Higher. Differentiation 28

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1 Higher Mathematics UNIT OUTCOME Differentiation Contents Differentiation 8 Introduction to Differentiation 8 Finding the Derivative 9 Differentiating with Respect to Other Variables 4 Rates of Change 4 5 Equations of Tangents 5 6 Increasing and Decreasing Curves 8 7 Stationar Points 8 8 Determining the Nature of Stationar Points 9 9 Curve Sketching 4 0 Closed Intervals 4 Graphs of Derived Functions 45 Optimisation 45 HSN00 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For more details about the copright on these notes, please see

2 Higher Mathematics Unit Mathematics OUTCOME Differentiation Introduction to Differentiation Differentiation belongs to a branch of Mathematics called calculus an area which we have not covered before. Calculus provides a tool for solving problems involving motion (e.g. the orbits of planets or flight path of a rocket). To introduce this Outcome, we will look at two distance/time graphs which describe the journe of a bugg. distance (m) distance (m) time (s) time (s) The first graph implies that the bugg could travel at a constant speed, then change instantl to a different speed. This is impossible the speed actuall changes graduall, as shown in the second graph. Instantaneous speed is the speed of an object at an eact moment in time. It is also known as the rate of change of distance with respect to time. Problems that involve rates of change are studied in differential calculus. The calculation of instantaneous speed can be a ver long process but differentiation provides a quick method. Page 8 HSN00

3 Higher Mathematics Unit Mathematics Finding the Derivative The basic rule for differentiating f ( ) = with respect to is: If n n f ( ) = then f ( ) = n where n R Stated simpl: the power (n) multiplies to the front of the term, and the power lowers b one ( n ). EXAMPLES 4. A function f is defined for R b f ( ) =. Find f ( ). f ( ) = 4. Differentiate f ( ) = with respect to. f ( ) = 4 For an epression in the form =, the derivative with respect to is epressed as d. EXAMPLE. Differentiate = d 4 = with respect to. To find the derivative of an epression in with respect to, the notation d is used. d EXAMPLE 4. Find the derivative of ( ) d d = Preparing to differentiate with respect to. It is important that before ou differentiate, all brackets are multiplied out, and there are no fractions with an term in the denominator (bottom line), for eample: = = = n 4 5 = = 5 Page 9 HSN00

4 Higher Mathematics Unit Mathematics EXAMPLES. Differentiate with respect to. d d ( ) = = =. Find the derivative of = with respect to. = = d = Note It is good practice to ti up our answer Terms with a coefficient The rule for differentiating f ( ) = a with respect to is as follows: If n n f ( ) = a then f ( ) = an where n R, a is a constant Stated simpl: the power (n) multiplies to the front of the term, then the power lowers b one ( n ). EXAMPLES. A function f is defined for R b f ( ) =. Find f ( ). f ( ) = 6. Differentiate = 4 with respect to. = 8 d 8 =. Differentiate d d ( ) = = 4 with respect to. n Page 0 HSN00

5 Higher Mathematics Unit Mathematics 4. Given =, find d. = = d 4 = 4 Differentiating more than one term The following rule allows us to differentiate epressions with several terms: If f ( ) = g ( ) + h ( ) then f ( ) = g ( ) + h ( ) Stated simpl: differentiate each term separatel. EXAMPLES. A function f is defined for R b f ( ) = + 5. Find f ( ). f ( ) = Differentiate d = = with respect to Note The derivative of an term (e.g.,, 0 ) is alwas a constant. For eample: d ( 6 ) = 6 d d ( d ) = The derivative of a constant (e.g., 0, π ) is alwas zero. For eample: d d = d ( ) ( ) 0 0 d = Page HSN00

6 Higher Mathematics Unit Mathematics Differentiating more comple epressions We will now consider more comple eamples were we will have to use several of the rules we have met. EXAMPLES. Find when ( )( ) d = +. = ( )( + ) = + = = d 6 6. A function f is defined for R, 0 b f ( ) = +. Find f ( ). 5 ( ) f = f ( ) = = 5. Differentiate d d ( ) = = = 5 5 with respect to. 4. Differentiate = with respect to. ( ) f = = f ( ) = Page HSN00

7 Higher Mathematics Unit Mathematics 5. Differentiate d d + 6 ( ) with respect to = + = = = = 6. Find the derivative of ( ) ( ) = + = = + 6 d 5 5 = = + with respect to. Remember a ab = b Remember a b a+ b = Differentiating with Respect to Other Variables So far we have differentiated functions and epressions with respect to. However, the rules we have been using still appl if we differentiate with respect to an variable. Variables such as t (for time) are used commonl to model real-life problems. EXAMPLES. Differentiate d d ( ) t t t = 6t t with respect to t.. Given A( r ) = πr, find A ( r ). A( r ) = πr A ( r ) = πr When differentiating with respect to a certain variable, all other variables are treated as constants. EXAMPLE. Differentiate ( ) d dp p = p with respect to p. Remember π is just a constant Note Since we are differentiating with respect to p, we treat as a constant Page HSN00

8 Higher Mathematics Unit Mathematics 4 Rates of Change The derivative of a function describes its rate of change. This can be evaluated for specific values b substituting them into the derivative. EXAMPLES 5. Given f ( ) = for R, find the rate of change of f when =. f ( ) 4 4 f ( ) ( ) 8 = 80. If =, calculate the rate of change of when = 8. = At = 8, = d 8 5 = d = = 5 = 96 = = Remember 48 5 a b b a = Displacement, Velocit and Acceleration The velocit ( v ) of an object is defined as the rate of change of displacement ( s ) with respect to time ( t ). That is: v ( t ) = s ( t ) Also, acceleration ( a ) is defined as the rate of change of velocit with respect to time: a( t ) = v ( t ) EXAMPLE. A ball is thrown so that its displacement s after t seconds is given b s ( t ) = t 5t. Find its velocit after seconds. v ( t ) = s ( t ) = 0 t b differentiating s ( t ) = t 5 t with respect to t Substitute t = into v ( t ) : v ( ) = 0( ) = After seconds, the ball has velocit metres per second. 5 5 Page 4 HSN00

9 Higher Mathematics Unit Mathematics 5 Equations of Tangents As we alrea know, the gradient of a straight line is constant. We can determine the gradient of a curve, at a particular point, b considering a straight line which touches the curve at the point. This line is called a tangent. tangent The gradient of the tangent to a curve = f ( ) at = a is given b f ( a). This is the same as finding the rate of change at = a. To work out the equation of a tangent we use b = m( a). Therefore we need to know two things about the tangent: A point, of which at least one coordinate will be given. The gradient, which is calculated b differentiating and substituting in the value of at the required point. EXAMPLES. Find the equation of the tangent to the curve with equation at the point (, ). We know the tangent passes through (, ). = To find its equation, we need the gradient at the point where = : = = d At =, m = ( ) = 4 Now we have the point (, ) and the gradient m = 4, so we can find the equation of the tangent: b = m( a) = 4( ) = Page 5 HSN00

10 Higher Mathematics Unit Mathematics. Find the equation of the tangent to the curve with equation at the point where =. = We need a point on the tangent. Using the given -coordinate, we can find the -coordinate of the point where the tangent meets the curve: = = ( ) ( ) = + = So the point is (, ) We also need the gradient at the point where = : = = d At =, m = ( ) = Now we have the point (,) and the gradient m =, so: b = m( a) = ( + ) +. A function f is defined for > 0 b f ( ) =. Find the equation of the tangent to the curve = f ( ) at P. P = f ( ) O We need a point on the tangent. Using the given -coordinate, we can find the -coordinate of the point where the tangent meets the curve: f ( ) = Remember a = a = b b c c = So the point is (, ) Page 6 HSN00

11 Higher Mathematics Unit Mathematics We also need the gradient at the point where = : f ( ) = f ( ) = = At =, m = 4 = 4 Now we have the point ( ) b = m( a) ( ) = 4 = , and the gradient m = 4, so: 4. Find the equation of the tangent to the curve where = 8. = at the point We need a point on the tangent. Using the given -coordinate, we can work out the -coordinate: = 8 = ( ) ( ) = 4 So the point is 8, 4 We also need the gradient at the point where = 8: = = = d = At = 8, m = 8 = = Now we have the point ( 8, 4) and the gradient m =, so: b = m( a) 4 = ( 8) = Page 7 HSN00

12 Higher Mathematics Unit Mathematics 6 Increasing and Decreasing Curves For the curve with equation = f ( ), if increases as increases, then the curve is said to be increasing. When the curve is increasing, tangents will slope upwards from left to right (i.e. their gradients are positive) so 0 d >. Similarl, when decreases as increases, the curve is said to be decreasing and 0 d <. increasing 0 d > O decreasing 0 d < increasing 0 d > 7 Stationar Points Some points on a curve ma be neither increasing nor decreasing we sa that the curve is stationar at these points. This means that the gradient of the tangent to the curve is zero at stationar points, so we can find them b solving f ( ) or. d The four possible stationar points are: Turning point Horizontal point of inflection Maimum Minimum Rising Falling A stationar point s nature (tpe) is determined b the behaviour of the graph to its left and right. This is done using a nature table. Page 8 HSN00

13 Higher Mathematics Unit Mathematics 8 Determining the Nature of Stationar Points To illustrate the method used to find stationar points and determine their nature, we will consider the graph of f ( ) = Step Differentiate the function. f ( ) = Step Find the stationar values b f ( ) solving f ( ) ( ) 6 + ( 6) ( )( ) = or = Step Find the stationar points. f ( ) = 9 so (, 9 ) is a stat. pt. f ( ) = 8 so (, 8 ) is a stat. pt. Step 4 Write the stationar values in the top row of the nature table, with arrows leading in f ( ) and out of them. Graph Step 5 Calculate f ( ) for the values in the table, and record the results. This gives the f ( ) 0 0 gradient at these values, so zeros confirm Graph that stationar points eist here. Step 6 Calculate f ( ) for values slightl lower and higher than the stationar values and record the sign in the second row, e.g.: f ( 0.8) > 0 so enter + in the first cell. Step 7 We can now sketch the graph near the stationar points: + means the graph is increasing and means the graph is decreasing. Step 8 The nature of the stationar points can then be concluded from the sketch. f ( ) Graph f ( ) Graph (, 9 ) is a ma. turning point (, 8 ) is a min. turning point Page 9 HSN00

14 Higher Mathematics Unit Mathematics EXAMPLES. A curve has equation = Find the stationar points on the curve and determine their nature. Given = = + 9 d Stationar points eist where 0 d = : + 9 ( ) 4 + ( ) 4 + ( )( ) or = = When =, = ( ) 6( ) + 9( ) 4 = Therefore the point is (, 0 ) Nature: d Graph So (, 0 ) is a maimum turning point, (, 4) is a minimum turning point. When =, = ( ) 6( ) + 9( ) 4 = = 4 Therefore the point is (, 4) Page 40 HSN00

15 Higher Mathematics Unit Mathematics. Find the stationar points of 4 Given = 4 = 8 d Stationar points eist where 0 d = : 8 ( ) 6 4 or 6 4 = When, = 4( 0) ( 0) 4 Therefore the point is ( 0, 0 ) Nature: 0 d Graph So ( 0, 0 ) is a rising point of inflection, 7 ( ), is a maimum turning point. 8 4 = 4 and determine their nature. When =, = 4 = = 7 8 ( ) ( ) 4 Therefore the point is 7 (, ) 8 9 Curve Sketching In order to sketch a curve, we need to first find the following: -ais intercepts (roots) solve -ais intercept find for 0 = stationar points and their nature. Page 4 HSN00

16 Higher Mathematics Unit Mathematics EXAMPLE Sketch the graph of =. -ais intercept: = ( 0) ( 0) Therefore the point is ( 0, 0 ) Given = = 6 6 d Stationar points eist where 0 d = : ( ) 6 or When, = ( 0) ( 0) Therefore the point is ( 0, 0 ) = Nature: 0 d Graph ais intercepts (roots): ( 0, 0) When =, ( ) or = ( ) ( ) = = Therefore the point is (, ) = (, 0) ( 0, 0 ) is a maimum turning point (, ) is a minimum turning point = O (, ) Page 4 HSN00

17 Higher Mathematics Unit Mathematics 0 Closed Intervals Sometimes it is necessar to restrict the part of the graph we are looking at using a closed interval (also called a restricted domain). The maimum and minimum -values can either be at stationar points or at the end points of the closed interval. Below is a sketch of a curve, with the closed interval 6 shaded. maimum O minimum EXAMPLE A function f is defined for R b f ( ) = Find the maimum and minimum value of f ( ) where 4. Given f ( ) = f ( ) = Stationar points eist where f ( ) : ( ) 5 ( )( + ) or + = = To find coordinates of stationar points: f ( ) = ( ) 5( ) 4( ) + = = Therefore the point is (, ) 6 ( ) ( ) ( ) ( ) = ( ) 5 ( ) 4( ) + f = = = 46 7 Therefore the point is (, 46 ) 7 Page 4 HSN00

18 Higher Mathematics Unit Mathematics Nature: f ( ) Graph 46 (, ) 7 is a ma. turning point (, ) is a min. turning point Points at etremities of closed interval: f ( ) = ( ) 5( ) 4( ) + = = Therefore the point is (, ) ( ) ( ) ( ) ( ) f 4 = = = Therefore the point is ( 4, ) Now we can make a sketch: ( 4, ) (, 46 ) 7 O (, ) (, ) The maimum value is which occurs when = 4 The minimum value is which occurs when = Page 44 HSN00

19 Higher Mathematics Unit Mathematics Graphs of Derived Functions n n The derivative of an function is an function the degree lowers b one. For eample the derivative of a cubic ( ) function, is a quadratic ( ) function. When drawing a derived graph: All stationar points of the original function become roots (i.e. lie on the -ais) on the derived graph The sign (+ or ) of the original graph s gradient becomes the value of the derived graph between the roots. For eample, if the gradient between two stationar points of the original graph is negative, then the graph between the roots of the derived graph will lie below the -ais (i.e. it will be negative). Quadratic Cubic Quartic O O O O + Linear + + O Quadratic + + O Cubic Optimisation Differentiation can be used to help solve real-life problems. However, often we need to set up an equation before we can differentiate. Page 45 HSN00

20 Higher Mathematics Unit Mathematics EXAMPLE Small wooden tras, with open tops and square bases, are being designed. The must have a volume of 08 cm. The internal length of one side of the base is cm, and the internal height of the tra is h cm. (a) Show that the total internal surface area A of one tra is given b 4 A = + (b) Find the dimensions of the tra which use the smallest amount of wood. (a) Volume = area of base height = h We are told that the volume is 08 cm, so: Volume 8 h 8 08 h = h Let A be the surface area for a particular value of : A = + 4h 08 We have h =, so: A = = + 4 ( ) (b) The smallest amount of wood is used when the surface area is minimised. da 4 = d da Stationar points occur when 0 d = Nature: 6 4 da d 0 + = 6 Graph = 6 So the minimum surface area occurs when = 6. For this value of : 08 h = = 6 So a length and depth of 6 cm and a height of cm uses the smallest amount of wood. Page 46 HSN00

Mathematics. Mathematics 1. hsn.uk.net. Higher HSN21000

Mathematics. Mathematics 1. hsn.uk.net. Higher HSN21000 Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For