14 Further Calculus 14 FURTHER CALCULUS
Chapter 14 Further Calculus 14 FURTHER CALCULUS
Objectives
After studying this chapter you should • be able to find the second derivatives of functions; • be able to use calculus to find maximum or minimum of functions; • be able to differentiate composite functions or 'function of a function'; • be able to differentiate the product or quotient of two functions; • be able to use calculus in curve sketching.
14.0 Introduction
You should have already covered the material in Chapters 8, 11 and 12 before starting this chapter. By now, you will already have met the ideas of calculus, both differentiation and integration, and you will have used the techniques developed in the earlier chapters to solve problems. This section extends the range of problems you can solve, including finding the greatest or least value of a function and differentiating complicated functions.
y 14.1 Rate of change of the y = x 3 + 3x 2 − 9x − 4 gradient
The sketch opposite shows the graph of
3 2 0 x y = x + 3x − 9x − 4 .
You have already seen that the derivative of this function is dy given by dx
dy = 2 + − dy 3x 6x 9. = 3x 2 + 6x − 9 dx dx 0 x This is also illustrated opposite.
265 Chapter 14 Further Calculus
dy The gradient function, , is also a function of x, and can be d 2 y
dx 2 differentiated again to give the second differential dx d 2 y = 6x + 6 dx 2 2 d y 0 x = 6x + 6 . dx 2
Again this is illustrated opposite.
Exercise 14A d 3 y 1. Find the second derivative of these functions : 2. Find if y is dx 3 3 4 2 (a) y = x (b) y = x (c) y = x 3 4 2 + x (a) x (b) 5x 2 (c) e 1 3 2 x (d) y = x (e) y = (f) y = 4x −12x + 5 x (d) ln x (e) 2x x (g) y = 3x +1 (h) y = e (i) y = ln x
14.2 Stationary points
In this section you will consider the curve with equation y 3 2 y = 2x + 3x −12x.
This is cubic, and a rough sketch of its graph is shown opposite. It has two stationary points (sometimes called turning points) at which the gradient is zero. 0 x
How can you find the coordinates of the stationary points? 3 2 y = 2x + 3x −12x
Activity 1
Find the coordinates of the stationary points for the curve with equation
3 2 y = 2x + 3x −12x.
In Chapter 8 you saw that the nature of the stationary points can be determined by looking at the gradient on each side of the stationary point. Here an alternative more formal method is developed, based on using second derivatives.
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Activity 2
Draw an accurate sketch of the curve with equation
3 2 y = 2x + 3x −12x between x =−3 and + 2. Choose the y axis to show values between −10 and + 20. y 12.96 18.24 For every x value –3, –2.8, –2.6, –2.4, ..., 2, note the gradient on the diagram. 24 10
Plot a graph of the gradient function and note how it behaves near the stationary point of the function. -3 -2 -1 012x
For a maximum value of a function, note that the gradient is decreasing in value as it passes through the value zero at the stationary point, whereas for a minimum value this gradient is increasing.
The result can be summarised as
For stationary points of a function y(x)
dy = 0 . dx
d 2 y If < 0 at a stationary point, it corresponds to a dx 2 maximum value of y.
d 2 y If > 0 at a stationary point, it corresponds to a dx 2 minimum value of y.
Example Find maxima and minima of the curve with equation
1 1 y = x 4 + x3 − 6x2 + 3. 4 3
Hence sketch the curve.
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Solution dy For stationary points, = 0 , which gives dx dy = x3 + x2 −12x . dx
dy So = 0 dx
⇒ x 3 + x2 −12x = 0
2 x (x + x −12) = 0
x (x + 4)(x − 3) = 0.
Hence there are stationary points at x = 0, − 4 and 3. To find out their nature, the second derivative is used. Now
d 2 y = 3x2 + 2x −12. dx 2
d2y At x = 0 , =−12 < 0 dx 2 ⇒ a maximum at x = 0 of value y = 3.
d2y At x =− 4, = 3(−4)2 + 2(−4) − 12 = 28 > 0 dx 2 ⇒ minimum at x =− 4 of value 1 1 151 y = (− 4)4 + (−4)3 − 6(−4)2 + 3 =− . 4 3 3
2 d y 2 At x = 3, = 3(3) + 2(3) − 12 = 21 > 0 dx 2 ⇒ minimum at x = 3 of value 1 1 87 y = (3)4 + (3)3 − 6(3)2 + 3 =− . 4 3 4 y The information so far found can be sketched on a graph (not to scale). You also know that as x →±∞,y→+∞. It is now clear -4–4 3 how to sketch the shape - shown dashed on the diagram. 0 x
There is one further type of stationary point to be considered and that is a point of inflection. An example is given in the next activity.
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Activity 3
3 Find the stationary point of y = x . d 2 y What is the value of at the stationary point? dx 2 Sketch the graph of y = x3. dy > 0 dx dy dy = dy > 0 0 For a horizontal point of inflection, not only does = 0, but dx dx dx 2 3 dy d y d y < 0 also = 0, and ≠ 0, at the point. These are sufficient but dx dx 2 dx 3 = 5 dy not necessary conditions, as can be seen by considering y x . = 0 dx dy < 0 dx Activity 4
5 Find the stationary point of y = x . d 2 y d 3y What is the value of and at its stationary point? dx 2 dx3 5 Sketch the graph of y = x .
Example Find the nature of the stationary points of the curve with equation
4 3 y = x + 4x − 6
Sketch a graph of the curve.
Solution Now dy = 4x3 +12x2 dx
= 0 when 4x3 +12x2 = 0
⇒ 4x2 (x + 3) = 0
⇒ x = 0, − 3 for stationary points
d 2 y But = 12x2 + 24x dx 2
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d 2 y d 3y At x = 0, = 0, but = 24x + 24 = 24 > 0 at x = 0 . dx2 dx3
So there is a point of inflection at (0, – 6).
d 2 y At x =−3, = 12(−3)2 + 24(−3) dx2 = 108 − 72 = 36 > 0
So there is a minimum at x =−3 of value
y(−3) = (−3)4 + 4(−3)3 − 6 = 81− 4 × 27 − 6 =−33 y To sketch the curve, also note that
y →∞as x →±∞. x (0, –6) 6) You can then deduce the form of curve, shown dashed opposite.
Finally, in this section, it should be noted that the conditions (-3, -33) given earlier are sufficient conditions to guarantee max/min values, but not necessary.
Activity 5
4 Draw the graph of y = x . Does it have any maximum or minimum values?
Do the conditions hold?
By now you are probably getting confused since there is a rule to determine the nature of stationary points, yet not all functions satisfy it. So let it be stressed that dy (a) Stationary points are always given by = 0 . dx
d 2 y (b) If, at the stationary point, > 0, there is a minimum, dx 2 d 2 y whereas if < 0, there is a maximum. dx 2
270 Chapter 14 Further Calculus
d 2 y (c) If, at the stationary point, = 0, then there is a point of dx 2 d 3y inflection provided ≠ 0. dx 3 Conditions (b) and (c) are sufficient to guarantee the nature of the stationary point, but, as you have already seen not necessary. When this analysis does not hold, that is when d 2 y d 3y = = 0 at a stationary point, it is easier to consider the dx 2 dx3 sign of the gradient each side of the stationary point.
y ′ > 0 y ′ < 0 y ′ = 0 y ′ < 0 y ′ > 0 y ′ = 0 y ′ = 0 y ′ > 0 y ′ < 0 ′ y ′ > 0 y < 0 y ′ = 0 gradients < M 0 M > > M 0 M < > M 0 M > < M 0 M < minimum maximum point of inflection point of inflection
Example
4 Show that y = x has a minimum at x = 0 .
Solution dy For stationary points = 4x3 dx = 0 ⇒ x = 0
d 2 y But, if x = 0, = 12x2 = 0 dx2
d 3y and = 24x = 0 dx3 So you cannot use the usual results, but at, for example,
dy x =−0.1 ⇒ = 4(− 0.1)3 =−0.004 < 0 dx dy x = 0.1 ⇒ = 4(0.1)3 = 0.004 > 0 dx
So there is a minimum at x = 0 .
271 Chapter 14 Further Calculus Exercise 14B
1. Find the turning points of these curves using 2. Investigate the nature of the stationary points of differentiation. In each case, find out whether the curve the points found are maxima, minima or points of 2 inflection. y = (x −1)( x − a) > < < = 2 + − = 3 − + when (a) a 1 (b) 0 a 1. In each case give a (a) y 2x 3x 1 (b) y x 12x 6 sketch of the curve.
3 2 2 (c) y = x + 2x − 5x − 6 (d) y = x − 4x + 2ln(x)
x x (e) y = e − 4x (f) y = e + 5
14.3 Differentiating composite functions
The crucial result needed here is the ‘function of a function’ rule. It can be justified by noting that if y is a function of u, and u is a function of x i.e. y = f (u) and u = g(x) then a small change, δ x say, in x will result in a small change, say δ u in u, which in turn results in a small change, δ y in y. Now
δy δy δu = × δ x δu δx and if you let δ x → 0, then δu → 0 and δy → 0, and the equation becomes
dy dy du = × dx du dx For example, suppose
2 y = (3x + 2) then one way to differentiate this function is to multiply out the brackets and differentiate term by term;
y = (3x + 2)(3x + 2)
= 9x2 +12x + 4
dy and = 18x +12. dx
272 Chapter 14 Further Calculus
Another method, which will be even more useful as the functions get more complicated, is to introduce a new variable u defined by
u = 3x + 2
2 so that y = u .
dy du Now = 2u and = 3, so using the result above du dx
dy = (2u) × 3 = 6(3x+2), as before. dx
Activity 6
Use both methods described above to differentiate the functions:
2 3 (a) y = (5x −1) (b) y = (3 − 2x) .
Example
2 dy If y = ex , find . dx
Solution As before, introduce a new variable u defined by u = x2 so that u y = e . dy du Now = eu and = 2x du dx
dy dy du 2 giving = × = (eu )2x = 2xex . dx du dx
2 xex dx What is the value of ∫ ?
Another important result needed is given by
dy dx = 1/ dx dy but note that this is only true for first derivatives.
273 Chapter 14 Further Calculus
2 For example, if y = x
dy then = 2x dx
But, expressing x as a function of y,
1 x = y 2
dx 1 − 1 1 1 = 2 = = and y 1 (generalising the result that dy 2 2y 2 2x d − ()xn =nxn 1 for any n) dx dy dx So, as expected = 1/ dx dy Activity 7
dy dx Verify the result = 1/ when dx dy
= 3 = 1 + (a) y x (b) y 2 (x 3)
The proof of the result is based on using small increments (increase) δ x and δy, noting that
δy δx = 1/ δ x δy
and taking the limit as δ x → 0 (and δy → 0).
This section is concluded by using this result in finding the derivative of the function x y = a To achieve this, you must first take ‘logs’ of both sides to give
x ln y = ln(a ) = x ln a (properties of logs)
du Defining u = (lna)x ⇒ = ln a dx and u = ln y
274 Chapter 14 Further Calculus
du 1 Hence = dy y
dy dy du giving = dx du dx du = 1/ ln a dy = y ln a d (ax ) = (lna)ax dx so, for example d (2x ) = (ln2)2x dx
What happens if a = ein the formula above? Exercise 14C
1. Differentiate these functions : 3. A kettle contains water which is cooling −0.04t 8 according to the equation W = 80e + 20 (a) y = (2x −5) (u = 2x − 5) where W is the temperature of the water in ° C at 2 3 10 2 3 (b) y = (x + x ) (u = x + x ) time t (in minutes) after the kettle was switched off. Find the rate at which the water is cooling in 1 −1 ° = = (c) y = (u = x − 2, y = u ) C /min when t 30 and when t 60 . − x 2 4. The population of a country is modelled by the function 1 − = = + = 2 (d) y 2 (u 3x 1, y u ) t (3x +1) P = 12 × (1.03)
1 where P is the population in millions and t is the (e) y = x +1 (u = x +1, y = u 2 ) time in years after the start of 1990. Find the rate at which the population is increasing in x 6 x (f) y = (e − x) (u = e − x) millions per year at the start of the year 2000.
(g) y = ln(3x + 4) (u = 3x + 4) 5. (a) Differentiate e 2x . Hence write down e2x dx. ∫ x − − (h) y = e (u =− x) (b) Differentiate e x , and so write down e x dx . ∫
2. Find any turning points on these curves : 2 6. Differentiate e x , and use your result to find x2 (a) y = e x2 xe dx . ∫ −()x+4 2 2 (b) y = e (u =−(x+4) ) 1 7. The derivative of ln x is . x = 2 = (c) y (ln(x)) (u ln(x)) By differentiating a suitable logarithmic 1 1 4 1 function, find ∫ dx . (d) y = x + u = x + x + 2 x x
275 Chapter 14 Further Calculus 14.4 Integration again
The results that have been developed in the last section are, as you will see, very useful in integration. For example, if
4 y = (x + 5)
dy then = 4(x + 5)3 (using the methods in the last section). dx
Hence 4(x + 5)3 dx = (x + 5)4 + C (C is arbitrary constant) ∫ 1 1 or ∫ (x + 5)3 dx = (x + 5)4 + C′ (C′ = C). 4 4
Activity 8
n+1 Differentiate y = (ax + b) and hence deduce the value of
(ax + b)n dx . ∫ Use your results to find:
1 − 9 − (a) ∫ (3x 1) dx (b) ∫ 2 dx (c) ∫ x 1 dx . (2x + 4)
Similarly, it can be seen that if
y = ln(ax + b) dy a then = . dx (ax + b) ⌠ adx Hence = ln(ax + b) + C ⌡ (ax + b)
⌠ dx 1 1 or = ln (ax + b) + C′ C′ = C ⌡ (ax + b) a a
Activity 9
Use the result above to evaluate:
⌠ dx ⌠ dx ⌠ dx (a) (b) (c) . ⌡ (x +1) ⌡ 4 − x ⌡ 1− 2x
276 Chapter 14 Further Calculus
Another important method of integration is based on the result that if y = ln f (x) then defining u = f (x), y = ln u and
dy dy du = × dx du dx 1 = × f ′(x) u f ′(x) = f (x)
⌠ f ′(x) Hence dx = ln f (x) + C ⌡ f (x)
Example
⌠ (3x2 + x) Find 3 2 dx ⌡ (2x + x )
Solution
3 2 ′ 2 If f (x) = 2x + x , then f (x) = 6x + 2x, so the integral can be written as
⌠ (3x2 + x) I = dx ⌡ (2x3 + x2 ) 1 ⌠ 2(3x2 + x) = dx 2 ⌡ (2x3 + x2 ) ′ 1 ⌠ f (x) 3 2 = dx (when f (x) = 2x + x ) 2 ⌡ f (x) 1 = ∫ ln f (x) + C 2 1 = ln(6x2 + 2x) + C 2
277 Chapter 14 Further Calculus Exercise 14D
1. Find 3. Find
4 ⌠ 1 ⌠ dx ⌠ x (a) (2x +1) dx (b) dx (a) (b) dx ∫ ⌡ − 2 ⌡ 2 (x 5) ⌡ (3x + 2) x +1
⌠ 1 x (c) dx (d) ∫ 4x −1 dx ⌠ e ⌡ x +1 (c) x dx ⌡ 1+ e 2. Evaluate 1 2 −x ⌠1 dx ⌠ 2 dx 4. Evaluate ∫ xe dx (a) (b) 0 + − ⌡ 0 (x 1) ⌡ 1 (2x 1)
14.5 Differentiating products and quotients
These are very useful formulae for differentiating both products and quotients. For example, if y is defined as the product of the functions, u and v of x, then
y (x) = u(x) v (x)
Using the basic definition of a derivative, let δ x be a small change in x, then consider y(x + δx) − y(x) = u(x + δx) v (x + δx) − u(x) v (x) = []u(x + δx) − u(x) v(x+δx) +u(x)[]v(x+δx)−v(x) (the middle two terms cancel)
Diving both sides by δ x gives y(x + δx) − y(x) (u(x + δx) − u(x)) (v(x + δx) − v(x)) = v (x + δx) + u(x) δx δx δx and taking the limit as δ x → 0, gives
d du dv ()uv = v + u dx dx dx
(since v (x + δx) → v (x) as δx → 0)
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Example
2 2 dy If y = x (x −1) , find by dx (a) using the formula above; (b) multiplying out and differentiating term by term.
Solution 2 2 (a) Here u = x , v = (x −1) ,
dy so = 2x (x −1)2 + x2 2(x−1) dx
= 2x (x −1)( x − 1 + x)
= 2x (x −1) (2 x − 1) .
2 2 4 3 2 (b) y = x (x − 2x +1) = x − 2x + x .
dy = 4x3 − 6x2 + 2x dx
2 = 2x (2x − 3x +1)
= 2x (x −1) (2 x − 1) (as before).
Example
− x dy If y = (x +1) e , find . dx
Solution
− x This time, you must use the formula, with u = x +1, v = e .
dy − − So = 1 × ()e x +(x+1)()−e x dx
−x =−xe .
Turning to the equivalent formula for a quotient, with
u 1 y = = u v v
1 which is the product of u and . v 279 Chapter 14 Further Calculus
Now d 1 d 1 dv = ('function of a function') dx v dv v dx
=− 1 dv v2 dx
dy du 1 d 1 and = × + u dx dx v dx v
= du × 1 + − 1 dv u 2 . dx v v dx
du dv v − u d u dx dx i.e. = dx v v2
Example
d − Use the quotient formula to find ((1+ x)e x ) . dx
Solution
x Here u = (1 + x), v = e , so that
u (1 + x) − = = = + x y x (1 x)e . v (e )
dy ex ×1− (1 + x)e x d Thus = , since (ex ) = ex dx (e x )2 dx x =− xe (ex )2 =− x ex − =−xe x (as in the previous example).
So you can see that the quotient formula is just another form of the product formula. It is though sometimes very convenient to use.
280 Chapter 14 Further Calculus
Example x −1 Differentiate y = with respect to x. 2x − 3
Solution
Here u = x −1, v = 2 x − 3 and using the quotient formula × − − − dy = 1 (2x 3) 2(x 1) 2 dx (2x − 3)
− − + = 2x 3 2x 2 2 (2x − 3)
−1 = . 2 (2x − 3)
*Activity 10
Develop a formula for differentiating the product of three functions . i.e. y (x) = u(x) v(x) w(x) .
Example Find any stationary points for the curve with equation
x3 y = . (1 + x)
Sketch the curve.
Solution u Here u = x3 , v = 1+ x , so y = and using the quotient v formula
2 + − 3 × dy = 3x (1 x) x 1 2 dx (1 + x)
3x2 + 2x3 = . 2 (1 + x)
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For stationary points,
3x2 + 2x3 = 0
x2 (3 + 2x) = 0
⇒ = − 3 x 0, 2 To determine the nature of these stationary points, you can dy check the sign of each side of the points. dx
So, for x = 0
dy 3(−0.1)2 + 2(−0.1)3 0.03 − 0.002 (−0.1) = = dx (1 − 0.1)2 (0.9)2 0.028 = > 0 0.81
dy 3(0.1)2 + 2(0.1)3 0.03 + 0.002 andand (0.1) = = dx (1 + 0.1)2 (1.1)2 0.032 = > 0 1. 21
So there is a point of inflection at x = 0 and value of function here is y = 0.
For x =−1. 5,
dy 3(−1. 6) 2 + 2(−1. 6) 3 −0.512 (−1. 6) = = < 0 dx (1 − 1. 6) 2 0.36
dy 3(−1. 4) 2 + 2(−1. 4) 3 0.392 − = = > ( 1. 4) 2 0 dx (1 − 1. 4) 0.16
27 Hence there is a minimum at x =−1.5 of value y = . 4 y Also note that the function has an asymptote at x =−1, that is
y →±∞as x →−1. 3 27 (− , ) 2 4 3 3 x x -1 0 x As x →±∞,y = ≈ = x2 + 1 x x asymptote x = -1 All this information is marked on the diagram opposite, and the curve is sketched in dotted.
282 Chapter 14 Further Calculus Exercise 14E ln(x − 2) 1. Differentiate the following functions with 6. Differentiate respect to x : x
x 2 7. Sketch the curve with equation (a) y = xe (b) y = x ln(x) 3 3 2 x y = x + x − 6x 2x +1 e 2 (c) y = (d) y = x 2 − 3 1+ x by first finding all stationary points and their nature. 2. Find any stationary points on the curves in Question 1. 8. Sketch the curve with equation
3. Find any stationary points on the curve − 1 1 x2 y = e ln( x +1) x 2 y = e x 9. Find any stationary points of the curve with equation x +1 4. Differentiate x 2 2x −1 y = 1− x x 5. Differentiate e (1 + x) Hence sketch the curve, indicating any asymptotes.
14.6 Miscellaneous Exercises − − 1. Find the maxima and/or minima of these 3. Differentiate e 3x 1. Hence write down ∫ e3x 1dx functions : = 2 − + = − 2 + 4. Find (2x − 4)5 dx . (a) y x 5x 6 (b) y 3x 2x 8 ∫
(c) y = x 3 + 2x 2 + x − 4 (d) y = 6x 2 − 2x 3 + 48x 6 1 5. Calculate ∫ dx . 3 3x −1 e x = 1 x (e) y 2 + x 6. Differentiate ln( x 1). Hence find ∫ 2 dx . 0 x +1 2. Differentiate these functions with respect to x
8 1 x (a) (5 −3x) (b) (c) 3 ( + ) x 1
− 1+ x x −x 2 (d) (e) ln(2 ) (f) e 1− x
283 Chapter 14 Further Calculus
284