map of Chapter

On a straight line the gradient remains constant, however with curves the gradient changes continually, and the gradient at any point is in fact the same as the gradient of the tangent at that point. The sides of the half-pipe are very steep(S) but it is not very steep near the base(B). B S Gradients & Curves Higher Demo

A Gradient of tangent = gradient of curve at A B Gradient of tangent = gradient of curve at B Gradients & Curves Higher

Gradients & Curves Higher For the function y = f(x) we do this by taking the point (x, f(x)) and another “very close point” ((x+h), f(x+h)). Then we find the gradient between the two. (x, f(x)) ((x+h), f(x+h)) True gradient Approx gradient To find the gradient at any point on a curve we need to modify the gradient formula

The gradient is not exactly the same but is quite close to the actual value We can improve the approximation by making the value of h smaller This means the two points are closer together. (x, f(x)) ((x+h), f(x+h)) True gradient Approx gradient Gradients & Curves Higher

We can improve upon this approximation by making the value of h even smaller. (x, f(x)) ((x+h), f(x+h)) True gradient Approx gradient So the points are even closer together. Gradients & Curves Higher Demo

Higher Derivative We have seen that on curves the gradient changes continually and is dependant on the position on the curve. ie the x-value of the given point. The process of finding the gradient is called DIFFERENTIATING or FINDING THE DERIVATIVE (Gradient) Differentiating Finding the GRADIENT Finding the rate of change

If the formula/equation of the curve is given by f(x) Then the derivative is called f '(x) - “f dash x” There is a simple way of finding f '(x) from f(x). f(x) f '(x) 2x 2 4x 4x 2 8x 5x 10 50x 9 6x 7 42x 6 x 3 3x 2 x 5 5x 4 x 99 99x 98 Derivative Higher

If f(x) = ax n n - 1 n Rule for Differentiating It can be given by this simple flow diagram... multiply by the power reduce the power by 1 then f '(x) = NB: the following terms & expressions mean the same GRADIENT,DERIVATIVE,RATE OF CHANGE,f '(x) Derivative Higher ax

Rule for Differentiating To be able to differentiate it is VERY IMPORTANT that you are comfortable using Fractions and Surds & Indices rules Derivative Higher

Surds & Indices x m. x n = x m+n

Surds & Indices HHM page 340 Ex8 HHM page 342 Ex9 Do YOU need extra help or revision then do

Special Points (I) f(x) = ax (Straight line function) If f(x) = ax = ax 1 then f '(x) = 1 X ax 0 = a X 1 = a Index Laws x 0 = 1 So if g(x) = 12x then g '(x) = 12 Also using y = mx + c The line y = 12x has gradient 12, and derivative = gradient !! Higher

(II) f(x) = a, (Horizontal Line) If f(x) = a = a X 1 = ax 0 then f '(x) = 0 X ax -1 = 0 Index Laws x 0 = 1 So if g(x) = -2 then g '(x) = 0 Also using formula y = c, (see outcome 1 !) The line y = -2 is horizontal so has gradient 0 ! Special Points Higher

Name : Gradient = Rate of change = Differentiation Differentiation techniques

Calculus Revision Differentiate

Calculus Revision Differentiate

Calculus Revision Differentiate

HHM Ex6D, Ex6E and Ex6F Even Numbers only Derivative Higher

Example 1 A curve has equation f(x) = 3x 4 Its gradient is f '(x) = 12x 3 f '(2) = 12 X 2 3 =12 X 8 = 96 Example 2 A curve has equation f(x) = 3x 2 Find the formula for its gradient and find the gradient when x = 2 Its gradient is f '(x) = 6x At the point where x = -4 the gradient is f '(-4) = 6 X -4 =-24 Derivative Higher Find the formula for its gradient and find the gradient when x = -4

Example 3If g(x) = 5x 4 - 4x 5 then find g '(2). g '(x) = 20x x 4 g '(2) = 20 X X 2 4 = = -160 Derivative Higher

Example 4 h(x) = 5x 2 - 3x + 19so h '(x) = 10x - 3 and h '(-4) = 10 X (-4) - 3 = = -43 Example 5 k(x) = 5x 4 - 2x x - 8, find k '(10). k '(x) = 20x 3 - 6x So k '(10) = 20 X X = Derivative Higher

Example 6 : Find the points on the curve f(x) = x 3 - 3x 2 + 2x + 7 where the gradient is 2. NB: gradient = derivative = f '(x) We need f '(x) = 2 ie 3x 2 - 6x + 2 = 2 or 3x 2 - 6x = 0 ie 3x(x - 2) = 0 ie 3x = 0 or x - 2 = 0 so x = 0 or x = 2 Now using original formula f(0) = 7 f(2) = = 7 Points are (0,7) & (2,7) Derivative Higher

Calculus Revision Differentiate

Calculus Revision Differentiate Straight line form Differentiate

Calculus Revision Differentiate Straight line form Differentiate

Calculus Revision Differentiate Chain Rule Simplify Straight line form

Calculus Revision Differentiate Straight line form Differentiate

Calculus Revision Differentiate Straight line form

Calculus Revision Differentiate Straight line form

Name : Gradient = Rate of change = Differentiation Differentiation techniques

Calculus Revision Differentiate Multiply out Differentiate

Calculus Revision Differentiate multiply out differentiate

Calculus Revision Differentiate Straight line form multiply out Differentiate

Calculus Revision Differentiate multiply out Differentiate

Calculus Revision Differentiate multiply out Simplify Differentiate Straight line form

Calculus Revision Differentiate Straight line form Multiply out Differentiate

Calculus Revision Differentiate Split up Straight line form Differentiate

If y is expressed in terms of x then the derivative is written as dy / dx. Leibniz Notation Leibniz Notation is an alternative way of expressing derivatives to f'(x), g'(x), etc. eg y = 3x 2 - 7x so dy / dx = 6x - 7. Example 19 Find dQ / dR NB: Q = 9R R -3 So dQ / dR = 18R + 45R -4 = 18R + 45 R 4 Q = 9R R 3 Higher

Example 20 A curve has equation y = 5x 3 - 4x Find the gradient where x = -2 ( differentiate ! ) gradient = dy / dx = 15x 2 - 8x if x = -2 then gradient = 15 X (-2) X (-2) = 60 - (-16) = 76 Leibniz Notation Higher

HHM Ex6H Q1 – Q3 HHM Ex6G Q1,4,7,10,13,16,19,22,25 Derivative Higher

Newton’s 2 nd Law of Motion s = ut + 1 / 2 at 2 where s = distance & t = time. Finding ds / dt means “diff in dist”  “diff in time” ie speed or velocity so ds / dt = u + at but ds / dt = v so we getv = u + at and this is Newton’s 1st Law of Motion Real Life Example Physics Higher

HHM Ex6H Q4 – Q6 Derivative Higher

A(a,b) y = mx +c y = f(x) Equation of Tangents tangent NB: at A(a, b) gradient of line = gradient of curve gradient of line = m (from y = mx + c ) gradient of curve at (a, b) = f (a) it follows that m = f (a) Higher Demo

Higher Equation of Tangents Example 21 Find the equation of the tangent line to the curve y = x 3 - 2x + 1 at the point where x = -1. Point: if x = -1 then y = (-1) 3 - (2 X -1) + 1 = -1 - (-2) + 1 = 2point is (-1,2) Gradient: dy / dx = 3x when x = -1 dy / dx = 3 X (-1) = = 1 m = 1 Straight line so we need a point plus the gradient then we can use the formula y - b = m(x - a). Demo

Now using y - b = m(x - a) we gety - 2 = 1( x + 1) or y - 2 = x + 1 or y = x + 3 point is (-1,2) m = 1 Equation of Tangents Higher

Example 22 Find the equation of the tangent to the curve y = 4 x 2 at the point where x = -2. (x  0) Also find where the tangent cuts the X-axis and Y-axis. Point:when x = -2 then y = 4 (-2) 2 = 4 / 4 = 1 point is (-2, 1) Gradient:y = 4x -2 so dy / dx = -8x -3 = -8 x 3 when x = -2 then dy / dx = -8 (-2) 3 = -8 / -8 = 1 m = 1 Equation of Tangents Higher

Now using y - b = m(x - a) we get y - 1 = 1( x + 2) or y - 1 = x + 2 or y = x + 3 AxesTangent cuts Y-axis when x = 0 so y = = 3at point (0, 3) Tangent cuts X-axis when y = 0 so 0 = x + 3 or x = -3at point (-3, 0) Equation of Tangents Higher

Example 23 - (other way round) Find the point on the curve y = x 2 - 6x + 5 where the gradient of the tangent is 14. gradient of tangent = gradient of curve dy / dx = 2x - 6 so2x - 6 = 14 2x = 20x = 10 Put x = 10 into y = x 2 - 6x + 5 Givingy = = 45 Point is (10,45) Equation of Tangents Higher

HHM Ex6J Derivative Higher

Increasing & Decreasing Functions and Stationary Points Consider the following graph of y = f(x) ….. X y = f(x) abcdef Higher

In the graph of y = f(x) The function is increasing if the gradient is positive i.e. f (x) > 0 when x < b or d < x < f or x > f. The function is decreasing if the gradient is negative and f (x) < 0 when b < x < d. The function is stationary if the gradient is zero and f (x) = 0 when x = b or x = d or x = f. These are called STATIONARY POINTS. At x = a, x = c and x = e the curve is simply crossing the X-axis. Increasing & Decreasing Functions and Stationary Points Higher

Example 24 For the function f(x) = 4x x + 19 determine the intervals when the function is decreasing and increasing. f (x) = 8x - 24 f(x) decreasing when f (x) < 0 so 8x - 24 < 0 8x < 24 x < 3 f(x) increasing when f (x) > 0so 8x - 24 > 0 8x > 24 x > 3 Check: f (2) = 8 X 2 – 24 = -8 Check: f (4) = 8 X 4 – 24 = 8 Increasing & Decreasing Functions and Stationary Points Higher

Example 25 For the curve y = 6x – 5/x 2 Determine if it is increasing or decreasing when x = 10. = 6x - 5x -2 so dy / dx = x -3 when x = 10 dy / dx = / 1000 = 6.01 Since dy / dx > 0 then the function is increasing. Increasing & Decreasing Functions and Stationary Points y = 6x - 5 x 2 = x 3 Higher

Example 26 Show that the function g(x) = 1 / 3 x 3 -3x 2 + 9x -10 is never decreasing. g (x) = x 2 - 6x + 9 = (x - 3)(x - 3)= (x - 3) 2 Since (x - 3) 2  0 for all values of x then g (x) can never be negative so the function is never decreasing. Squaring a negative or a positive value produces a positive value, while 0 2 = 0. So you will never obtain a negative by squaring any real number. Increasing & Decreasing Functions and Stationary Points Higher

Example 27 Determine the intervals when the function f(x) = 2x 3 + 3x x + 41 is (a) Stationary (b) Increasing (c) Decreasing. f (x) = 6x 2 + 6x - 36 = 6(x 2 + x - 6) = 6(x + 3)(x - 2) Function is stationary when f (x) = 0 ie 6(x + 3)(x - 2) = 0 ie x = -3 or x = 2 Increasing & Decreasing Functions and Stationary Points Higher

We now use a special table of factors to determine when f (x) is positive & negative. x-32 f’(x) + Function increasing when f (x) > 0ie x < -3 or x > 2 Function decreasing when f (x) < 0ie -3 < x < 2 Increasing & Decreasing Functions and Stationary Points Higher

HHM Ex6K and Ex6L Derivative Higher

Stationary Points and Their Nature Consider this graph of y = f(x) again X y = f(x) ab c Higher

This curve y = f(x) has three types of stationary point. When x = a we have a maximum turning point (max TP) When x = b we have a minimum turning point (min TP) When x = c we have a point of inflexion (PI) Each type of stationary point is determined by the gradient ( f(x) ) at either side of the stationary value. Stationary Points and Their Nature Higher

Maximum Turning point xa f(x) Minimum Turning Point xb f(x) Stationary Points and Their Nature Higher

Rising Point of inflexion xc f(x) Other possible type of inflexion xd f(x) Stationary Points and Their Nature Higher

Example 28 Find the co-ordinates of the stationary point on the curve y = 4x and determine its nature. SP occurs when dy / dx = 0 so 12x 2 = 0 x 2 = 0 x = 0 Using y = 4x if x = 0 then y = 1 SP is at (0,1) Stationary Points and Their Nature Higher

Nature Table x 0 dy / dx + So (0,1) is a rising point of inflexion. Stationary Points and Their Nature dy / dx = 12x 2 Higher + 0

Example 29 Find the co-ordinates of the stationary points on the curve y = 3x x and determine their nature. SP occurs when dy / dx = 0 So 12x x 2 = 0 12x 2 (x - 4) = 0 12x 2 = 0 or (x - 4) = 0 x = 0 or x = 4 Using y = 3x x if x = 0 then y = 24 if x = 4 then y = -232 SPs at (0,24) & (4,-232) Stationary Points and Their Nature Higher

Nature Table x0 4 dy / dx So (0,24) is a Point of inflexion and (4,-232) is a minimum Turning Point Stationary Points and Their Nature dy / dx =12x x 2 Higher

Example 30 Find the co-ordinates of the stationary points on the curve y = 1 / 2 x 4 - 4x and determine their nature. SP occurs when dy / dx = 0 So 2x 3 - 8x = 0 2x(x 2 - 4) = 0 2x(x + 2)(x - 2) = 0 x = 0 or x = -2 or x = 2 Using y = 1 / 2 x 4 - 4x if x = 0 then y = 2 if x = -2 then y = -6 SP’s at(-2,-6), (0,2) & (2,-6) if x = 2 then y = -6 Stationary Points and Their Nature Higher

Nature Table x0 dy / dx So (-2,-6) and (2,-6) are Minimum Turning Points and (0,2) is a Maximum Turning Points Stationary Points and Their Nature Higher

Curve Sketching Note: A sketch is a rough drawing which includes important details. It is not an accurate scale drawing. Process (a)Find where the curve cuts the co-ordinate axes. for Y-axis put x = 0 for X-axis put y = 0 then solve. (b)Find the stationary points & determine their nature as done in previous section. (c)Check what happens as x  + / - . This comes automatically if (a) & (b) are correct. Higher

Dominant Terms Suppose that f(x) = -2x 3 + 6x x - 99 As x  + / -  (ie for large positive/negative values) The formula is approximately the same as f(x) = -2x 3 As x  +  then y  -  As x  -  then y  +  Graph roughly Curve Sketching Higher

Example 31 Sketch the graph of y = -3x x + 15 (a) AxesIf x = 0 then y = 15 If y = 0 then -3x x + 15 = 0(  -3) x 2 - 4x - 5 = 0 (x + 1)(x - 5) = 0 x = -1 or x = 5 Graph cuts axes at (0,15), (-1,0) and (5,0) Curve Sketching Higher

(b) Stationary Points occur where dy / dx = 0 so -6x + 12 = 0 6x = 12 x = 2 If x = 2 then y = = 27 Nature Table x2 dy / dx So (2,27) is a Maximum Turning Point Stationary Point is (2,27) Curve Sketching Higher

(c) Large values using y = -3x 2 as x  +  then y  -  as x  -  then y  -  Sketching X Y y = -3x x + 15 Curve Sketching Cuts x-axis at -1 and 5 Summarising Cuts y-axis at 15 5 Max TP (2,27)(2,27) 15 Higher

Example 32 Sketch the graph of y = -2x 2 (x - 4) (a) Axes If x = 0 then y = 0 X (-4) = 0 If y = 0 then -2x 2 (x - 4) = 0 x = 0 or x = 4 Graph cuts axes at (0,0) and (4,0). -2x 2 = 0 or (x - 4) = 0 (b) SPs y = -2x 2 (x - 4) = -2x 3 + 8x 2 SPs occur where dy / dx = 0 so -6x x = 0 Curve Sketching Higher

-2x(3x - 8) = 0 -2x = 0 or (3x - 8) = 0 x = 0 or x = 8 / 3 If x = 0 then y = 0 (see part (a) ) If x = 8 / 3 then y = -2 X ( 8 / 3 ) 2 X ( 8 / 3 -4) = 512 / 27 nature x0 8/38/3 dy / dx - Curve Sketching Higher

Summarising (c) Large values using y = -2x 3 as x  +  then y  -  as x  -  then y  +  Sketch X y = -2x 2 (x – 4) Curve Sketching Cuts x – axis at 0 and Max TP’s at ( 8 / 3, 512 / 27 ) ( 8 / 3, 512 / 27 ) Y Higher

HHM Ex6M Derivative Higher

Max & Min on Closed Intervals In the previous section on curve sketching we dealt with the entire graph. In this section we shall concentrate on the important details to be found in a small section of graph. Suppose we consider any graph between the points where x = a and x = b (i.e. a  x  b) then the following graphs illustrate where we would expect to find the maximum & minimum values. Higher

y =f(x) X a b (a, f(a)) (b, f(b)) max = f(b) end point min = f(a) end point Max & Min on Closed Intervals Higher

x y =f(x) (b, f(b)) (a, f(a)) max = f(c ) max TP min = f(a) end point a b (c, f(c)) c NB: a < c < b Max & Min on Closed Intervals Higher

y =f(x) x a b c (a, f(a)) (b, f(b)) (c, f(c)) max = f(b) end point min = f(c) min TP NB: a < c < b Max & Min on Closed Intervals Higher

From the previous three diagrams we should be able to see that the maximum and minimum values of f(x) on the closed interval a  x  b can be found either at the end points or at a stationary point between the two end points Example 34 Find the max & min values of y = 2x 3 - 9x 2 in the interval where -1  x  2. End pointsIf x = -1 then y = = -11 If x = 2 then y = = -20 Max & Min on Closed Intervals Higher

Stationary points dy / dx = 6x x= 6x(x - 3) SPs occur where dy / dx = 0 6x(x - 3) = 0 6x = 0 or x - 3 = 0 x = 0 or x = 3 in interval not in interval If x = 0 then y = = 0 Hence for -1  x  2, max = 0 & min = -20 Max & Min on Closed Intervals Higher

Extra bit Using function notation we can say that Domain = {x  R: -1  x  2 } Range = {y  R: -20  y  0 } Max & Min on Closed Intervals Higher

Derivative Graphs Higher Demo

Optimization Note: Optimum basically means the best possible. In commerce or industry production costs and profits can often be given by a mathematical formula. Optimum profit is as high as possible so we would look for a max value or max TP. Optimum production cost is as low as possible so we would look for a min value or min TP. Higher

Optimization Higher Problem Practical exercise on optimizing volume. Graph

Example 35 Higher Optimization Q. What is the maximum volume We can have for the given dimensions A rectangular sheet of foil measuring 16cm X 10 cm has four small squares each x cm cut from each corner. 16cm 10cm x cm NB: x > 0 but 2x < 10 or x < 5 ie 0 < x < 5 This gives us a particular interval to consider ! x cm

(16 - 2x) cm (10 - 2x) cm x cm The volume is now determined by the value of x so we can write V(x) = x(16 - 2x)(10 - 2x) = x( x + 4x 2 ) = 4x x x We now try to maximize V(x) between 0 and 5 Optimization By folding up the four flaps we get a small cuboid Higher

Considering the interval 0 < x < 5End Points V(0) = 0 X 16 X 10 = 0 V(5) = 5 X 6 X 0 = 0 SPsV '(x) = 12x x = 4(3x x + 40) = 4(3x - 20)(x - 2) Optimization Higher

SPs occur when V '(x) = 0 ie 4(3x - 20)(x - 2) = 0 3x - 20 = 0 or x - 2 = 0 ie x = 20 / 3 or x = 2 not in interval in interval When x = 2 thenV(2) = 2 X 12 X 6 = 144 We now check gradient near x = 2 Optimization Higher

x2 V '(x) + Hence max TP when x = 2 So max possible volume = 144cm 3 Nature Optimization Higher - 0

Example 36 When a company launches a new product its share of the market after x months is calculated by the formula So after 5 months the share is S(5) = 2 / 5 – 4 / 25 = 6 / 25 Find the maximum share of the market that the company can achieve. (x  2) Optimization Higher

End pointsS(2) = 1 – 1 = 0 There is no upper limit but as x   S(x)  0. SPs occur where S (x) = 0 Optimization Higher

8x 2 = 2x 3 8x 2 - 2x 3 = 0 2x 2 (4 – x) = 0 x = 0 or x = 4 Out with interval In interval We now check the gradients either side of 4 rearrange Optimization Higher

x  4  S (x) S (3.9 ) = … S (4.1) = … Hence max TP at x = 4 And max share of market = S(4) = 2 / 4 – 4 / 16 = 1 / 2 – 1 / 4 = 1 / 4 Optimization Nature Higher + - 0

HHM Ex6Q and Ex6R Derivative Higher

Differentiation of Polynomials f(x) = ax n then f’x) = anx n-1 Derivative = gradient = rate of change Graphs f’(x)=0 f’(x)=0 Stationary Pts Max. / Mini Pts Inflection Pt Nature Table x f’(x) Max Gradient at a point Equation of tangent line Straight Line Theory Leibniz Notation

Are you on Target ! Update you log book Make sure you complete and correct ALL of the Differentiation 1Differentiation 1 questions in the past paper booklet.