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Introduction to Differentiation I

Introduction to Differentiation I. Basic Functions. Instantaneous rate of change. dy/dx. f'(x). Gradient of tangent. F Servello. Table of contents. (1) Differentiating Polynomial Expressions (Basic): ax n. GO!. (2) The Chain Rule: ( f ( x )) n. GO!.

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Introduction to Differentiation I

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  1. Introduction to Differentiation I Basic Functions Instantaneous rate of change dy/dx f'(x) Gradient of tangent F Servello

  2. Table of contents (1) Differentiating Polynomial Expressions (Basic): axn GO! (2) The Chain Rule: (f(x))n GO! (3) The Product Rule: f(x).g(x) GO! (4) The Quotient Rule: f(x)/g(x) GO!

  3. Section 1 Differentiating Polynomials

  4. If the Original function is called….. Then the Derived function should be called….. y y’ or f ’ or f ’ (x) f(x) Background info slide #1 Differentiating, or finding the derivative, means finding the instantaneous rate of changei.e. gradient of the tangent of a function y = f(x). It is important to know thoroughly the notations associated with this operation. Study this closely!

  5. (4x – 7)8 3 sin x 4x 5x2 – 3 tan2x Background info slide #2 For the purposes of Section 1 of this powerpoint, we will define a POLYNOMIAL* as any expression containing terms only of the format axn, where a and n are constants and x(or any other letter) is the variable. It may be a single term, or a string of terms connected with addition and/or subtraction signs. Examples of “polynomials” (for our purposes*) include: 3x4 5t – 7 5x – 3 + ½ xx 4x6 + 2x 4x1/2 – π 4w4 – 2w– ¼8 Examples of non-polynomials (for our purposes*)include: *NOTE! The definition of “Polynomial” has been adjusted for convenience here. The strict definition of “Polynomial” excludes terms with powers that are negative and/or integers. Also, factored expressions such as (4x – 7)8 can be expanded to make expressions that are, by definition, polynomials.

  6. 1.1 1.2 1.3 Section 1. Differentiating Polynomials: axn This is the most basic form of differentiation. The rules you must learn are as follows: x is the variable; whereas a and n are constants This is just a special case of 1.1 with n = 1. So 3x becomes 3; – 2x becomes – 2 etc… Another special case of 1.1 with n = 0. Constants (numbers on their own) always just become 0.

  7. 1 2 3 Note – the alternative phrasing of this question. 4 If y = 3x2 – ½ x4 – x , find y’. y’ merely means Section 1: Examples to try Click to check your answers! Note the 3 and 5 are multiplied, and the original power (5) is lowered by 1 to give 4 15x4 2x – 5 Note x2 is treated as 1x2 so it becomes 2x1 i.e. 2x, and 5x just becomes 5. Note the 3x becomes 3, and the – 7 disappears because it’s a constant. 4x – 3 Ans: y’ = 6x – 2x3 – 1 , using previous rules 1.1, 1.2

  8. "Disguised" Polynomials

  9.  x  ±  x ±  x ….etc Where the squares represent numbers. See Preliminary Notes on Slide # 4 Important note Many expressions don’t look like polynomials at first, but you can turn them into polynomials by removing brackets or cancelling fractions, as in the following examples. To differentiate these “disguised” polynomials successfully, you should aim to get your function looking like….

  10. “Disguised polynomials” : Expressions which can be put into polynomial format, i.e. axn. Example 1.1 CAUTION! If the power were higher than a ”square”, expanding might be too onerous! There is an alternative we learn later (the Chain Rule) Given f(x) = (3x – 5)2, find f’(x). Solution Aim to get the (3x – 5)2 looking like a string of terms of the form axn(as on previous slide) Expanding (3x – 5)2, we get f(x) = 9x2 – 30x + 25 Did you remember the 30x in the middle? Which is now in polynomial format and so it’s differentiable! (a – b)2 = a2 – 2ab + b2 Using our rules from before! Answer: f ’(x) = 18x – 30

  11. If y = (x – 7)(x + 7), find Expressions which can be put into polynomial format, i.e. axn. Example 1.2 Solution Aim to get the (x – 7)(x + 7)looking like a string of terms of the form axn.(See Slide 9) Expanding (x – 7)(x + 7), we get y = x2 – 49 Did you remember the difference of 2 squares? Which is now in polynomial format and hence differentiable! (a – b)(a + b) = a2 – b2 Using our rules from before! Answer: y’ = 2x

  12. Note Differentiate Solution Use (3) above to rewrite as Ans! Differentiating gives = Background info slide #3 When multiplying a whole number or letter by a fraction, the whole number multiplies into the TOP of the fraction! Ability to manipulate these expressions will be very handy as we move into more difficult examples in later slides. Example

  13. Given y = , find We need to get the into the format axn.. This requires a simple manipulation from which is now of the format axn [Note a = & n = 1] Answer: Expressions which can be put into polynomial format, i.e. axn. Example 1.3 Solution to

  14. Given y = find Ans: Expressions which can be put into polynomial format, i.e. axn. Example 1.4 Solution Decompose the fraction as follows: Now we can differentiate! which is now in diffrentiable format!

  15. Background info slide #4 Negative and Fractional Powers can be very useful in differentiating! Learn these rules and refer to the INDICES Powerpoint for further examples Make sure you can do manipulations like these….. Study this one closely!

  16. Differentiate Differentiate Rewrite as Rewrite as So So or Ans Ans Example 1.5 Solution or Example 1.6 Solution

  17. Important note Refer to the previous slide….. You will note that there are TWO correct answers for each example. You shouldpractise manipulating your answers to get them out of the first format (with negative and/or fractional powers) and into the second format (without negative/fractional powers). Practising this process will greatly strengthen your understanding of the algebra required for manipulation of indices.

  18. Example 1.7 Differentiate Solution Rewrite as i.e. Ans So Note #1 Again here, the second of these two answers is the preferred one although both are correct – make sure you get plenty of practice at swapping in and out of negative & fractional power format. Note #2 Subtracting 1 from ½ gives – ½ , which is the new power. Also 6 × ½ = 3. Remember also that the negative in the power means that term (and that term only – in this case the x but not the 3 – is in the BOTTOM of the fraction)

  19. Example 1.8 Differentiate Rewrite as NOTE!! It may be timely here to revise how to multiply fractions. Just multiply across the top and then across the bottom. So DID YOU KNOW? There’s a fast way to subtract 1 from any fraction (in your head) ! SHOW ME! Solution Which is now in differentiable axn format Ans OR… Ans Make sure you can follow this!

  20. (a) (b) (c) Background info slide #5 It is helpful to be able to convert between SURDS and FRACTIONAL POWERS. For the most part the rules required stem from Year Ten work on indices. For example Add powers when multiplying: 2 + ½ = 5/2 Multiply powers when removing brackets Subtract powers when dividing

  21. Example 1.9 Differentiate Solution Split first: Change surds to powers: Simplify each top & bottom by adding powers: Extract fractions to front & subtract powers Which is now differentiable!

  22. Ans Remember – it’s a good idea to practise turning this index notation back into surd format Ans

  23. Point (– 1, y1 ) Normal Parabola Tangent Find the equation of the tangent and normal to the parabola y = x2 – 2x at the point where x = – 1. Example 1.10 It may be useful to know what the question is really asking! Study this graph. The normal and tangent are at right angles to each other, so the gradient of the normal is found by first finding the gradient of the tangent then taking the NEGATIVE RECIPROCAL. So, if the tangent’s gradient is 2/3, then the normal’s gradient will be – 3/2. If tangent’s gradient is – 2, then normal’s will be ½ …etc

  24. Important note Questions requiring equations of TANGENTS AND NORMALS will almost always involve the formula y – y1 = m(x – x1) You will need to substitute 3 constants into this equation…… • x1, which will generally be given • y1, which will sometimes be given. If not, you can find it by subst. x1 into the original equation y = f(x) • m, which usually is not given. You can find it by subst. x1 into the derived equation y = f ‘ (x), (i.e. subst into dy/dx)

  25. i.e. y = – 4x – 1 Eq of tangent 1. TANGENT Remember we need 3 constants: x1, y1 and m. We already know x1 = – 1, so we only have to find y1 and m. Step 1 is to find y1 at the required point (-1, y1 ). Step 2 is to find m (i.e. y’) at the required point (-1, y1 ). y = x2 – 2x, soy’ = 2x – 2. Now subst x = – 1 into 2x – 2 to get y’ = – 4 Subst. x = – 1 into y = x2 – 2x to get y = (– 1)2 – 2(– 1) = 3 y1 = 3 m = – 4 y – y1 = m(x – x1) y – 3 = – 4 (x + 1)

  26. Point (– 1, 3) Normal Parabola Tangent y = – 4x – 1 Some observations so far……. The ultimate formula you ALWAYS use in questions involving tangents and normals is y – y1 = m(x – x1). To use this effectively, you always need to find m, which you do by finding On some occasions (like this example) you may need to find y1 as well as m. This is done by substituting x1 (which you’re given) into the equation for y.

  27. i.e. y = ¼ x + 3 ¼ Eq of normal 2. NORMAL We use the 3 constants we found before, namely x1, y1 and m. But we have to take the NEGATIVE RECIPROCAL of m and use this as our new m. Step 1 is as before: find y1 at the required point (-1, y1 ). Step 2 is to first find mTAN(i.e. y’) at the required point (-1, y1 ) as we did before, then take negative reciprocal. Subst. x = – 1 into y = x2 – 2x to get y1 = (– 1)2 – 2(– 1) = 3 mtan = – 4 y1 = 3 So mnorm = ¼ y – y1 = m (x – x1) y – 3 = ¼ (x + 1) End of Section Click to return to Menu

  28. Section 2 The Chain Rule

  29. 2.1 Section 2. The Chain Rule – For differentiating expressions of the form y = (f [x])n This form of differentiation enables you to deal with expressions that can’t be turned into polynomials and for which our formulae in Section 1 can’t be used. The rule you need to learn (at this present moment) is: On Powerpoint #2 you will learn how to differentiate other functions including trigonometric and logarithmic functions. In that work, the chain rule is the dominant rule and it will assume different formats to 2.1 above.

  30. Differentiate y = (2x – 7)6 Example 2.1 If we chose to use the method of Section 1 we could expand this and then differentiate. But what a pain! So….enter theCHAIN RULE!! Solution Here, f(x) = (2x – 7) and n = 6. Differentiating 2x – 6 gives f’(x) = 2. So using the formula on the previous slide, i.e. Ans

  31. Differentiate y = Example 2.2 Solution First, rewrite as y = (4x – 5)1/2 Here, f(x) = (4x – 5) and n = ½ . Differentiating 4x – 5 gives f’(x) = 4. So using the formula from 2 slides ago, i.e. Ans or better still

  32. Differentiate y = NOTE the 2 in front just multiplies the n when you apply the formula. This gives the – 2 . Example 2.3 Solution First, rewrite as y = 2(x2 + 1)– 1 Here, f(x) = (x2 + 1) and n = – 1. Differentiating x2 + 1 gives f’(x) = 2x. So using the formula from 3 slides ago, i.e. Ans or better still

  33. First find . Check that m is found by subst. x = 1 into y1 is found by subst. x = 1 into Example 2.4 Find the equation of the tangent to the curve at the point where x = 1 Solution Remember we need to use the formula y – y1 = m(x – x1) and have to substitute 3 numbers: x1, y1 and m. x1 = 1 x1is already given : it is 1. y1 = 1 m = -¼

  34. Now we substitute these 3 constants into y – y1 = m(x – x1) y – 1 = – ¼ (x – 1) y – 1 = – ¼ x + ¼ y= – ¼ x + 1¼ is the equation of the tangent Note (for interest only!) that the tangent line has a gradient of – ¼ and a y-intercept of 1 ¼ ! End of Section Click to return to Menu

  35. Section 3 The Product Rule

  36. 3.1 Section 3. The Product Rule - for differentiating expressions of the form y = f(x) × g(x) This method enables you to deal with expressions that are a multiplication of two other variablefunctions, i.e. of the form f(x) × g(x). For convenience, the f(x) and g(x) are often written simply as f and g. The product rule only need be used when the f(x) and g(x) are two distinct functions being multiplied together. Check first whether expanding might be a quicker and easier option! Also look out for invisible × signs !! BEWARE!!

  37. Which of these would you need to use the product rule to differentiate? Easier to expand first then differentiate using basic polynomial rules. Product Rule could also be used with f = x – 1 and g = x2 + 1. MUST use product rule with f = 2x and g = (x3 – 5)1/2 Cannot easily expand because of the power 5. Use Product Rule with f = 4x and g = (x2 + 1)5. To find g’ you will need to use the Chain Rule. No need for Product Rule here as 4 is a constant. Use Chain Rule. Ans is 40x(x2 + 1)4 MUST use product rule with f = (x + 3)4 and g = (x – 1)6 . You will need to use Chain Rule to find both f’ and g’. Note – There are invisible × signs in (ii), (iii) and (v)!

  38. Example 3.1 Differentiate y = (x – 1)(x2 + 1) using the Product Rule Noting that f = x – 1 and g = x2 + 1 f = x – 1 f ‘ = 1 We differentiate each of these and set it out as follows. Easiest to set aside a workspace like this and get your four expressions ready to substitute: g = x2 + 1 g‘ = 2x Check that f ’ and g ’ are correct! Note it would be quicker in this case to expand then differentiate! Expanding the expression gives x3 – x2 + x – 1, and its derivative is equal to 3x2 – 2x + 1 !!

  39. Example 3.2 Differentiate y = (3x – 1)5 (x2 – 2)3 using the Product Rule Noting that f = (2x – 1)5 and g = (x2 – 2)3 we differentiate these as follows using the Chain Rule in each case! f = (3x – 1)5 f ‘ = 15(3x – 1)4 g = (x2 – 2)3 g‘ = 6x(x2 – 2)2 You should now inspect these two terms to see if common factors can be taken out…..

  40. ( ) (x2 – 2)2 2x (3x – 1) (1) +5 (1) (x2 – 2) 3 (3x – 1)4 To factorise We “pair” the terms up and take out the highest common factor of each and position in front of the brackets below: HCF of 6x and 15 is 3. Put 3 outside brackets and 2x & 5 inside HCF of (3x – 1)5 and (3x – 1)4 is (3x – 1)4. Put (3x – 1)4 outside brackets and (3x – 1) & 1 inside HCF of (x2 – 2)2 and (x2 – 2)3 is (x2 – 2)2. Put (x2 – 2)2 outside brackets and (1) and (x2 – 2) inside. Now expand and clean up the expressions in the large brackets: = 3(3x – 1)4(x2 – 2)2 (11x2 – 2x – 10) ANS

  41. Differentiate using the Product Rule Example 3.3 Noting that f = 2x and g = (x2 – 1)1/2 f = 2x f ‘ = 2 We differentiate each of these as follows. g = (x2 – 1)1/2 g‘ = x(x2 – 1) – 1/2 g ’ is obtained by Chain Rule – REMEMBER?? Ans….but see over

  42. It is worth noting that expressions like can be further simplified a c b Noting that the square root signs multiply out. i.e. a a= a We use the result Check this!! End of Section Click to return to Menu

  43. Section 4 The Quotient Rule

  44. Section 4. The Quotient Rule – for differentiating expressions of the form This method enables you to deal with expressions that are FRACTIONS with variablefunctions in top & bottom, i.e. of the form f(x) /g(x). NOTE! The order of terms in the top is important. The format is like the product rule but because of the minus you must begin with gf’ For convenience, again the f(x) and g(x) can be written as f and g. The Quotient Rule need only be used when the fraction cannot be simplified and put into axnformat (like Example 1.4). Usually fractions with denominators containing more than one term would indicate the need for the Quotient Rule to be used.

  45. No need for QR as it can be broken into and differentiated easily using axn rules. Which of these would you need to use the Quotient Rule to differentiate? QR necessary here. See example 4.1 next slide Hmmm…Depends whether you’re awake or not! The x2 – 9 factorises into (x – 3)(x + 3) so the whole thing just is equal to x – 3 which differentiates to 1 ! Otherwise use QR. QR necessary here. See example 4.2 next slides

  46. Differentiate using the Quotient Rule Example 4.1 f = x2 – 1 f ‘ = 2x Again like the Product Rule, set out your component parts as follows: g = x + 2 g‘ = 1 Note– the numerator is usually expanded & simplified, but the denominator is left in factorised format.

  47. Differentiate using the Quotient Rule Example 4.2 f = x f ‘ = 1 Setting out parts as follows: g = x + 1 g‘ = 1 Note– again the numerator is expanded & simplified, but the denominator is left in factorised format.

  48. Differentiate Example 4.3 Setting out parts as follows: Chain rule here See slides 15, 17, 18 to revisit differentiation of surds! Now work on simplifying the numerator (see over)

  49. Note! This last step uses the rule This becomes the top of the original expression The whole expression becomes simpler if we can first simplify the numerator! To simplify this expression, we use basic fraction subtraction rules i.e. Now the original…. Ans

  50. You might find this helpful for remembering the Quotient Rule When you begin writing the Quotient Rule, begin with f’ – g’ f Now just complete the rest of the top as if it were the Product Rule, but REMEMBER TO PUT A MINUS BETWEEN THE TERMS, NOT A PLUS!! END SHOW

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