STAT 1000 Study Guide - Midterm Guide: Memorial University Of Newfoundland, Indeterminate Form, Asymptote
10 Nov 2016
School
Department
Course
Professor
MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
Sample Midterm 2 MATHEMATICS 1000 Solutions
1. Use logarithmic differentiation to find the derivatives of the following functions.
(a)
y=sin3x
(5x+ 3)√x+ 4
ln y= 3 ln(sin x)−ln(5x+ 3) −1
2ln(x+ 4)
1
yy0= 3 1
sin x·cos x−1
5x+ 3 ·5−1
2·1
x+ 4
1
yy0= 3 cot x−5
5x+ 2 −1
2x+ 8
y0=y3 cot x−5
5x+ 2 −1
2x+ 8
y0=sin3x
(5x+ 3)√x+ 4 3 cot x−5
5x+ 2 −1
2x+ 8
(b)
y= (x3−3x)e2x
ln y= e2xln(x3−3x)
1
yy0= e2x·2·ln(x3−3x)+e2x·1
x3−3x·(3x2−3)
1
yy0= 2e2xln(x3−3x)+e2x·3x2−3
x3−3x
y0=y2e2xln(x3−3x)+e2x·3x2−3
x3−3x
y0= (x3−3x)e2x2e2xln(x3−3x)+e2x·3x2−3
x3−3x
2. (a) Find the derivative of the following function with respect to x:
exy = 3x2−y
exy (1 ·y+xy0)=6x−y0
yexy +xy0exy = 6x−y0
xy0exy +y0= 6x−yexy
y0=6x−yexy
xexy + 1
–2–
(b) Find the second derivative of the following function with respect to x:
x3+y3= 5
3x2+ 3y2y0= 0
3y2y0=−3x2
y0=−x2
y2
y00 =(−2x)(y2)−(−x2)(2yy0)
(y2)2
y00 =−2xy2+ 2x2y·−x2
y2
y4
y00 =−2xy2−2x4
y
y4·y
y
y00 =−2xy3−2x4
y5
3. Evaluate the following limits. If the limit does not exist but approaches ±∞, indicate that. You
may not use L’Hˆopital’s rule.
(a) Plugging in x= 0 gives the indeterminate form 0
0, so we simplify by factoring.
lim
x→1
3x2−x−2
x3−1= lim
x→1
(3x+ 2)(x−1)
(x−1)(x2+x+ 1) = lim
x→1
3x+ 2
x2+x+ 1 =3(1) + 2
(1)2+ (1) + 1 =5
3
(b) Plugging in x= 0 gives the indeterminate form 0
0, so we simplify by factoring and use a
trig limit.
lim
x→0
cos x−cos2x
x= lim
x→0cos x·1−cos x
x= cos(0) ·0=1·0 = 0
(c) Note: There was a typo in the question in the first version of the sample midterm posted.
The question with the error would give the answer DNE.
Plugging in x= 0 gives the indeterminate form ∞−∞, so we simplify by finding a common
denominator.
lim
x→2x
x−2−2
x2−3x+ 2= lim
x→2x(x−1)
(x−2)(x−1) −2
(x−2)(x−1)
= lim
x→2
x2−x−2
(x−2)(x−1)
= lim
x→2
(x−2)(x+ 1)
(x−2)(x−1)
= lim
x→2
x+ 1
x−1=2+1
2−1= 3
