2003 Test 2 solution

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10 Apr 2012
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MAT 137Y, 2003-2004 Term Test 2 Solutions
1.
(6%) (i) Let f(x) = 1
x2. Find f0(x)using the definition of derivative.
f0(x) = lim
h0
f(x+h)f(x)
h=f0(x) = lim
h0
1
(x+h)21
x2
h=lim
h0
x2(x+h)2
hx2(x+h)2
=lim
h02xhh2
hx2(x+h)2=lim
h02xh
x2(x+h)2=2x
x4=2
x3.
(6%) (ii) Differentiate g(x) = 3secx+xtan(πx)
x5+1. Do not simplify your answer.
g0(x) =
x5+1[3secxtanx+tan(πx)xsec2(πx)] 5x4
2x5+1[3secx+xtan(πx)]
x5+1.
(6%) (iii) Find dy
dt when t=1 if y=u2
u+1,u= (s3+1)2,s=t1/2.
Differentiating we get
dy
du =(u+1)(u2)
(u+1)2=3
(u+1)2,du
ds =2(s3+1)·3s2=6s2(s3+1),ds
dt =1
2t3/2.
When t=1, then s=1 and u=4. By the chain rule, dy
dt =dy
du ·du
ds ·ds
dt , so
dy
dt
t=1=dy
du
u=4·du
ds
s=1·ds
dt
t=1=3
25 ·12·1
2=18
25.
(6%) (iv) Evaluate lim
x0+1
sin2xcosx
xsinx.
Let Lbe the limit above. Then
L=lim
x0+
xcosxsinx
xsin2x
H
=lim
x0+
1(cos2xsin2x)
sin2x+2xsinxcosx=lim
x0+
2sin2x
sin2x+2xsinxcosx
=lim
x0+
2sinx
sinx+2xcosxH
=lim
x0+
2cosx
cosx+2cosx2xsinx=2
3.
(6%) (v) Let x0=2 be the initial approximation of a root of f(x) = x424
x+1.
Using Newton’s Method, find x1.
We have f(2) = 8
3and f0(x) = 4x3(x+1)x4+24
(x+1)2=f0(2) = 104
9. By Newton’s Method,
x1=x0f(x0)
f0(x0)=2f(2)
f0(2)=2+8
3·9
104 =2+3
13 =29
13.
1
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