INTRODUCTION TO DIFFERENTIATION

INTRODUCTION TO DIFFERENTIATION GRADIENT OF A CURVE We have looked at the process needed for finding the gradient of a curve (or the rate of change of a curve). We have defined the gradient of a curve at a particular point as being the gradient of the tangent to the curve at that point. To find the gradient of the tangent we start with looking at the gradient of a secant to the curve through the required point and another point close by. We used the idea of a limit as we investigated what happened as the second point moved closer to the required point i.e. as the gradient of the secant became a better and better approximation for the gradient of the tangent. http://www.youtube.com/watch?v=y4dlco6pchg&feature=player_detailpage http://www.khanacademy.org/math/precalculus/v/introduction-to-limits : will go over the ideas involved in finding a limit http://www.youtube.com/watch?v=ayf9gkwjxly : looks at how to find the derivative of a function from first principles You ll be pleased to know that we don t always find a derivative from first principles. We simply use it to develop some rules which allow us to find these derivatives very quickly. We can also draw a graph which describes the shape of the gradient function of a curve - see http://www.youtube.com/watch?v=gbtma_uqpro&feature=related. We never differentiate from first principles unless instructed to do so in the question the rule is always faster. Date edited: 22/9/12 P a g e 1

LIMIT LAWS There are a number of limit laws which we tend to use in limit questions without writing them into our calculations in a formal manner. 1. The limit of a sum or difference is the same as the sum or difference of the limits of each part. 2. The limit of a constant multiple of a function is equal to a constant multiple of the limit of a function. 3. The limit of a product/quotient is the same as the product/quotient of the limits of each function. 4. For the constant function, CONTINUITY Also important in calculus is the idea of continuity. Basically, if we can trace the length of a function without taking our pencil off the paper, the function is continuous. More formally: A function is continuous at x = a if (i) f(a) exists (ii) exists and equals f(a). DIFFERENTIABILITY There are certain conditions which must exist for a function to be differentiable at x = a: (i) The function is continuous at x = a (ii) the gradient function is also continuous at x = a. Date edited: 22/9/12 P a g e 2

RULES FOR DIFFERENTIATING POWER RULE SOME NOTATION These phrases all require the same process to be carried out: Differentiate Find the derivative of... Find the gradient function for... Find the differential coefficient of... These notations are all used for the result of differentiating: y f (x) ( ) D x y (not seen very often) Over the next 2 semesters, you will learn a number of different rules for different groups of functions. The first one we deal with is the rule for functions of the type. We will use differentiation from first principles and also the fact that x n a n = (x a)(x n-1 +a.x n 2 +a 2.x n-3 +a 3.x n-4 +... + a n-1 ) which allows us to factorise (x + h) n x n =[ (x +h) x][(x+h) n-1 +x.(x+h) n 2 +x 2.(x+h) n-3 +x 3.(x+h) n-4 +... + x n-1 ] You may like to view http://www.youtube.com/watch?v=ayf9gkwjxly to see how this is done with small values for n. Date edited: 22/9/12 P a g e 3

General proof: So if, then Then, and from the factors above: So if f(x) = x n, then f (x) nx n-1 n can be positive or negative, integer or fraction. roots etc must be rewritten with fractional indices. must be rewritten as x -n and square roots, cube Date edited: 22/9/12 P a g e 4

ASSOCIATED LAWS If f(x) = k, then f (x) = 0 (can you explain this graphically?) If f(x) = a.x n, then f (x) = anx n-1 If f(x) = g(x) h(x) then f (x) = g (x) h (x) RATIONAL FUNCTIONS WHERE THE DENOMINATOR IS A SINGLE TERM We are all very familiar with expanding expressions like 2x(3x 7). We know that every term in the bracket is multiplied by the term outside the brackets i.e 2x(3x 7) = 6x 2 14x. We are not so familiar with simplifying expressions like. The same rules apply as the fraction bar (vinculum) acts like brackets, so we need to divide every term in the numerator by the denominator. So = and we must change this to ½x 2 + 4x -1 before differentiating using the rule we have just learnt. (Power Rule) PRACTICE AND CONSOLIDATION Hawker 11 Math Methods 9A-9D Melba Maths Quest Exercise 8A-8D Date edited: 22/9/12 P a g e 5

PRODUCT RULE So far we have differentiated the sum or difference of functions involving terms which can be written in the form y = ax n. The derivative of the sum of functions is the same as the sum of the derivatives of the parts. (Remember: we have to change reciprocals to negative indices and irrational terms to fractional powers). Now for products: If we consider y = x 5 and its equivalent y = x 2.x 3. We know the derivative of x 5 is 5x 4, can we differentiate the factors of the product and get this answer? Does 2x times 3x 2 give us the same answer? No it gives us 3x 2, so it looks as though the derivative of a product is not the same as the product of the derivatives. Of course we will simplify the product if possible and then differentiate the terms, however, sometimes we can t expand a product or it is too long and open to errors. Let s investigate this from first principles. Once we have derived the formula, or process, you will use the rule only. Let us consider y = u.v where u and v are both functions of x. If we increase the x values by a small amount,, then u will change by a small amount and v will change by a small amount. Using our formula for differentiating from first principles: Then dividing each term in the numerator by δx we get In abbreviated form this looks like: The rhyme we can use to remember this is: First times the derivative of the second PLUS the second times the derivative of the first. Date edited: 22/9/12 P a g e 6

Now, for an example: Find the derivative of f(x) = (2x+3)(x 2-2x -5) we could expand this, but let s use the product rule Let u = 2x+3 and v = x 2 2x 5 So u = 2 v = 2x 2 Since You could check this by expanding the function before differentiating. We will soon meet some functions where we don t have the option of expanding. WATCH NOW http://youtu.be/lhr7u32xw0c Date edited: 22/9/12 P a g e 7

CHAIN RULE Differentiating a function of a function using the chain rule i.e. differentiating Consider y = (3x 2 2) 5 and y = [f(x)] n Let u = 3x 2 2 Let u = f(x) So now y = u 5 y = u n From the last two statements we can differentiate to get We really need to find so let s create a chain :. Algebraically this looks OK but it needs ideas from University Maths to prove it. At the moment we will just accept the statement. Therefore, So the chain rule is: = 5u 4.6x = n.u n-1.f (x) = 30x(3x 2 2) 4 If and then And also If then The rhyme to remember this one is.. the derivative of the outside, times the derivative of the inside WATCH NOW http://youtu.be/dhucsxzbtcm http://youtu.be/mx_ltejesnm Date edited: 22/9/12 P a g e 8

EXAMPLE Differentiate y = (2x 3) Let u = 2x 3 Let Then then Date edited: 22/9/12 P a g e 9

EXAMPLE Or another one: differentiate f(x) = (3x 7) 6 (x 2 +5x) 8 Let Let Then then As you can see there is always a bit of algebraic simplification, one piece of advice I can give you is to make sure you keep things as factorised as possible for as long as possible. Often with the product rule, you can factorise out common parts from both pieces. WATCH NOW http://youtu.be/cngzq5xioby Date edited: 22/9/12 P a g e 10

PRACTICE AND CONSOLIDATION Hawker 12 Math Methods 7D: Q1, 3, 7, 12, 17 Cambridge Yr 11 3 unit: 7D: Q1, 2, 6, 8, 16, 18, (SM -13, 22) 7E:Q1, 2, 4, 6, 11, (SM -8, 9, 10, 15) 7F: Q1, 2, 3, 4, (SM -8, 13) QUOTIENT RULE We use the quotient rule to differentiate functions which look a bit like fractions. That is, they are rational expressions where we have one function divided by another. The quotient rule states that if: where u and v are functions in the same variable, then We can also write this using function notation: And also write it using Leibniz notation: Date edited: 22/9/12 P a g e 11

EXAMPLE Find the derivative of Let Then state And Let and Then write the rule: And substitute in all the values And then algebraically manipulate to simplify We leave the answer in this form (it is as factorised as possible). Notice I didn t expand the bottom term, mostly this is the best approach. WATCH NOW http://youtu.be/o6m4o7zy5ea http://youtu.be/n29vczbhgti PRACTICE AND CONSOLIDATION Hawker 12 Math Methods 7I: Q2, Q4 a, b, d, j, l Cambridge Yr 11 3 unit: 7G: Q1, 3, 5, 6, 9b, SM10 Date edited: 22/9/12 P a g e 12

IMPLICIT DIFFERENTIATION (SM ONLY) Implicit differentiation is used instead of rearranging equations like or into a y= form to be able to use our previous 4 differentiation laws. Implicit differentiation is the process of differentiating EACH term with respect to x. Look at the following example. EXAMPLE Differentiate Differentiate each term Complete each term (the easy ones) Now look at the term We use the chain rule on this term Now rearrange to get its own on Substitute y= into the equation Date edited: 22/9/12 P a g e 13

WATCH NOW http://youtu.be/ximf06lmqpm (Patrick) http://youtu.be/sl6mc-lkorw (Khan) http://youtu.be/_vhy63og-f4 (Mathmeeting) PRACTICE AND CONSOLIDATION Hawker 12 Specialist Maths 5F: 2, 3, 5, 7, 9 Cambridge Yr 11 3 unit: 7K: Q1, 2, 3, 5, 7, 13 PARAMETRIC EQUATIONS (SM ONLY) Sometimes functions can be defined giving x and y in some third variable called t. This t is a parameter. For example, this actually specifies the parabola. To differentiate a parametric equation, we apply the chain rule. Date edited: 22/9/12 P a g e 14

So for the example above: So WATCH NOW http://youtu.be/k5qnagvk1ji (Patrick) http://youtu.be/w2ohnus7fxk (Maths247) PRACTICE AND CONSOLIDATION Hawker Cambridge Yr 11 3 unit: 7E Q3, 7, 7G Q8, Date edited: 22/9/12 P a g e 15

CURVE PROPERTIES TANGENTS AND NORMALS DEFINITIONS Tangent: A straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point. The gradient of a tangent at a point, is equal to the instantaneous rate of change at that point - ie the gradient of the tangent is equal to the value of the derivative at that point. Normal: A normal line at point P is a line that is perpendicular to the tangent at the point P. The gradient of a normal is the negative reciprocal of the gradient of the tangent. If is the gradient of the tangent and is the gradient of the normal, then FINDING EQUATIONS Finding equations of tangents and normals requires knowing how to find the equation of a straight line. Remember that we need a point (x ), and the gradient m. Knowing these two things we can formulate and equation using the point gradient formula. EXAMPLE Find the equation of the tangent and normal to the curve with equation, at the point. Step 1: Find the gradient function Step 2: Evaluate the gradient at the point. Step 3: Identify the gradient of the tangent and gradient of the normal. the value of the gradient at the point Step 4: Find the equation of the tangent using the point gradient formula with point and gradient 7 ( ) This is the equation of the tangent Step 5: Find the equation of the tangent using the point gradient formula with point and gradient ( ) Date edited: 22/9/12 P a g e 16

This is the equation of the normal You can also calculate the equation of the tangent and normal directly on your CAS. READ method one http://www.classpad.com.au/pdf/cp141_equation_of_tangent_to_curve.pdf method two http://www.classpad.com.au/pdf/cp343_equation_of_tangent_to_curve.pdf WATCH method one http://www.classpad.com.au/ go to intermediate, working in main and watch the equation of tangent to curve. method two http://www.classpad.com.au/ go to intermediate, graph and table and watch the equation of tangent to the curve. WATCH NOW http://youtu.be/aq1-_d-yur4 PRACTICE AND CONSOLIDATION Hawker - 12 Math Methods Book 8A 1, 2, 3, 4, 6, 7, 8, 13 If you still want more try these - Cambridge Yr 11 3 unit: Some of these you should have completed over the previous weeks. 7C Q8, 9, 10, 12 SM 17 23 26 SMChallenge 31 7D 17, 20 7E 5, 6, 7, 13 7F 3 7G 4, 8 Date edited: 22/9/12 P a g e 17

CURVE PROPERTIES We can use the gradient function to identify important properties on the original curve. The gradient function can be used to identify increasing and decreasing sections, turning points and stationary points. Increasing function An increasing function is one whose y-values increase as it's x-values increase. That is - it is going up from left to right. An increasing function has a positive gradient. Decreasing function An decreasing function is one whose y-values decrease as it's x-values increase. That is - it is going down from left to right. An decreasing function has a negative gradient. Stationary point A stationary point is a place on the curve where the function is stationary at that point. The function is neither increasing or decreasing at this point. A stationary point has gradient equal to zero. Turning point A turning point is a place on the curve where it 'turns'. That is - it changes increasing to decreasing, its gradient changes from positive to negative. It is a special form of stationary point. A turning point has gradient equal to zero, and it changes from positive to negative, or negative to positive. Date edited: 22/9/12 P a g e 18

HERE ARE SOME PICTURES TO HELP DEMONSTRATE. increasing graph f'(x)>0 positive gradient decreasing graph f'(x)<0 negative gradient stationary point f'(x)=0 tangent is horizontal turning point f'(x)=0 tangent is horizontal gradient changes from positive to negative Date edited: 22/9/12 P a g e 19

PRACTICE AND CONSOLIDATION Hawker - 12 Math Methods Book 8B 1 (half of), 2, 3, 5, 7, 8a, d, i, 13, 20 Cambridge 3 Unit - 10A 1, 3, 7, 8, 9, 14, 17, 18, 10B 1, 4, 6, 8, 9, SM 13, 14 10C 1, 5, 7, 11 SM 3, 8, 9 SECOND DERIVATIVE We are having so much fun with differentiation that we needn't stop just there. All that we have done so far is actually discussing the FIRST DERIVATIVE. We can also take the derivative again, and we call this the SECOND DERIVATIVE. We use the following notation to describe the second derivative: The second derivative can tell us about concavity. Concavity is a word used to describe the curve of the curve, concave up concave down Date edited: 22/9/12 P a g e 20

Increasing and decreasing functions could be either concave up or down, here are some examples: image from http://tutorial.math.lamar.edu/classes/calci/shapeofgraphptii_files/image001.gif If then the curve is concave up. If then the curve is concave down. If then the curve COULD have a point of inflection at that point. A point of inflection is a point where the concavity changes from positive to negative, or negative to positive. at the point where, further clarification is necessary to fully identify. What follows from this is the idea that we can now identify if a turning point is a maximum or minimum. If then the point c is a minimum If then the point c is a maximum If then the point c could be a relative maximum, minimum or neither. Date edited: 22/9/12 P a g e 21

The following table summarizes our first derivative and second derivative curve analysis. Decreasing graph Flat spot Increasing graph Concave down Possible POI Or Concave up WATCH NOW http://youtu.be/o2i3lajjkle http://youtu.be/gnwz94njcby http://youtu.be/die22el6q90 (Khan) PRACTICE AND CONSOLIDATION Hawker Cambridge 3 Unit - 10D 1 a,b 2a,b 3a,c 5b,f 5, 68, 10b 10E 1, 2, 4, 5, 7, 8, 10, 15 SM 18 10F 1a,d 3, 4, 7b, SM10, 12 10G 1, 2a,b h(sm), 3c,d,f Date edited: 22/9/12 P a g e 22

APPLICATIONS MAXIMUM AND MINIMUM PROBLEMS PRACTICE AND CONSOLIDATION Hawker 12 Methods Book 8C 1a,b,c 4, 6, 7, 12 8D 1, 2, 3, 4, 5, 6, 9, 11 Cambridge 3 Unit - 10H 1, 3, 4, 6, 7, 8, 9 any 3 of 10-17 SM 18, 25, 26, 27 10I 2, 3, 7 any 2 of 10-14 RELATED RATES PRACTICE AND CONSOLIDATION Hawker 12 Methods Book 8E 1, 3, 5, 6, 14 8F 1, 2, 3, 4, 5, 8, 11 Cambridge 3 Unit 7H any 3 of 1-6 any 2 of 7-11 Date edited: 22/9/12 P a g e 23