Differentiation s5 Unit 1 Outcome 3CAPE

Surds & Indicesxm . xn = xm+n

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Special Points(I) f(x) = ax (Straight line function)If f(x) = ax = ax1then f '(x) = 1 X ax0 = a X 1 = aIndex Laws x0 = 1So if g(x) = 12x then g '(x) = 12 Also using y = mx + cThe line y = 12x has gradient 12,and derivative = gradient !!HigherOutcome 3



(II) f(x) = a, (Horizontal Line)If f(x) = a = a X 1 = ax0then f '(x) = 0 X ax-1= 0Index Laws x0 = 1So 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 PointsHigherOutcome 3


Name : Gradient=Rate of change=DifferentiationDifferentiationDifferentiation techniques

Calculus RevisionDifferentiate



Example 1A curve has equation f(x) = 3x4Its gradient is f '(x) = 12x3 f '(2) = 12 X 23 = 12 X 8 = 96Example 2A curve has equation f(x) = 3x2Find the formula for its gradient and find the gradient when x = 2Its gradient is f '(x) = 6xAt the point where x = -4 the gradient isf '(-4) = 6 X -4 =-24DerivativeHigherOutcome 3Find the formula for its gradient and find the gradient when x = -4



Example 3If g(x) = 5x4 - 4x5 then find g '(2) .g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24 = 160 - 320 = -160DerivativeHigherOutcome 3



Example 4h(x) = 5x2 - 3x + 19so h '(x) = 10x - 3and h '(-4) = 10 X (-4) - 3 = -40 - 3 = -43Example 5k(x) = 5x4 - 2x3 + 19x - 8, find k '(10) .k '(x) = 20x3 - 6x2 + 19So k '(10) = 20 X 1000 - 6 X 100 + 19 = 19419DerivativeHigherOutcome 3



Example 6 : Find the points on the curve f(x) = x3 - 3x2 + 2x + 7 where the gradient is 2. NB: gradient = derivative = f '(x) We need f '(x) = 2ie 3x2 - 6x + 2 = 2or 3x2 - 6x = 0 ie 3x(x - 2) = 0ie 3x = 0 or x - 2 = 0so x = 0 or x = 2Now using original formulaf(0) = 7f(2) = 8 -12 + 4 + 7 = 7Points are (0,7) & (2,7)DerivativeHigherOutcome 3


Calculus RevisionDifferentiateStraight line formDifferentiate

Calculus RevisionDifferentiateChain RuleSimplifyStraight line form

Calculus RevisionDifferentiateStraight line formDifferentiate

Calculus RevisionDifferentiateDifferentiateStraight line form

Name : Gradient=Rate of change=DifferentiationDifferentiationDifferentiation techniques

Calculus RevisionDifferentiateMultiply outDifferentiate

Calculus RevisionDifferentiatemultiply outdifferentiate

Calculus RevisionDifferentiateStraight line formmultiply outDifferentiate

Calculus RevisionDifferentiatemultiply outDifferentiate

Calculus RevisionDifferentiatemultiply outSimplifyDifferentiateStraight line form

Calculus RevisionDifferentiateStraight line formMultiply outDifferentiate

Calculus RevisionDifferentiateSplit upStraight line formDifferentiate


If y is expressed in terms of x then the derivative is written as dy/dx .Leibniz NotationLeibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc. eg y = 3x2 - 7x so dy/dx = 6x - 7 .Example 19Find dQ/dR NB: Q = 9R2 - 15R-3So dQ/dR = 18R + 45R-4HigherOutcome 3



Example 20A curve has equation y = 5x3 - 4x2 + 7 . Find the gradient where x = -2 ( differentiate ! )gradient = dy/dx = 15x2 - 8xif x = -2 thengradient = 15 X (-2)2 - 8 X (-2)= 60 - (-16) = 76Leibniz NotationHigherOutcome 3



Newtons 2ndLaw of Motions = ut + 1/2at2 where s = distance & t = time.Finding ds/dt means diff in dist diff in timeie speed or velocityso ds/dt = u + at but ds/dt = v so we getv = u + atand this is Newtons 1st Law of MotionReal Life ExamplePhysicsHigherOutcome 3



A(a,b)y = mx +cy = f(x)Equation of TangentstangentNB: 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) HigherOutcome 3Demo



HigherOutcome 3Equation of TangentsExample 21Find the equation of the tangent line to the curvey = x3 - 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 = 3x2 - 2when x = -1 dy/dx = 3 X (-1)2 - 2 = 3 - 2 = 1 m = 1Straight 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 get y - 2 = 1( x + 1)or y - 2 = x + 1or y = x + 3point is (-1,2)m = 1Equation of TangentsHigherOutcome 3



Example 22Find the equation of the tangent to the curve y = 4 x2 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 x3when x = -2 then dy/dx = -8 (-2)3= -8/-8 = 1m = 1Equation of TangentsHigherOutcome 3



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



Example 23 - (other way round)Find the point on the curve y = x2 - 6x + 5 where the gradient of the tangent is 14.gradient of tangent = gradient of curve dy/dx = 2x - 6so2x - 6 = 14 2x = 20x = 10Put x = 10 into y = x2 - 6x + 5 Giving y = 100 - 60 + 5 = 45Point is (10,45)Equation of TangentsHigherOutcome 3



Increasing & Decreasing Functions and Stationary PointsConsider the following graph of y = f(x) ..Xy = f(x)abcdef+++++--000HigherOutcome 3



In the graph of y = f(x)The function is increasing if the gradient is positivei.e. f (x) > 0 when x < b or d < x < f or x > f . The function is decreasing if the gradient is negativeand f (x) < 0 when b < x < d .The function is stationary if the gradient is zeroand 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 PointsHigherOutcome 3



Example 24For the function f(x) = 4x2 - 24x + 19 determine the intervals when the function is decreasing and increasing.f (x) = 8x - 24f(x) decreasing when f (x) < 0so 8x - 24 < 0 8x < 24x < 3f(x) increasing when f (x) > 0

so 8x - 24 > 08x > 24x > 3Check: f (2) = 8 X 2 24 = -8 Check: f (4) = 8 X 4 24 = 8Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3



Example 25For the curve y = 6x 5/x2 Determine if it is increasing or decreasing when x = 10. = 6x - 5x-2so dy/dx = 6 + 10x-3when x = 10 dy/dx = 6 + 10/1000 = 6.01Since dy/dx > 0 then the function is increasing. Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3



Example 26Show that the function g(x) = 1/3x3 -3x2 + 9x -10 is never decreasing. g (x) = x2 - 6x + 9 = (x - 3)(x - 3)= (x - 3)2Since (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 02 = 0. So you will never obtain a negative by squaring any real number.Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3



Example 27Determine the intervals when the functionf(x) = 2x3 + 3x2 - 36x + 41is (a) Stationary (b) Increasing (c) Decreasing.f (x) = 6x2 + 6x - 36 = 6(x2 + x - 6)= 6(x + 3)(x - 2)Function is stationary when f (x) = 0ie 6(x + 3)(x - 2) = 0 ie x = -3 or x = 2Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3



We now use a special table of factors to determine when f (x) is positive & negative. x-32f(x)

+Function increasing when f (x) > 0ie x < -3 or x > 2Function decreasing when f (x) < 0ie -3 < x < 2Increasing & Decreasing Functions and Stationary PointsHigherOutcome 3-0 +0



Stationary Points and Their NatureConsider this graph of y = f(x) againXy = f(x)abc+++++--000HigherOutcome 3



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 NatureHigherOutcome 3



Maximum Turning pointxaf(x) + 0 -Minimum Turning Pointxbf(x) - 0 +Stationary Points and Their NatureHigherOutcome 3



Rising Point of inflexion xcf(x) + 0 +Other possible type of inflexion xdf(x) - 0 -Stationary Points and Their NatureHigherOutcome 3



Example 28Find the co-ordinates of the stationary point on the curve y = 4x3 + 1 and determine its nature.SP occurs when dy/dx = 0so 12x2 = 0 x2 = 0 x = 0Using y = 4x3 + 1 if x = 0 then y = 1SP is at (0,1)Stationary Points and Their NatureHigherOutcome 3



Nature Tablex 0dy/dx +So (0,1) is a rising point of inflexion.Stationary Points and Their Naturedy/dx = 12x2HigherOutcome 3 + 0



Example 29Find the co-ordinates of the stationary points on the curve y = 3x4 - 16x3 + 24 and determine their nature.SP occurs when dy/dx = 0So 12x3 - 48x2 = 0 12x2(x - 4) = 012x2 = 0 or (x - 4) = 0x = 0 or x = 4Using y = 3x4 - 16x3 + 24 if x = 0 then y = 24if x = 4 then y = -232SPs at (0,24) & (4,-232)Stationary Points and Their NatureHigherOutcome 3



Nature Tablex04dy/dx - 0 - 0 +So (0,24) is a Point of inflexion and (4,-232) is a minimum Turning PointStationary Points and Their Naturedy/dx=12x3 - 48x2HigherOutcome 3



Example 30Find the co-ordinates of the stationary points on the curve y = 1/2x4 - 4x2 + 2 and determine their nature.SP occurs when dy/dx = 0So 2x3 - 8x = 0 2x(x2 - 4) = 02x(x + 2)(x - 2) = 0x = 0 or x = -2 or x = 2Using y = 1/2x4 - 4x2 + 2if x = 0 then y = 2if x = -2 then y = -6SPs at(-2,-6), (0,2) & (2,-6)if x = 2 then y = -6Stationary Points and Their NatureHigherOutcome 3



Nature Tablex0dy/dx-22 - 0 + 0 - 0 +So (-2,-6) and (2,-6) are Minimum Turning Pointsand (0,2) is a Maximum Turning PointsStationary Points and Their NatureHigherOutcome 3



Curve SketchingNote: 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 = 0for 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.HigherOutcome 3



Dominant TermsSuppose that f(x) = -2x3 + 6x2 + 56x - 99 As x +/- (ie for large positive/negative values) The formula is approximately the same as f(x) = -2x3 As x + then y - As x - then y +Graph roughlyCurve SketchingHigherOutcome 3



Example 31Sketch the graph of y = -3x2 + 12x + 15(a) AxesIf x = 0 then y = 15If y = 0 then -3x2 + 12x + 15 = 0( -3)x2 - 4x - 5 = 0(x + 1)(x - 5) = 0x = -1 or x = 5Graph cuts axes at (0,15) , (-1,0) and (5,0)Curve SketchingHigherOutcome 3



(b) Stationary Points occur where dy/dx = 0so -6x + 12 = 06x = 12x = 2If x = 2 then y = -12 + 24 + 15 = 27Nature Tablex2dy/dx + 0 -So (2,27) is a Maximum Turning PointStationary Point is (2,27)Curve SketchingHigherOutcome 3



(c) Large valuesusing y = -3x2as x + then y - as x - then y -SketchingXYy = -3x2 + 12x + 15Curve SketchingCuts x-axis at -1 and 5SummarisingCuts y-axis at 15-15Max TP (2,27)(2,27)15HigherOutcome 3



Example 32Sketch the graph of y = -2x2 (x - 4)(a) AxesIf x = 0 then y = 0 X (-4) = 0If y = 0 then -2x2 (x - 4) = 0x = 0 or x = 4Graph cuts axes at (0,0) and (4,0) .-2x2 = 0 or (x - 4) = 0(b) SPsy = -2x2 (x - 4)= -2x3 + 8x2SPs occur where dy/dx = 0

so -6x2 + 16x = 0Curve SketchingHigherOutcome 3



-2x(3x - 8) = 0-2x = 0 or (3x - 8) = 0x = 0 or x = 8/3If x = 0 then y = 0 (see part (a) ) If x = 8/3 then y = -2 X (8/3)2 X (8/3 -4) =512/27naturex08/3dy/dx - Curve SketchingHigherOutcome 3 - + 0 0



Summarising(c) Large valuesusing y = -2x3as x + then y -as x - then y +SketchXy = -2x2 (x 4)Curve SketchingCuts x axis at 0 and 404Max TPs at (8/3, 512/27) (8/3, 512/27)YHigherOutcome 3



Example 33Sketch the graph of y = 8 + 2x2 - x4(a) AxesIf x = 0 then y = 8 (0,8)If y = 0 then 8 + 2x2 - x4 = 0

Graph cuts axes at (0,8) , (-2,0) and (2,0)Let u = x2 so u2 = x4

Equation is now 8 + 2u - u2 = 0 (4 - u)(2 + u) = 0(4 - x2)(2 + x2) = 0or (2 + x) (2 - x)(2 + x2) = 0So x = -2 or x = 2 but x2 -2Curve SketchingHigherOutcome 3



(b) SPsSPs occur where dy/dx = 0So 4x - 4x3 = 0 4x(1 - x2) = 0 4x(1 - x)(1 + x) = 0 x = 0 or x =1 or x = -1 Using y = 8 + 2x2 - x4when x = 0 then y = 8when x = -1 then y = 8 + 2 - 1 = 9 (-1,9) when x = 1 then y = 8 + 2 - 1 = 9 (1,9)Curve SketchingHigherOutcome 3



naturex-101dy/dx+So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .Curve SketchingHigherOutcome 3+--000



Cuts y axis at 8Cuts x axis at -2 and 2(c) Large valuesUsing y = - x4as x + then y -as x - then y -Sketch isXY-228(-1,9)(1,9)y = 8 + 2x2 - x4Max TPs at (-1,9)(1,9)Curve SketchingSummarisingHigherOutcome 3



Max & Min on Closed IntervalsIn 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.HigherOutcome 3



y =f(x)Xa b(a, f(a))(b, f(b))max = f(b) end pointmin = f(a) end pointMax & Min on Closed IntervalsHigherOutcome 3



xy =f(x)(b, f(b))(a, f(a))max = f(c ) max TP min = f(a) end pointa b(c, f(c))cNB: a < c < bMax & Min on Closed IntervalsHigherOutcome 3



y =f(x)xa b c(a, f(a))(b, f(b))

(c, f(c))max = f(b) end pointmin = f(c) min TPNB: a < c < bMax & Min on Closed IntervalsHigherOutcome 3



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 pointsExample 34Find the max & min values of y = 2x3 - 9x2 in the interval where -1 x 2. End pointsIf x = -1 then y = -2 - 9 = -11If x = 2 then y = 16 - 36 = -20 Max & Min on Closed IntervalsHigherOutcome 3



Stationary pointsdy/dx = 6x2 - 18x = 6x(x - 3)SPs occur where dy/dx = 0 6x(x - 3) = 06x = 0 or x - 3 = 0x = 0 or x = 3in interval not in intervalIf x = 0 then y = 0 - 0 = 0Hence for -1 x 2 , max = 0 & min = -20Max & Min on Closed IntervalsHigherOutcome 3



Extra bitUsing function notation we can say thatDomain = {xR: -1 x 2 } Range = {yR: -20 y 0 }Max & Min on Closed IntervalsHigherOutcome 3



Derivative GraphsHigherOutcome 3Demo



OptimizationNote: 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.HigherOutcome 3



OptimizationHigherOutcome 3ProblemPractical exercise on optimizing volume. Graph



Example 35HigherOutcome 3OptimizationQ. What is the maximum volumeWe can have for the given dimensionsA rectangular sheet of foil measuring 16cm X 10 cm has four small squares each x cm cut from each corner. 16cm10cmx cmNB: x > 0 but 2x < 10 or x < 5ie 0 < x < 5This gives us a particular interval to consider ! x cm



(16 - 2x) cm(10 - 2x) cmx cmThe volume is now determined by the value of x so we can writeV(x) = x(16 - 2x)(10 - 2x)= x(160 - 52x + 4x2)= 4x3 - 52x2 +160xWe now try to maximize V(x) between 0 and 5OptimizationBy folding up the four flaps we get a small cuboid HigherOutcome 3



Considering the interval 0 < x < 5End PointsV(0) = 0 X 16 X 10 = 0V(5) = 5 X 6 X 0 = 0SPsV '(x) = 12x2 - 104x + 160 = 4(3x2 - 26x + 40)= 4(3x - 20)(x - 2)OptimizationHigherOutcome 3



SPs occur when V '(x) = 0 ie 4(3x - 20)(x - 2) = 03x - 20 = 0 or x - 2 = 0ie x = 20/3 or x = 2not in intervalin intervalWhen x = 2 thenV(2) = 2 X 12 X 6 = 144We now check gradient near x = 2OptimizationHigherOutcome 3



x2V '(x)+ Hence max TP when x = 2So max possible volume = 144cm3NatureOptimizationHigherOutcome 3- 0



Example 36When a company launches a new product its share of the market after x months is calculated by the formulaSo after 5 months the share isS(5) = 2/5 4/25 = 6/25Find the maximum share of the market that the company can achieve.(x 2)OptimizationHigherOutcome 3



End pointsS(2) = 1 1 = 0There is no upper limit but as x S(x) 0. SPs occur where S (x) = 0OptimizationHigherOutcome 3



8x2 = 2x38x2 - 2x3 = 02x2(4 x) = 0x = 0 or x = 4Out with intervalIn intervalWe now check the gradients either side of 4rearrangeOptimizationHigherOutcome 3



x 4 S (x)S (3.9 ) = 0.00337S (4.1) = -0.0029Hence max TP at x = 4And max share of market = S(4)= 2/4 4/16= 1/2 1/4= 1/4OptimizationNatureHigherOutcome 3+-0


Differentiationof Polynomialsf(x) = axnthen fx) = anxn-1Derivative = gradient = rate of change Graphsf(x)=0f(x)=0Stationary PtsMax. / Mini PtsInflection PtNature TableGradient at a pointEquation of tangent lineStraight LineTheoryLeibniz Notation



Are you on Target ! Update you log book Make sure you complete and correct ALL of the Differentiation 1questions in the past paper booklet.Outcome 3


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Differentiation s5 Unit 1 Outcome 3CAPE