Maths with Maple — Week 5 tutorial

Differentiation 1 Solution 1. The quotient rule gives  2  d 2x ln(x) − x2 .x−1 x 2x. ln(x) − x2 . ln0 (x) = = 2 dx ln(x) ln(x) ln(x)2 2x x = . − ln(x) ln(x)2 (Note here that ln(x)2 is not the same as ln(x2 ) = 2 ln(x). For example, ln(e) = 1, so ln(e)2 = 12 = 1, but ln(e2 ) = 2.) Solution 2. Put u = ax + b and v = cx + d, so u0 = a and v 0 = c. We then have y 0 = (u/v)0 = (u0 v − uv 0 )/v 2 = ((acx + ad) − (acx + bc))/(cx + d)2 =

ad − bc . (cx + d)2

Solution 3. Put u = ax + b/x and v = cx + d/x and y = u/v; we must find y 0 . Note that u0 = a − b/x2 v 0 = c − d/x2 u0 v − uv 0 = (a − b/x2 )(cx + d/x) − (ax + b/x)(c − d/x2 ) = acx + ad/x − bc/x − bd/x3 − acx + ad/x − bc/x + bd/x3 = 2(ad − bc)/x, y0 =

u0 v − uv 0 2(ad − bc) = . v2 x(cx + d/x)2

Solution 4. First, recall that cos0 (t) = − sin(t), so the quotient rule, we get

d dt (1

+ a cos(t)) = −a sin(t). Using this and

dx (− sin(t))(1 + a cos(t)) − cos(t)(−a sin(t)) − sin(t) − a sin(t) cos(t) + a sin(t) cos(t) = = dt (1 + a cos(t))2 (1 + a cos(t))2 sin(t) =− . (1 + a cos(t))2 Next, we have cos(t)2 + sin(t)2 1 = , (1 + a cos(t))2 (1 + a cos(t))2 = (1 + a cos(t))2 . Multiplying these two results together, we get x2 + y 2 =

so (x2 + y 2 )−1

(x2 + y 2 )−1

dx = − sin(t). dt

Solution 5. We use the product rule, and the fact that

4 d dx ((x

d dx ((x 3

2 d dx ((x 3 2

d x dx e

= ex . This gives

− 2x + 2)ex ) = (2x − 2)ex + (x2 − 2x + 2)ex = x2 ex

− 3x + 6x − 6)ex ) = (3x2 − 6x + 6)ex + (x3 − 3x2 + 6x − 6)ex = x3 ex

− 4x + 12x2 − 24x + 24)ex ) = (4x3 − 12x2 + 24x − 24)ex + (x4 − 4x3 + 12x2 − 24x + 24)ex = x4 ex .

The next thing in the sequence is 5 d dx ((x

− 5x4 + 20x3 − 60x2 + 120x − 120)ex ) = x5 ex . 1

2

The coefficients here are 1, −5, (−5) × (−4) = 20, (−5) × (−4) × (−3) = −60, (−5) × (−4) × (−3) × (−2) = 120 and (−5) × (−4) × (−3) × (−2) × (−1) = −120. Solution 6. The most efficient method is as follows: y 0 /y = ln(y)0 = (ln(p) + ln(q) − ln(r) − ln(s))0 = p0 /p + q 0 /q − r0 /r − s0 /s. Alternatively, we have (pq)0 rs − pq(rs)0 (rs)2 p0 qrs + pq 0 rs − pqr0 s − pqrs0 = , r2 s2

y0 =

so y0 p0 qrs + pq 0 rs − pqr0 s − pqrs0 rs = y r2 s2 pq 0 0 0 0 p qrs + pq rs − pqr s − pqrs = pqrs 0 0 = p /p + q /q − r0 /r − s0 /s. Now take p = x, q = x + 3, r = x + 1 and s = x + 2, so y = x(x + 3)/((x + 1)(x + 2)). We then have p0 = q 0 = r0 = s0 = 1, so 1 1 1 1 y= − − + x x+1 x+2 x+3 1 1 (x + 2)(x + 3) − x(x + 1) = − = x(x + 1) (x + 2)(x + 3) x(x + 1)(x + 2)(x + 3) 4x + 6 . = x(x + 1)(x + 2)(x + 3) Solution 7. (a),(b) The left hand picture shows a typical odd function f (x). The tangent lines at x and −x have the same slope, so f 0 (−x) = f 0 (x), so f 0 (x) is even.

f (x)

g(x)=g(−x)

−x x

−x

x

f (−x)=−f (x)

The right hand picture shows a typical even function g(x). The tangent lines at x and −x have opposite slopes, so g 0 (−x) = −g 0 (x), so g 0 (x) is an odd function. (c),(d) Take f (x) = 1 + x. This is neither even nor odd (because f (1) = 2 and f (−1) = 0 which is neither +2 nor −2). However, f 0 (x) is the constant function 1, which is even. Thus this f (x) answers both (c) and (d). Solution 8. Probably the simplest answer is f (x) = − cos(πx), so f 0 (x) = π sin(πx), so f 0 (n) = 0 for all integers n. We also have f (0) = −1 and f (1) = 1, so f (0) < f (1) as required. Another approach is to take f (x) = x2 /2 − x4 /4, so f 0 (x) = x − x3 = x(1 + x)(1 − x), so again f 0 (0) = f 0 (1) = f 0 (−1) = 0. In this case we have f (0) = 0 and f (1) = 1/4, so f (0) < f (1).

3

Solution 9. The obvious example is f (x) = e−x (so f 0 (x) = −e−x < 0). Other examples include x 1 − √1+x and 1 − tanh(x). 2 Solution 10. We have f 0 (x) = 2ax + b for all x. As f 0 (x√ 0 ) = 0, we must have 2ax0 + b = 0, so x0 = −b/(2a). We also have f (x0√ ) = 0, so x0 = −b/(2a) ± b2 − 4ac/(2a). These two expressions for x0 can only be compatible if b2 − 4ac/(2a) = 0, so b2 − 4ac = 0, so c = b2 /(4a). Solution 11. g 0 (v) = − 21 = − 12

v2 c2

−3/2

1−

v2 c2

−3/2

1−

 

−2



= vc

v2 1− 2 c

d dv

 1−

v2 c2

 (power rule)

(−2v/c2 )

−3/2 . 2

Solution 12. Put t = −a2 x2 and u = exp(t) = e−ax and v = sin(ωx) and y = uv = 2 2 e−a x sin(ωx); we must find dy/dx. We have dt dx

= −2a2 x

du dt

=

d t dt e

du dx dv dx dy dx

=

dt du dx dt

2

= et = e−a

x2 2

= −2a2 xe−a

x2

= ω cos(ωx) dv = u dx + 2

x2

2

2

= e−a

= e−a

x

du dx v 2

ω cos(ωx) − 2a2 xe−a

x2

sin(ωx)

(ω cos(ωx) − 2a2 x sin(ωx)).

(It is a common mistake for things like 2a to creep in. If a were a variable then we would have d d d 2 da (a ) = 2a. However, a is in fact a constant, and we are using dx rather than da , so the equation d 2 da (a ) = 2a is not relevant.) Solution 13. Put u = xp − xq , so y = u1/pq . Then du/dx = pxp−1 − qxq−1 = x−1 (pxp − qxq ) and

1 1/pq−1 1 p dy = u = (x − y q )1/pq−1 . du pq pq

We therefore have x(xp − xq )

dy dy du = xu dx du dx = x(xp − xq )

1 p (x − xq )1/pq−1 x−1 (pxp − qxq ) pq

= (xp − xq )1/pq (pxp − qxq )/(pq) = (xp − xq )1/pq (xp /q − xq /p). Solution 14. Put u = (x − a)/(x − b) and y = f (x) = un . Then du 1.(x − b) − (x − a).1 a−b = = , 2 dx (x − b) (x − b)2 so dy du f (x) = = nun−1 = n(a − b) dx dx 0



x−a x−b

n−1

(x − b)−2 = n(a − b)(x − a)n−1 (x − b)−n−1 .

4 dz Solution 15. It is convenient to introduce the notation Lz = x dx . The question then asks us to find Ly, LLy and LLLy. We have

dy = x(a1 + 2a2 x + 3a3 x2 + 4a4 x3 ) dx = a1 x + 2a2 x2 + 3a3 x3 + 4a4 x4

Ly = x

LLy = x(a1 + 2 × 2a2 x + 3 × 3a3 x+ 4 × 4a4 x3 ) = a1 x + 4a2 x2 + 9a3 x3 + 16a4 x4 = a1 x + 22 a2 x2 + 32 a3 x3 + 42 a4 x4 LLLy = x(a1 + 2 × 4a2 x + 3 × 9a3 x2 + 4 × 16a4 x3 ) = a1 x + 8a2 x2 + 27a3 x3 + 64a4 x4 = a1 x + 23 a2 x2 + 33 a3 x3 + 43 a4 x4 . P P The general rule is clearly that if y = k ak xk , then Ln y = k k n ak xk . Solution 16. We first calculate the successive derivatives: d x (e y) = ex y + ex y 0 dx d2 x (e y) = (ex y + ex y 0 ) + (ex y 0 + ex y 00 ) dx2 = ex y + 2ex y 0 + ex y 00 d3 x (e y) = (ex y + ex y 0 ) + 2(ex y 0 + ex y 00 ) + (ex y 00 + ex y 000 ) dx3 = ex y + 3ex y 0 + 3ex y 00 + ex y 000 . It follows that d x (e y) = y + y 0 dx d2 e−x 2 (ex y) = y + 2y 0 + y 00 dx 3 −x d e (ex y) = y + 3y 0 + 3y 00 + y 000 . dx3 You should recognize the numbers here as binomial coefficients; they are the same as in the formulae e−x

(1 + t)1 = 1 + t (1 + t)2 = 1 + 2t + t2 (1 + t)3 = 1 + 3t + 3t2 + t3 . The pattern seems to be that e−x

    2 dn x dy d y dn y n n (e y) = y + + + ··· + n n 2 2 1 dx dx dx dx  k n  X d y n = . k dxk k=0

One way to prove this is by induction; we omit the details. Here is another, more abstract approach; you can ignore it if you are not interested. Consider d (ex z). We have seen that Lz = z + z 0 = (1 + D)z, so the operators Dz = z 0 and Lz = e−x dx L = 1 + D, so  n  X n n n Dk , L = (1 + D) = k k=0

5

so

X n  X  n  dk y k L y= . D y= k k dxk n

k

k

On the other hand, we have d L2 y = L(Ly) = e−x (ex Ly) dx   −x d x −x d x =e e e (e y) dx dx d2 = e−x 2 (ex y) dx    d d x 3 −x d L y=e ex e−x ex e−x (e y) dx dx dx 3 d = e−x 3 (ex y) dx and so on. It follows that e−x

X dn x (e y) = Ln y = n dx



k

n k



dk y , dxk

as claimed. Solution 17. d dx

 √

x 1 + x2

=

1.(1 + x2 )1/2 − x. 12 (1 + x2 )−1/2 .2x 1 + x2

=

(1 + x2 )1/2 − x2 (1 + x2 )−1/2 , 1 + x2

=

1 (1 + x2 ) − x2 = = (1 + x2 )−1 1 + x2 1 + x2



so d dx



d dx



x 1 + x2



x 1 + x2 In other words, we have c = −3/2.



(1 + x2 )1/2

√

so √

= (1 + x2 )−3/2 .

Solution 18. f 0 (x) = 1 + x + x2 + x3 + x4 = (x5 − 1)/(x − 1) d d (x + x3 ) dx (1 + x2 + x4 ) − (1 + x2 + x4 ) dx (x + x3 ) (x + x3 )(2x + 4x3 ) − (1 + x2 + x4 )(1 + 3x2 ) = 3 2 (x + x ) (x + x3 )2 2x2 + 4x4 + 2x4 + 4x6 − 1 − 3x2 − x2 − 3x4 − x4 − 3x6 x6 + 2x4 − 2x2 − 1 = = 3 2 (x + x ) (x + x3 )2

g 0 (x) =

h0 (x) = 7 tan(x)6 tan0 (x) = 7 tan(x)6 (1 + tan(x)2 )  6 sin(x) 1 =7 = 7 sin(x)7 / cos(x)9 . cos(x) cos(x)2 k 0 (x) = 1/ sin0 (arcsin(x)) = 1/ cos(arcsin(x)) = (1 − x2 )−1/2 m0 (x) = ln0 (cos(x)) cos0 (x) =

1 sin(x) .(− sin(x)) = − = − tan(x). cos(x) cos(x)