Higher Unit 1 Finding the gradient for a polynomial Differentiating a polynomial Differentiating Negative Indices Differentiating Roots Differentiating Brackets Differentiating Fraction Terms Differentiating with Leibniz Notation Exam Type Questions Equation of a Tangent Line Increasing / Decreasing functions Max / Min and Infection Points Curve Sketching Max & Min Values on closed Intervals Optimization Higher Outcome 3 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 Outcome 3

A Gradient of tangent = gradient of curve at A B Gradient of tangent = gradient of curve at B Gradients & Curves Higher Outcome 3

Gradients & Curves Higher Outcome 3 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 Outcome 3

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 Outcome 3

Higher Outcome 3 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. We can use the formula for the curve to produce a formula for the gradient. This process is called DIFFERENTIATING or FINDING THE DERIVATIVE 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 Outcome 3

Rules for Differentiating These can be given by the simple flow diagram... multiply by the power reduce the power by 1 Or If f(x) = ax n then f '(x) = nax n-1 NB: the following terms & expressions mean the same GRADIENT,DERIVATIVE,RATE OF CHANGE,f '(x) Derivative Higher Outcome 3

Example 1 A curve has equation f(x) = 3x 4 Its gradient is f '(x) = 12x 3 At the point where x = 2 the gradient is 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 = -4. Its gradient is f '(x) = 6x At the point where x = -4 the gradient is f '(-4) = 6 X -4 =-24 Derivative Higher Outcome 3

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 Outcome 3

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 Outcome 3

(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 Outcome 3

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 Outcome 3

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 Outcome 3

Negative Indices Index Law 1 = x -m xmxm NB: Before we can differentiate a term it must be in the form ax n Bottom line terms get negative powers !! Higher Outcome 3 YOU need to be good with FRACTIONS and INDICES

Example 7 f(x) = 1 = x 2 x -2 So f '(x) = -2x -3 = Example 8 g(x) = -3 x 4 = -3x -4 So g '(x) = 12x -5 = 12 x 5 Negative Indices Higher Outcome 3 -2 x 3

h(y) = 4 3y 3 Example 9 = 4 / 3 y -3 So h '(y) = -12 / 3 y -4 = -4 y 4 Example 10 k(t) = 3 4t 2/3 = 3 / 4 t -2/3 So k '(t) = - 6 / 12 t -5/3 = -1 2t 5/3 Negative Indices Higher Outcome 3

Example 11 The equation of a curve is f(x) = 8 (x  0) x 4 Find the gradient at the point where x = -2. f(x) = 8 x 4 = 8x -4 so f '(x) = -32x -5 = -32 x 5 Required gradient =f '(-2) = -32 (-2) 5 = = 1 Negative Indices Higher Outcome 3

Differentiating Roots Fractional Indices n  x = x 1/n Higher Outcome 3 From the credit or Intermediate 2 course n  x m or ( n  x) m = x m/n

Check the following using the power button on your calc. 64 1/2 =  64 = /3 = 3  64 = /2 = 1 = 1 or 0.2  /4 = ( 4  16) 3 = 2 3 = /3 = ( 3  125) 5 = 5 5 = 3125 Fractional Powers top line - power bottom line - root Higher Outcome 3 n  x m or ( n  x) m = x m/n

so h'(y) = -7 / 3 y -10/3 =  y 10 Example 12 f(x) =  x so f '(x) = 1 / 2 x -1/2 = 1 2  x Example 13 so g'(t) = 5 / 2 t 3/2 = 5  t 3 2 Example 14 h(y) = 1 3  y 7 = y -7/3 g(t) =  t 5 Differentiating Roots Higher Outcome 3 = x 1/2 = t 5/2

Example 15 Find the rate of change at the point where x = 4 on the curve with equation g(x) = 4.  x = 4x -1/2 NB: rate of change = gradient = g'(x). g'(x) = -2x -3/2 so g'(4) = -2 / 8 = -1 / 4 Differentiating Roots Higher Outcome 3 g(x) = 4  x = -2 (  x) 3 = -2 (  4) 3

Brackets Basic Rule: Break brackets before you differentiate ! Example 16h(x) = 2x(x + 3)(x -3) = 2x(x 2 - 9) = 2x x So h'(x) = 6x Higher Outcome 3

Fractions Reversing the above we get the following “rule” ! This can be used as follows ….. Higher Outcome 3

Example 17 f(x) = 3x 3 - x + 2 x 2 = 3x - x x -2 f '(x) = 3 + x x -3 Fractions = 3x 3 - x + 2 x 2 x 2 x 2 = x 2 x 3 Higher Outcome 3

Example 18 (tricky) g(y) = (y +  y)(y + 1) y  y = (y + y 1/2 )(y + 1) y X y 1/2 = y 2 + y 3/2 + y + y 1/2 y 3/2 = y 2 + y 3/2 + y + y 1/2 = y 1/ y -1/2 + y -1 y 3/2 Change to powers √=½ Brackets Single fractions Correct form Indices Fractions Higher Outcome 3

g'(y) = 1 / 2 y -1/2 - 1 / 2 y -3/2 - y -2 = 1 / / / 16 = 1 / 8 = y 1/2 2y 3/2 y 2 Also g'(4) = X 4 1/2 2 X 4 3/2 4 2 = X  4 2 X (  4) 3 16 Fractions Higher Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

Higher Outcome 3 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).

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

Increasing & Decreasing Functions and Stationary Points Consider the following graph of y = f(x) ….. X y = f(x) abcdef Higher Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome

Stationary Points and Their Nature Consider this graph of y = f(x) again X y = f(x) ab c Higher Outcome 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 Nature Higher Outcome 3

Maximum Turning point xa f(x) Minimum Turning Point xb f(x) Stationary Points and Their Nature Higher Outcome 3

Rising Point of Inflection xc f(x) Other possible type of infection xd f(x) Stationary Points and Their Nature Higher Outcome 3

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 Outcome 3

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 Outcome 3 + 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 Outcome 3

Nature Table x0 4 dy / dx So (0,24) is a Point of Infection and (4,-232) is a minimum Turning Point Stationary Points and Their Nature dy / dx =12x x 2 Higher Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

(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 Outcome 3

(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 Outcome 3

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 Outcome 3

-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 Outcome

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 Outcome 3

Example 33Sketch the graph of y = 8 + 2x 2 - x 4 (a) Axes If x = 0 then y = 8 (0,8) If y = 0 then 8 + 2x 2 - x 4 = 0 Graph cuts axes at (0,8), (-2,0) and (2,0) Let u = x 2 so u 2 = x 4 Equation is now 8 + 2u - u 2 = 0 (4 - u)(2 + u) = 0 (4 - x 2 )(2 + x 2 ) = 0 or (2 + x) (2 - x)(2 + x 2 ) = 0 So x = -2 or x = 2 but x 2  -2 Curve Sketching Higher Outcome 3

(b) SPs SPs occur where dy / dx = 0 So 4x - 4x 3 = 0 4x(1 - x 2 ) = 0 4x(1 - x)(1 + x) = 0 x = 0 or x =1 or x = -1 Using y = 8 + 2x 2 - x 4 when x = 0 then y = 8 when x = -1 then y = = 9 (-1,9) when x = 1 then y = = 9 (1,9) Curve Sketching Higher Outcome 3

nature x01 dy / dx + So (0,8) is a min TP while (-1,9) & (1,9) are max TPs. Curve Sketching Higher Outcome

Cuts y – axis at 8 Cuts x – axis at -2 and 2 (c) Large values Using y = - x 4 as x  +  then y  -  as x  -  then y  -  Sketch is X Y (-1,9)(1,9) y = 8 + 2x 2 - x 4 Max TP’s at (-1,9) (1,9) Curve Sketching Summarising Higher Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 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 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 Outcome 3

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 Outcome 3

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 Outcome 3

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 Outcome 3

Example 35 Higher Outcome 3 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 Outcome 3

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 Outcome 3

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 Outcome 3

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

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 Outcome 3

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 Outcome 3

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 Outcome

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

Differentiation Higher Mathematics Next

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Split up Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate multiply out differentiate

Calculus Revision Back Next Quit Differentiate Straight line form Differentiate

Calculus Revision Back Next Quit Differentiate Straight line form multiply out Differentiate

Calculus Revision Back Next Quit Differentiate multiply out Simplify Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate Chain rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify Straight line form

Calculus Revision Back Next Quit Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate Straight line form

Calculus Revision Back Next Quit Differentiate Straight line form Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

Calculus Revision Back Next Quit Differentiate

Calculus Revision Back Next Quit Differentiate Multiply out Differentiate

Calculus Revision Back Next Quit Differentiate Chain Rule Simplify

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