Differentiation and Antidifferentiation of Polynomials PDF

17 Differentiation and antidifferentiation of polynomials Objectives I To understand the concept of limit. I To understand the definition of the derivative of a function. I To unders...

17 Differentiation and antidifferentiation of polynomials Objectives I To understand the concept of limit. I To understand the definition of the derivative of a function. I To understand and use the notation for the derivative of a polynomial function. I To find the gradient of a tangent to a polynomial function by calculating its derivative. I To apply the rules for differentiating polynomials to solving problems. I To be able to differentiate expressions of the form xn where n is a negative integer. I To understand and use the notation for the antiderivative of a polynomial function. It is believed that calculus was discovered independently in the late seventeenth century by two great mathematicians: Isaac Newton and Gottfried Leibniz. Like most scientific breakthroughs, the discovery of calculus did not arise out of a vacuum. In fact, many mathematicians and philosophers going back to ancient times made discoveries relating to calculus. In the previous chapter, we investigated the rate of change of one quantity with respect to another quantity. In this chapter, we will develop a technique for calculating the rate of change for polynomial functions. To illustrate the idea, we start with an introductory example: On planet X, an object falls a distance of y metres in t seconds, where y = 0.8t2. Can we find a general expression for the speed of such an object after t seconds? (Note that, on Earth, the commonly used model is y = 4.9t2.) ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  Chapter 17: Differentiation and antidifferentiation of polynomials 567 In the previous chapter, we found that y we could approximate the gradient of a y = 0.8t 2 curve at a given point P by finding the gradient of a secant PQ, where Q is a point on the curve as close as possible Q to P. The gradient of PQ approximates the speed of the object at P. The closer we P make the point Q to the point P, the 0 5 (5 + h) t better the approximation. Let P be the point on the curve where t = 5. Let Q be the point on the curve corresponding to h seconds after t = 5. That is, Q is the point on the curve where t = 5 + h. 0.8(5 + h)2 − 0.8 × 52 Gradient of PQ = (5 + h) − 5
0.8 (5 + h)2 − 52 = h = 0.8(10 + h) The table gives the gradient of PQ for different values of h. h Gradient of PQ Use your calculator to check these. 0.7 8.56 If we take values of h with smaller and smaller magnitude, 0.6 8.48 then the gradient of PQ gets closer and closer to 8. So, at the 0.5 8.40 point where t = 5, the gradient of the curve is 8. 0.4 8.32 Thus the speed of the object at the moment t = 5 is 8 m/s. 0.3 8.24 The speed of the object at the moment t = 5 is the limiting 0.2 8.16 value of the gradients of PQ, as Q approaches P. 0.1 8.08 We want to find a general formula for the speed of the object at any time t. Let P be the point with coordinates (t, 0.8t2 ) on the curve and let Q be the point with coordinates (t + h, 0.8(t + h)2 ). 0.8(t + h)2 − 0.8t2 Gradient of PQ = (t + h) − t = 0.8(2t + h) Now consider the limit as h approaches 0, that is, the value of 0.8(2t + h) as h becomes arbitrarily small. This limit is 1.6t. So the gradient of the tangent to the curve at the point corresponding to time t is 1.6t. Hence the speed at time t is 1.6t m/s. This technique can be used to investigate the gradient of the tangent at a given point for similar functions. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  568 Chapter 17: Differentiation and antidifferentiation of polynomials 17A The derivative y We first recall that a chord of a curve is a line segment joining points P and Q on the curve. A secant is a line through points P and Q on the curve. In the previous chapter we considered what Q1 happened when we looked at a sequence of secants Q2 PQ1 , PQ2 ,... , PQn ,... , where the points Qi get Q3 closer and closer to P. The idea of instantaneous rate Qn of change at P was introduced. P In this section we focus our attention on the gradient of the tangent at P. x 0 The tangent to a curve at a point Consider the function f : R → R, f (x) = x2. y f (x) = x2 (a + h, (a + h)2) Q P(a, a2) x 0 The gradient of the secant PQ shown on the graph is (a + h)2 − a2 gradient of PQ = a+h−a a2 + 2ah + h2 − a2 = h = 2a + h The limit of 2a + h as h approaches 0 is 2a, and so the gradient of the tangent at P is said to be 2a. The straight line that passes through the point P and has gradient 2a is called the tangent to the curve at P. It can be seen that there is nothing special about a here. The same calculation works for any real number x. The gradient of the tangent to the graph of y = x2 at any point x is 2x. We say that the derivative of x2 with respect to x is 2x, or more briefly, we can say that the derivative of x2 is 2x. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17A The derivative 569 Example 1 y By first considering the gradient of the secant PQ, find the gradient of the y = x2 – 2x tangent line to y = x2 − 2x at the point P (3 + h, (3 + h)2 – 2(3 + h)) with coordinates (3, 3). Q P(3, 3) x 0 2 Solution (3 + h)2 − 2(3 + h) − 3 Gradient of PQ = 3+h−3 9 + 6h + h2 − 6 − 2h − 3 = h 4h + h2 = h =4+h Now consider the gradient of PQ as h approaches 0. The gradient of the tangent line at the point P(3, 3) is 4. Example 2 y Find the gradient of the secant PQ and hence find y = x2 + x the derivative of x2 + x. Q(x + h, (x + h)2 + (x + h)) P(x, x2 + x) x –1 0 Solution (x + h)2 + (x + h) − (x2 + x) Gradient of PQ = x+h−x x2 + 2xh + h2 + x + h − x2 − x = h 2 2xh + h + h = h = 2x + h + 1 From this it is seen that the derivative of x2 + x is 2x + 1. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  570 Chapter 17: Differentiation and antidifferentiation of polynomials The expansion of (a + b)n You are already familiar with the identity (a + b)2 = a2 + 2ab + b2 By multiplying both sides by a + b, we can obtain new identities: (a + b)3 = a3 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 You may recognise the pattern of coefficients from Pascal’s triangle (see Section 10C). We will give the general expansion of (a + b)n in the next section. Limit notation The notation for the limit of 2x + h + 1 as h approaches 0 is lim (2x + h + 1) h→0 The derivative of a function with rule f (x) may be found by:

1 finding an expression for the gradient of the line through P x, f (x) and Q x + h, f (x + h) 2 finding the limit of this expression as h approaches 0. Example 3 Consider the function f (x) = x3. By first finding the gradient of the secant through P(2, 8)
and Q 2 + h, (2 + h)3 , find the gradient of the tangent to the curve at the point (2, 8). Solution (2 + h)3 − 8 Gradient of PQ = 2+h−2 8 + 12h + 6h2 + h3 − 8 = h 12h + 6h2 + h3 = h = 12 + 6h + h2 The gradient of the tangent line at (2, 8) is lim (12 + 6h + h2 ) = 12. h→0 The following example provides practice in determining limits. Example 4 Find: 3x2 h + 2h2 a lim (22x2 + 20xh) b lim c lim 4 h→0 h→0 h h→0 Solution 3x2 h + 2h2 a lim (22x2 + 20xh) = 22x2 b lim = lim (3x2 + 2h) c lim 4 = 4 h→0 h→0 h h→0 h→0 = 3x2 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17A The derivative 571 Using the TI-Nspire To calculate a limit, use menu > Calculus > Limit and complete as shown. Note: The limit template can also be accessed from the 2D-template palette t. When you insert the limit template, you will notice a superscript field (small box) on the template – generally this will be left empty. Using the Casio ClassPad  In M, enter and highlight the expression 3x2 h + 2h2 h Note: Use h from the Var keyboard.  Select ; from the Math2 keyboard.  Enter h and 0 in the spaces provided as shown. Tap EXE. Definition of the derivative In general, consider the graph y = f (x) of a y function f : R → R. y = f (x) f (x + h) − f (x) Gradient of secant PQ = x+h−x f (x + h) − f (x) Q = (x + h, f (x + h)) h P (x, f (x)) The gradient of the tangent to the graph of y = f (x) at the point P(x, f (x)) is the limit of x 0 this expression as h approaches 0. Derivative of a function The derivative of the function f is denoted f 0 and is defined by f (x + h) − f (x) f 0 (x) = lim h→0 h The tangent line to the graph of the function f at the point (a, f (a)) is defined to be the line through (a, f (a)) with gradient f 0 (a). Warning: This definition of the derivative assumes that the limit exists. For polynomial functions, such limits always exist. But it is not true that for every function you can find the derivative at every point of its domain. This is discussed further in Sections 17F and 17G. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  572 Chapter 17: Differentiation and antidifferentiation of polynomials Differentiation by first principles Determining the derivative of a function by evaluating the limit is called differentiation by first principles. Example 5 For f (x) = x2 + 2x, find f 0 (x) by first principles. Solution f (x + h) − f (x) f 0 (x) = lim h→0 h (x + h)2 + 2(x + h) − (x2 + 2x) = lim h→0 h x2 + 2xh + h2 + 2x + 2h − x2 − 2x = lim h→0 h 2xh + h2 + 2h = lim h→0 h = lim (2x + h + 2) h→0 = 2x + 2 ∴ f 0 (x) = 2x + 2 Example 6 For f (x) = 2 − x3 , find f 0 (x) by first principles. Solution f (x + h) − f (x) f 0 (x) = lim h→0 h 2 − (x + h)3 − (2 − x3 ) = lim h→0 h 2 − (x3 + 3x2 h + 3xh2 + h3 ) − (2 − x3 ) = lim h→0 h −3x2 h − 3xh2 − h3 = lim h→0 h = lim (−3x2 − 3xh − h2 ) h→0 = −3x2 ∴ f 0 (x) = −3x2 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17A The derivative 573 Using the TI-Nspire  Assign the function f (x) as shown. Note: The assign symbol := is accessed using ctrl t.  Use menu> Calculus > Limit or the 2D-template palette t, and complete as shown. Using the Casio ClassPad  In M, enter and highlight the expression 2 − x3. Select Interactive > Define and tap OK.  Now enter and highlight the expression f (x + h) − f (x) h Note: Select f from the Math3 keyboard and select x, h from the Var keyboard.  Select ; from the Math2 keyboard.  Enter h and 0 in the spaces provided as shown. Tap EXE. Approximating the value of the derivative From the definition of the derivative, we can y y = f (x) see that f (a + h) − f (a) Q(a + h, f (a + h)) f 0 (a) ≈ h for a small value of h. This is the gradient P(a, f (a)) of the secant through points P(a, f (a)) and Q(a + h, f (a + h)). x 0 We can often obtain a better approximation y y = f (x) by using f (a + h) − f (a − h) Q(a + h, f (a + h)) f 0 (a) ≈ 2h for a small value of h. This is the gradient of P(a, f (a)) the secant through points R(a − h, f (a − h)) R(a − h, f (a − h)) and Q(a + h, f (a + h)). x 0 In Exercise 17A, you are asked to compare these two approximations for several functions. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  574 Chapter 17: Differentiation and antidifferentiation of polynomials 17A Summary 17A  The derivative of the function f is denoted f 0 and is defined by f (x + h) − f (x) f 0 (x) = lim h→0 h  The tangent line to the graph of the function f at the point (a, f (a)) is defined to be the line through (a, f (a)) with gradient f 0 (a).  The value of the derivative of f at x = a can be approximated by f (a + h) − f (a) f (a + h) − f (a − h) f 0 (a) ≈ or f 0 (a) ≈ h 2h for a small value of h. Exercise 17A Example 1 1 Let f (x) = −x2 + 4x. y The graph of y = f (x) is shown opposite. P(3, 3) a Find the gradient of PQ. Q(3 + h, –(3 + h)2 + 4(3 + h)) b Find the gradient of the tangent to the curve at P x 0 by considering what happens as h approaches 0.
2 Let f (x) = x2 − 3x. Then the points P(4, 4) and Q 4 + h, (4 + h)2 − 3(4 + h) are on the curve y = f (x). a Find the gradient of the secant PQ. b Find the gradient of the tangent line to the curve at the point P by considering what happens as h approaches 0.
Example 2 3 The points P(x, x2 − 2x) and Q x + h, (x + h)2 − 2(x + h) are on the curve y = x2 − 2x. Find the gradient of PQ and hence find the derivative of x2 − 2x.
Example 3 4 By first considering the gradient of the secant through P(2, 16) and Q 2 + h, (2 + h)4 for the curve y = x4 , find the gradient of the tangent to the curve at the point (2, 16). Hint: (x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 5 A space vehicle moves so that the distance travelled over its first minute of motion is given by y = 4t4 , where y is the distance travelled in metres and t the time in seconds. By finding the gradient of the secant through the points where t = 5 and t = 5 + h, calculate the speed of the space vehicle when t = 5. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17A 17A The derivative 575 6 A population of insects grows so that the size of the population, P, at time t (days) is given by P = 1000 + t2 + t. By finding the gradient of the secant through the points where t = 3 and t = 3 + h, calculate the rate of growth of the insect population at time t = 3. Example 4 7 Find: 2x2 h3 + xh2 + h a lim (10x2 − 5xh) b lim (20 − 10h) c lim h→0 h→0 h→0 h 2 2 3x h − 2xh + h 30hx2 + 2h2 + h d lim e lim f lim 5 h→0 h h→0 h h→0 8 Find: (x + h)2 + 2(x + h) − (x2 + 2x) a lim i.e. the derivative of y = x2 + 2x h→0 h (5 + h)2 + 3(5 + h) − 40 b lim i.e. the gradient of y = x2 + 3x at x = 5 h→0 h (x + h)3 + 2(x + h)2 − (x3 + 2x2 ) c lim i.e. the derivative of y = x3 + 2x2 h→0 h 9 For the curve with equation y = 3x2 − x: a Find the gradient of the secant PQ, where P is the point (1, 2) and Q is the point
1 + h, 3(1 + h)2 − (1 + h). b Find the gradient of PQ when h = 0.1. c Find the gradient of the tangent to the curve at P. 2 10 For the curve with equation y = : x  2  a Find the gradient of the chord AB, where A = (2, 1) and B = 2 + h,. 2+h b Find the gradient of AB when h = 0.1. c Find the gradient of the tangent to the curve at A. For the curve with equation y = x2 + 2x − 3: CAS 11 a Find the gradient of the secant PQ, where P is the point (2, 5) and Q is the point
2 + h, (2 + h)2 + 2(2 + h) − 3. b Find the gradient of PQ when h = 0.1. c Find the gradient of the tangent to the curve at P. f (x + h) − f (x) Example 5 12 For each of the following, find f 0 (x) by finding lim : h→0 h Example 6 a f (x) = 3x2 b f (x) = 4x c f (x) = 3 2 d f (x) = 3x + 4x + 3 e f (x) = 2x3 − 4 f f (x) = 4x2 − 5x g f (x) = 3 − 2x + x2 h f (x) = 2x − x3 i f (x) = 2x − 3x2 13 By first considering the gradient of the secant through P(x, f (x)) and Q(x + h, f (x + h)) for the curve f (x) = x4 , find the derivative of x4. Hint: (x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  576 Chapter 17: Differentiation and antidifferentiation of polynomials 17A Consider the following two approximations for f 0 (a), where h is small: CAS 14 f (a + h) − f (a) f (a + h) − f (a − h) i f 0 (a) ≈ ii f 0 (a) ≈ h 2h Compare these two approximations for each of the following: a f (x) = x2 , a = 2 b f (x) = x3 , a = 2 c f (x) = x3 + 2x − 4, a = 2 17B Rules for differentiation The derivative of xn where n is a positive integer From your work in the first section of this chapter, you may have noticed that differentiating from first principles gives the following:  For f (x) = x, f 0 (x) = 1.  For f (x) = x2 , f 0 (x) = 2x.  For f (x) = x3 , f 0 (x) = 3x2. This suggests the following general result. For f (x) = xn , f 0 (x) = nxn−1 , where n = 1, 2, 3,... Proof In order to prove this result, we use the binomial theorem, which gives the general expansion of (a + b)n. The notation nCr for the number of combinations of n objects in groups of size r is introduced in Chapter 10. Binomial theorem For each n ∈ N, we have (a + b)n = an + nC1 an−1 b + nC2 an−2 b2 + · · · + nCr an−r br + · · · + nCn−1 abn−1 + bn The binomial theorem is proved in the polynomials appendix in the Interactive Textbook. Now we let f (x) = xn , where n ∈ N with n ≥ 2. Then f (x + h) − f (x) = (x + h)n − xn = xn + nC1 xn−1 h + nC2 xn−2 h2 + · · · + nCn−1 xhn−1 + hn − xn = nC1 xn−1 h + nC2 xn−2 h2 + · · · + nCn−1 xhn−1 + hn = nxn−1 h + nC2 xn−2 h2 + · · · + nCn−1 xhn−1 + hn f (x + h) − f (x) 1  n−1  and so = nx h + nC2 xn−2 h2 + · · · + nCn−1 xhn−1 + hn h h = nxn−1 + nC2 xn−2 h + · · · + nCn−1 xhn−2 + hn−1 f (x + h) − f (x)   ∴ lim = lim nxn−1 + nC2 xn−2 h + · · · + nCn−1 xhn−2 + hn−1 h→0 h h→0 = nxn−1 Hence f 0 (x) = nxn−1. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17B Rules for differentiation 577 The derivative of a polynomial function The following results are very useful when finding the derivative of a polynomial function.  Constant function: If f (x) = c, then f 0 (x) = 0.  Linear function: If f (x) = mx + c, then f 0 (x) = m.  Multiple: If f (x) = k g(x), where k is a constant, then f 0 (x) = k g0 (x). That is, the derivative of a number multiple is the multiple of the derivative. For example: if f (x) = 5x2 , then f 0 (x) = 5(2x) = 10x.  Sum: If f (x) = g(x) + h(x), then f 0 (x) = g0 (x) + h0 (x). That is, the derivative of the sum is the sum of the derivatives. For example: if f (x) = x2 + 2x, then f 0 (x) = 2x + 2.  Difference: If f (x) = g(x) − h(x), then f 0 (x) = g0 (x) − h0 (x). That is, the derivative of the difference is the difference of the derivatives. For example: if f (x) = x2 − 2x, then f 0 (x) = 2x − 2. You will meet rules for the derivative of products and quotients in Mathematical Methods Units 3 & 4. The process of finding the derivative function is called differentiation. Example 7 Find the derivative of x5 − 2x3 + 2, i.e. differentiate x5 − 2x3 + 2 with respect to x. Solution Explanation Let 5 f (x) = x − 2x + 2 3 We use the following results: Then f 0 (x) = 5x4 − 2(3x2 ) + 0  the derivative of xn is nxn−1 = 5x4 − 6x2  the derivative of a number is 0  the multiple, sum and difference rules. Example 8 Find the derivative of f (x) = 3x3 − 6x2 + 1 and thus find f 0 (1). Solution Let f (x) = 3x3 − 6x2 + 1 Then f 0 (x) = 3(3x2 ) − 6(2x) + 0 = 9x2 − 12x ∴ f 0 (1) = 9 − 12 = −3 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  578 Chapter 17: Differentiation and antidifferentiation of polynomials Using the TI-Nspire For Example 7:  Use menu > Calculus > Derivative and complete as shown. Note: The derivative template can also be accessed from the 2D-template palette t. Alternatively, using shift – will paste the derivative template to the screen. For Example 8:  Assign the function f (x) as shown.  To find the derivative, use menu > Calculus > Derivative.  To find the value of the derivative at x = 1, use menu > Calculus > Derivative at a Point. Using the Casio ClassPad For Example 7:  In M, enter and highlight the expression x5 − 2x3 + 2.  Go to Interactive > Calculation > diff and tap OK. Note: Alternatively, select the derivative template ] from the Math2 keyboard. Enter x5 − 2x3 + 2 in the main box and x in the lower box. Tap EXE. For Example 8:  In M, enter and highlight the expression 3x3 − 6x2 + 1.  Go to Interactive > Define and tap OK.  Go to Interactive > Calculation > diff. Enter f (x) and tap OK.  To find the value of the derivative at x = 1, tap the stylus at the end of the entry line. Select | from the Math3 keyboard and type x = 1. Then tap EXE.  Alternatively, define the derivative as g(x) and find g(1). ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17B Rules for differentiation 579 Finding the gradient of a tangent line We discussed the tangent line at a point on a graph in Section 17A. We recall the following: The tangent line to the graph of the function f at the point (a, f (a)) is defined to be the line through (a, f (a)) with gradient f 0 (a). Example 9 For the curve determined by the rule f (x) = 3x3 − 6x2 + 1, find the gradient of the tangent line to the curve at the point (1, −2). Solution Now f 0 (x) = 9x2 − 12x and so f 0 (1) = 9 − 12 = −3. The gradient of the tangent line at the point (1, −2) is −3. Alternative notations It was mentioned in the introduction to this chapter that the German mathematician Gottfried Leibniz was one of the two people to whom the discovery of calculus is attributed. A form of the notation he introduced is still in use today. Leibniz notation An alternative notation for the derivative is the following: dy dy If y = x3 , then the derivative can be denoted by , and so we write = 3x2. dx dx dy In general, if y is a function of x, then the derivative of y with respect to x is denoted by. dx dz Similarly, if z is a function of t, then the derivative of z with respect to t is denoted. dt Warning: In this notation, the symbol d is not a factor and cannot be cancelled. This notation came about because, in the y eighteenth century, the standard diagram for finding the limiting gradient was labelled Q as shown: δy  δx means a small difference in x  δy means a small difference in y P δx where δ (delta) is the lowercase Greek letter d. x 0 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  580 Chapter 17: Differentiation and antidifferentiation of polynomials Example 10 dy dx 1 3 dz a If y = t2 , find. b If x = t3 + t, find. c If z = x + x2 , find. dt dt 3 dx Solution 1 3 a y = t2 b x = t3 + t c z= x + x2 3 dy dx dz = 2t = 3t2 + 1 = x2 + 2x dt dt dx Example 11 dy dz a For y = (x + 3)2 , find. b For z = (2t − 1)2 (t + 2), find. dx dt x2 + 3x dy c For y = , find. d Differentiate y = 2x3 − 1 with respect to x. x dx Solution a It is first necessary to write y = (x + 3)2 b Expanding: in expanded form: z = (4t2 − 4t + 1)(t + 2) y = x2 + 6x + 9 = 4t3 − 4t2 + t + 8t2 − 8t + 2 dy = 4t3 + 4t2 − 7t + 2 ∴ = 2x + 6 dx dz ∴ = 12t2 + 8t − 7 dt c First simplify: d y = 2x3 − 1 y= x+3 (for x , 0) dy ∴ = 6x2 dy dx ∴ =1 (for x , 0) dx Operator notation d ‘Find the derivative of 2x2 − 4x with respect to x’ can also be written as ‘find (2x2 − 4x)’. dx d
In general: f (x) = f 0 (x). dx Example 12 Find: d d d a (5x − 4x3 ) b (5z2 − 4z) c (6z3 − 4z2 ) dx dz dz Solution d d d a (5x − 4x3 ) b (5z2 − 4z) c (6z3 − 4z2 ) dx dz dz = 5 − 12x2 = 10z − 4 = 18z2 − 8z ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17B Rules for differentiation 581 Example 13 For each of the following curves, find the coordinates of the points on the curve at which the gradient of the tangent line at that point has the given value: a y = x3 , gradient = 8 2 b y = x − 4x + 2, gradient = 0 c y = 4 − x3 , gradient = −6 Solution dy a y = x3 implies = 3x2 dx ∴ 3x2 = 8 r √ 8 2 6 ∴ x=± =± 3 3  2 6 16√6  √  2√6 16√6  The points are , and − ,−. 3 9 3 9 dy b y = x2 − 4x + 2 implies = 2x − 4 dx ∴ 2x − 4 = 0 ∴ x=2 The only point is (2, −2). dy c y = 4 − x3 implies = −3x2 dx ∴ −3x2 = −6 ∴ x2 = 2 √ ∴ x=± 2  1 3  1 3 The points are 2 2 , 4 − 2 2 and −2 2 , 4 + 2 2. Using the TI-Nspire  Assign the function f (x) as shown.  Use menu> Algebra > Zeros and menu > Calculus > Derivative to solve the equation d
f (x) = −6. dx  To find the y-coordinates, calculate the values √ √ of f (− 2) and f ( 2) as shown. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  582 Chapter 17: Differentiation and antidifferentiation of polynomials 17B Using the Casio ClassPad  In M, enter and highlight the expression 4 − x3.  Go to Interactive > Define and tap OK.  In the next entry line, type and highlight f (x).  Go to Interactive > Calculation > diff and tap OK. d
 Type = −6 after f (x). Highlight the equation dx and use Interactive > Equation/Inequality > solve. √ √  Enter f (− 2) and f ( 2) to find the y-coordinates. Note: Alternatively, you can select commands and templates from Math2 and Math3. Summary 17B  For f (x) = xn , f 0 (x) = nxn−1 , where n = 1, 2, 3,...  Constant function: If f (x) = c, then f 0 (x) = 0.  Linear function: If f (x) = mx + c, then f 0 (x) = m.  Multiple: If f (x) = k g(x), where k is a constant, then f 0 (x) = k g0 (x). That is, the derivative of a number multiple is the multiple of the derivative.  Sum: If f (x) = g(x) + h(x), then f 0 (x) = g0 (x) + h0 (x). That is, the derivative of the sum is the sum of the derivatives.  Difference: If f (x) = g(x) − h(x), then f 0 (x) = g0 (x) − h0 (x). That is, the derivative of the difference is the difference of the derivatives. For example, if f (x) = 5x3 − 10x2 + 7, then f 0 (x) = 5(3x2 ) − 10(2x) + 0 = 15x2 − 20x. Exercise 17B Example 7 1 Find the derivative of each of the following with respect to x: a x2 + 4x b 2x + 1 c x3 − x 1 2 d x − 3x + 4 e 5x3 + 3x2 f −x3 + 2x2 2 2 For each of the following, find f 0 (x): a f (x) = x12 b f (x) = 3x7 c f (x) = 5x d f (x) = 5x + 3 e f (x) = 3 f f (x) = 5x2 − 3x 1 3 1 2 g f (x) = 10x5 + 3x4 h f (x) = 2x4 − x − x +2 3 4 Example 8 3 For each of the following, find f 0 (1): a f (x) = x6 b f (x) = 4x5 c f (x) = 5x 2 d f (x) = 5x + 3 e f (x) = 3 f f (x) = 5x2 − 3x 1 3 g f (x) = 10x4 − 3x3 h f (x) = 2x4 − x i f (x) = −10x3 − 2x2 + 2 3 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17B 17B Rules for differentiation 583 4 For each of the following, find f 0 (−2): a f (x) = 5x3 b f (x) = 4x2 c f (x) = 5x3 − 3x d f (x) = −5x4 − 2x2 Example 9 5 Find the gradient of the tangent line to the graph of f at the given point: a f (x) = x2 + 3x,(2, 10) b f (x) = 3x2 − 4x, (1, −1) 2 3 c f (x) = −2x − 4x, (3, −30) d f (x) = x − x, (2, 6) dy dx 1 4 dz Example 10 6 a If y = t2 − 7, find. b If x = −5t3 + t, find. c If z = x − x2 , find. dt dt 2 dx dy Example 11 7 For each of the following, find : dx a y = −x b y = 10 c y = 4x3 − 3x + 2 1 3 d y= (x − 3x + 6) e y = (x + 1)(x + 2) f y = 2x(3x2 − 4) 3 5x − x2 10x5 + 3x4 g y = (2x − 1)2 h y= ,x,0 i y= ,x,0 x 2x2 dy dz 8 a For y = (x + 4)2 , find. b For z = (4t − 1)2 (t + 1), find. dx dt x3 + 3x dy c For y = , find. x dx 9 a For the curve with equation y = x3 + 1, find the gradient of the tangent line at points: i (1, 2) ii (a, a3 + 1) b Find the derivative of x3 + 1 with respect to x. dy dy 10 a Given that y = x3 − 3x2 + 3x, find. Hence show that ≥ 0 for all x, and interpret dx dx 3 2 this in terms of the graph of y = x − 3x + 3x. x2 + 2x dy b Given that y = , for x , 0, find. x dx c Differentiate y = (3x + 1)2 with respect to x. For each of the following curves, find the y-coordinate of the point on the curve with the CAS 11 given x-coordinate, and find the gradient of the tangent line at that point: a y = x2 − 2x + 1, x=2 b y = x2 + x + 1, x=0 2 c y = x − 2x, x = −1 d y = (x + 2)(x − 4), x = 3 e y = 3x2 − 2x3 , x = −2 f y = (4x − 5)2 , x = 12 12 a For each of the following, first find f 0 (x) and f 0 (1). Then, for y = f (x), find the set { (x, y) : f 0 (x) = 1 }. That is, find the coordinates of the points where the gradient of the tangent line is 1. 1 1 i f (x) = 2x2 − x ii f (x) = 1 + x + x2 2 3 3 4 iii f (x) = x + x iv f (x) = x − 31x b What is the interpretation of { (x, y) : f 0 (x) = 1 } in terms of the graphs? ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  584 Chapter 17: Differentiation and antidifferentiation of polynomials 17B Find: CAS Example 12 13 d 2 d d a (3t − 4t) b (4 − x2 + x3 ) c (5 − 2z2 − z4 ) dt dx dz d d d d (3y2 − y3 ) e (2x3 − 4x2 ) f (9.8t2 − 2t) dy dx dt Example 13 14 For each of the following curves, find the coordinates of the points on the curve at which the gradient of the tangent line has the given value: a y = x2 , gradient = 8 b y = x3 , gradient = 12 2 c y = x(2 − x), gradient = 2 d y = x − 3x + 1, gradient = 0 e y = x3 − 6x2 + 4, gradient = −12 f y = x2 − x3 , gradient = −1 17C Differentiating xn where n is a negative integer In the previous section we have seen how to differentiate polynomial functions. In this section we add to the family of functions that we can differentiate. In particular, we will consider functions which involve linear combinations of powers of x, where the indices may be negative integers. e.g. f : R \ {0} → R, f (x) = x−1 f : R \ {0} → R, f (x) = 2x + x−1 f : R \ {0} → R, f (x) = x + 3 + x−2 Note: We have reintroduced function notation to emphasise the need to consider domains. Example 14 1 Let f : R \ {0} → R, f (x) =. Find f 0 (x) by first principles. x Solution The gradient of secant PQ is given by y f (x + h) − f (x)  1 1 1 = − × x+h−x x+h x h P x, 1x x − (x + h) 1 = × 1 (x + h)x h Q x + h, –––– x+h −h 1 x = × 0 (x + h)x h −1 = (x + h)x So the gradient of the curve at P is −1 −1 lim = 2 = −x−2 h→0 (x + h)x x Hence f 0 (x) = −x−2. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17C Differentiating xn where n is a negative integer 585 Example 15 Let f : R \ {0} → R, f (x) = x−3. Find f 0 (x) by first principles. Solution The gradient of secant PQ is given by y (x + h)−3 − x−3 P(x, x–3) h x3 − (x + h)3 1 = × (x + h)3 x3 h Q (x + h, (x + h)–3) x3 − (x3 + 3x2 h + 3xh2 + h3 ) 1 = × x (x + h)3 x3 h 0 2 2 3 −3x h − 3xh − h 1 = 3 3 × (x + h) x h −3x2 − 3xh − h2 = (x + h)3 x3 So the gradient of the curve at P is given by −3x2 − 3xh − h2 −3x2 lim = 6 = −3x−4 h→0 (x + h)3 x3 x Hence f 0 (x) = −3x−4. We are now in a position to state the generalisation of the result we found in Section 17B. This result can be proved by again using the binomial theorem. For f (x) = xn , f 0 (x) = nxn−1 , where n is a non-zero integer. For f (x) = c, f 0 (x) = 0, where c is a constant. When n is positive, we take the domain of f to be R, and when n is negative, we take the domain of f to be R \ {0}. Note: We will consider rational indices in Chapter 20. Example 16 Find the derivative of: 5x3 + x a 2x−3 − x−1 + 2, x , 0 b , x,0 x2 Solution 5x3 + x a Let f (x) = 2x−3 − x−1 + 2 b Let f (x) = x2 Then f 0 (x) = 2(−3x−4 ) − (−x−2 ) + 0 = 5x + x−1 = −6x−4 + x−2 (for x , 0) Then f 0 (x) = 5 + (−x−2 ) 1 = 5 − 2 (for x , 0) x ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  586 Chapter 17: Differentiation and antidifferentiation of polynomials 17C Example 17 Find the derivative f 0 of f : R \ {0} → R, f (x) = 3x2 − 6x−2 + 1. Solution f 0 : R \ {0} → R, f 0 (x) = 3(2x) − 6(−2x−3 ) + 0 = 6x + 12x−3 Example 18 Find the gradient of the tangent to the curve determined by the function f : R \ {0} → R, 1 f (x) = x2 + at the point (1, 2). x Solution f 0 : R \ {0} → R, f 0 (x) = 2x + (−x−2 ) = 2x − x−2 Therefore f 0 (1) = 2 − 1 = 1. The gradient of the curve is 1 at the point (1, 2). Example 19 Show that the derivative of the function f : R \ {0} → R, f (x) = x−3 is always negative. Solution f 0 : R \ {0} → R, f 0 (x) = −3x−4 3 =− 4 x Since x4 is positive for all x , 0, we have f 0 (x) < 0 for all x , 0. Summary 17C For f (x) = xn , f 0 (x) = nxn−1 , where n is a non-zero integer. For f (x) = c, f 0 (x) = 0, where c is a constant. Skill- sheet Exercise 17C 1 Example 14 1 a Let f : R \ {3} → R, f (x) =. Find f 0 (x) by first principles. x−3 1 b Let f : R \ {−2} → R, f (x) =. Find f 0 (x) by first principles. x+2 Example 15 2 a Let f : R \ {0} → R, f (x) = x−2. Find f 0 (x) by first principles. b Let f : R \ {0} → R, f (x) = x−4. Find f 0 (x) by first principles. Hint: (x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17C 17D Graphs of the derivative function 587 Example 16 3 Differentiate each of the following with respect to x: 3 5 4 a 3x−2 + 5x−1 + 6 b 2 + 5x2 c + +1 x x3 x2 5 3x2 + 2 d 3x2 + x−4 + 2 e 6x−2 + 3x f 3 x 4 Find the derivative of each of the following: 3z2 + 2z + 4 3+z 2z2 + 3z a 2 , z,0 b , z,0 c , z,0 z z3 4z 5z − 3z2 d 9z2 + 4z + 6z−3 , z , 0 e 9 − z−2 , z , 0 f , z,0 5z Example 17 5 a Find the derivative f 0 of f : R \ {0} → R, f (x) = 3x4 − 6x−3 + x−1. b Find the derivative f 0 of f : R \ {0} → R, f (x) = 5x4 + 4x−2 + x−1. 1 Carefully sketch the graph of f (x) = , x , 0. CAS 6 x2
a For points P(1, f (1)) and Q 1 + h, f (1 + h) , find the gradient of the secant PQ. 1 b Hence find the gradient of the tangent line to the curve y = 2 at x = 1. x Example 18 7 For each of the following curves, find the gradient of the tangent line to the curve at the given point: x−2 a y = x−2 + x3 , x , 0, at (2, 8 41 ) b y= , x , 0, at (4, 12 ) x 1 c y = x−2 − , x , 0, at (1, 0) d y = x(x−1 + x2 − x−3 ), x , 0, at (1, 1) x 8 For the curve with equation f (x) = x−2 , find the x-coordinate of the point on the curve at which the gradient of the tangent line is: a 16 b −16 Example 19 9 Show that the derivative of the function f : R \ {0} → R, f (x) = x−1 is always negative. 17D Graphs of the derivative function Sign of the derivative y Consider the graph of y = f (x) shown R(a, f (a)) here. At a point (x, f (x)) on the graph, y = f (x) the gradient is f 0 (x). By noting whether the curve is sloping b upwards or downwards at a particular x a 0 point, we can tell the sign of the S(b, f (b)) derivative at that point: Values of x x 0 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  588 Chapter 17: Differentiation and antidifferentiation of polynomials Example 20 y For the graph of f : R → R, find: (5, 6) 0 a { x : f (x) > 0 } b { x : f 0 (x) < 0 } c { x : f 0 (x) = 0 } x 0 y = f (x) (–1, –7) Solution a { x : f 0 (x) > 0 } = { x : −1 < x < 5 } = (−1, 5) b { x : f 0 (x) < 0 } = { x : x < −1 } ∪ { x : x > 5 } = (−∞, −1) ∪ (5, ∞) c { x : f 0 (x) = 0 } = {−1, 5} Example 21 Sketch the graph of y = f 0 (x) for each of the following. (It is impossible to determine all features.) a y b y c y y = f (x) y = f(x) y = f (x) (–1.5, 16.25) 1 x x x 0 2 4 –1 0 0 (3, –1) (1, –15) Solution a f 0 (x) > 0 for x > 3 b f 0 (x) = 1 for all x c f 0 (x) > 0 for x > 1 f 0 (x) < 0 for x < 3 f 0 (x) < 0 for −1.5 < x < 1 f 0 (x) = 0 for x = 3 f 0 (x) > 0 for x < −1.5 f 0 (−1.5) = 0 and f 0 (1) = 0 y y y y = f ′(x) y = f'(x) 1 y = f ′(x) x x 0 3 x 0 −1.5 0 1 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17D Graphs of the derivative function 589 If the rule for the function is given, then a CAS calculator can be used to plot the graph of its derivative function. Using the TI-Nspire Plot the graphs of f 1(x) = x3 − 2x2 + x − 1 d
f 2(x) = f 1(x) dx Note: Access the derivative template from t. Increasing and decreasing functions We say a function f is strictly increasing on an interval y if a < b implies f (a) < f (b). For example:  The graph opposite shows a strictly increasing function.  A straight line with positive gradient is strictly increasing. x 0 2  The function f : [0, ∞) → R, f (x) = x is strictly increasing. We say a function f is strictly decreasing on an interval y if a < b implies f (a) > f (b). For example:  The graph opposite shows a strictly decreasing function.  A straight line with negative gradient is strictly decreasing. x 0 2  The function f : (−∞, 0] → R, f (x) = x is strictly decreasing. Note: The word strictly refers to the use of the strict inequality signs rather than ≤, ≥. Derivative tests We can sometimes establish that a function is strictly increasing or strictly decreasing on an interval by checking the sign of its derivative on that interval.  If f 0 (x) > 0, for all x in the interval, then the function is strictly increasing. (Think of the tangents at each point – they each have positive gradient.)  If f 0 (x) < 0, for all x in the interval, then the function is strictly decreasing. (Think of the tangents at each point – they each have negative gradient.) Warning: These two tests are not equivalent to the definitions of strictly increasing and strictly decreasing. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  590 Chapter 17: Differentiation and antidifferentiation of polynomials For example, the function f : R → R, f (x) = x3 is strictly increasing, but f 0 (0) = 0. This means that strictly increasing does not imply f 0 (x) > 0. We can see that f (x) = x3 is strictly increasing from its graph. y Alternatively, consider a3 − b3 = (a − b)(a2 + ab + b2 ) 1

1
2  = a − b a2 + ab + 12 b 2 + b2 − 2b

 x = a − b a + 21 b 2 + 43 b2 –1 0 1 We can see that if a < b, then a3 < b3. –1 An angle associated with the gradient of a curve at a point The gradient of a curve at a point is the gradient of the tangent at that point. A straight line, the tangent, is associated with each point on the curve. If θ is the angle that a straight line makes with the positive direction of the x-axis, then the gradient, m, of the straight line is equal to tan θ. That is: m = tan θ For example:  If θ = 45◦ , then tan θ = 1 and the gradient is 1.  If θ = 20◦ , then the gradient of the straight line is tan 20◦.  If θ = 135◦ , then tan θ = −1 and the gradient is −1. Example 22 Find the coordinates of the points on the curve with equation y = x2 − 7x + 8 at which the tangent line: a makes an angle of 45◦ with the positive direction of the x-axis b is parallel to the line y = −2x + 6. Solution dy a = 2x − 7 b The line y = −2x + 6 has gradient −2. dx 2x − 7 = 1 (as tan 45◦ = 1) 2x − 7 = −2 2x = 8 2x = 5 x=4 5 ∴ ∴ x= 2 y = 42 − 7 × 4 + 8 = −4 5 13  The coordinates are (4, −4). The coordinates are ,−. 2 4 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17D Graphs of the derivative function 591 Example 23 The planned path for a flying saucer leaving a planet is defined by the equation 1 2 y = x4 + x3 for x > 0 4 3 The units are kilometres. (The x-axis is horizontal and the y-axis vertical.) a Find the direction of motion when the x-value is: i 2 ii 3 b Find a point on the flying saucer’s path where the path is inclined at 45◦ to the positive x-axis (i.e. where the gradient of the path is 1). c Are there any other points on the path which satisfy the situation described in part b? Solution dy a = x3 + 2x2 dx dy dy i When x = 2, = 8 + 8 = 16 ii When x = 3, = 27 + 18 = 45 dx dx tan−1 16 = 86.42◦ (to the x-axis) tan−1 45 = 88.73◦ (to the x-axis) b, c When the flying saucer is flying at 45◦ to the positive direction of the x-axis, the gradient of the curve of its path is given by tan 45◦. Thus to find the point at which this happens we consider the equation dy = tan 45◦ dx x3 + 2x2 = 1 x3 + 2x2 − 1 = 0 (x + 1)(x2 + x − 1) = 0 √ −1 ± 5 ∴ x = −1 or x = 2 √ −1 + 5 The only acceptable solution is x = ≈ 0.62, as the other two possibilities 2 give negative values for x and we are only considering positive values for x. Summary 17D  A function f is strictly increasing on an interval if a < b implies f (a) < f (b).  A function f is strictly decreasing on an interval if a < b implies f (a) > f (b).  If f 0 (x) > 0 for all x in the interval, then the function is strictly increasing.  If f 0 (x) < 0 for all x in the interval, then the function is strictly decreasing. ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  592 Chapter 17: Differentiation and antidifferentiation of polynomials 17D Skill- sheet Exercise 17D dy 1 For which of the following curves is positive for all values of x? dx a y b y c y x x x 0 0 0 d y e y x x 0 0 dy 2 For which of the following curves is negative for all values of x? dx a y b y c y x 0 x x 0 0 d y e y f y x 0 x x 0 0 3 For the function f (x) = 2(x − 1)2 , find the values of x for which: a f (x) = 0 b f 0 (x) = 0 c f 0 (x) > 0 d f 0 (x) < 0 e f 0 (x) = −2 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17D 17D Graphs of the derivative function 593 Example 20 4 For each of the following, use the given graph of y = f (x) to find the sets: i { x : f 0 (x) > 0 } ii { x : f 0 (x) < 0 } iii { x : f 0 (x) = 0 } a y b y –3, 5 12 (4, 6) (1.5, 3) x 0 x 0 (–1, –2) 1 , –3 2 5 Which of the graphs labelled A–F correspond to each of the graphs labelled a–f? a y b y c y x x x 0 0 0 d y e y f y x 0 x x 0 0 A dy B dy C dy dx dx dx 0 x 0 x 0 x D dy E dy F dy dx dx dx 0 x x 0 x 0 ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  594 Chapter 17: Differentiation and antidifferentiation of polynomials 17D Example 21 6 Sketch the graph of y = f 0 (x) for each of the following: a y b y c y d y (3, 4) (3, 4) y = f (x) 3 y = f (x) 1 x x x x 0 1 5 0 1 0 0 (–1, –3) Example 22 7 Find the coordinates of the points on the curve y = x2 − 5x + 6 at which the tangent: a makes an angle of 45◦ with the positive direction of the x-axis b is parallel to the line y = 3x + 4. 8 Find the coordinates of the points on the parabola y = x2 − x − 6 at which: a the gradient of the tangent is zero b the tangent is parallel to the line x + y = 6. Use a calculator to plot the graph of y = f 0 (x) where: CAS 9 a f (x) = sin x b f (x) = cos x c f (x) = 2 x 1 2 Example 23 10 The path of a particle is defined by the equation y = x3 + x2 , for x > 0. The units are 3 3 metres. (The x-axis is horizontal and the y-axis vertical.) a Find the direction of motion when the x-value is: i 1 ii 0.5 b Find a point on the particle’s path where the path is inclined at 45◦ to the positive direction of the x-axis. c Are there any other points on the path which satisfy the situation described in part b? 11 A car moves away from a set of traffic lights so that the distance, S (t) metres, covered after t seconds is modelled by S (t) = 0.2 × t3. a Find its speed after t seconds. b What will its speed be when t = 1, 3, 5? 12 The curve with equation y = ax2 + bx has a gradient of 3 at the point (2, −2). a Find the values of a and b. b Find the coordinates of the point where the gradient is 0. 13 A rocket is launched from Cape York Peninsula so that after t seconds its height, h(t) metres, is given by h(t) = 20t2 , 0 ≤ t ≤ 150. After 2 12 minutes this model is no longer appropriate. a Find the height and the speed of the rocket when t = 150. b After how long will its speed be 1000 m/s? ISBN 978-1-009-11448-6 © Michael Evans et al 2022 Cambridge University Press Photocopying is restricted under law and this material must not be transferred to another party.  17E Antidifferentiation of polynomial functions 595 17E Antidifferentiation of polynomial functions The derivative of x2 with respect to x is 2x. Conversely, given that an unknown expression has derivative 2x, it is clear that the unknown expression could be x2. The process of finding a function from its derivative is called antidifferentiation. Now consider the functions f (x) = x2 + 1 and g(x) = x2 − 7. We have f 0 (x) = 2x and g0 (x) = 2x. So the two different functions have the same derivative function. Both x2 + 1 and x2 − 7 are said to be y antiderivatives of 2x. y = x2 + 1 y = x2 If two functions have the same derivative y = x2 – 1 function, then they differ by a constant. So the graphs of the two functions can be obtained from each other by translation 1 y = x2 – 7 parallel to the y-axis.