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OKLAHOMA STATE UNIVERSITY
Department of Mathematics
MATH 2144 (Calculus I)
Instructor: Dr. Mathias Schulze
MIDTERM 2
October 27, 2010
Duration: 50 minutes
No aids allowed.
This examination paper consists of 7 pages and 10 questions. Please bring any discrepancy
to the attention of an invigilator. The number in brackets at the start of each question is
the number of points the question is worth.
Answer 8 out of 10 questions.
Skip (or mark) two problems. Give arguments to support your answers.
For graders’ use:
page 2
Score
1
2
3
4
5
6
7
8
9
10
Total
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(5)
(50)
Page 2 of 7
page 3
1. [5] Find f 0 (x) and f 00 (x) where f (x) =
x
.
3+ex
Solution:
3 + ex − xex
(3 + ex )2
−xex (3 + ex )2 − 2ex (3 + ex )(3 + ex − xex )
f 00 (x) =
(3 + ex )4
−xex (3 + ex ) − 2ex (3 + ex − xex )
=
(3 + ex )3
xex − 2ex − 3x − 6 x
=
e
(3 + ex )3
f 0 (x) =
2. [5] Differentiate y =
ex −e−x
.
ex +e−x
Solution:
y = tanh(x)
y 0 = sech2 (x)
Page 3 of 7
page 4
3. [5] Find the 1000th derivative of f (x) = xe−x .
Solution: Since ((x − k)e−x )0 = −(x − (k + 1))e−x , we have f (k) = (−1)k (x − k)e−x ,
and hence f (1000) = (x − 1000)e−x .
x
4. [5] Find y 0 by implicit differentiation: e y = x − y.
Solution:
1 xy 0
e
− 2 = 1 − y0
y
y
1 x
x xy
1 − 2 e y0 = 1 − e y
y
y
x
y
x
0
y =y
y − ey
x
y 2 − xe y
Page 4 of 7
page 5
5. [5] Use implicit differentiation to find the equation of the tangent line to the curve
x2 + xy + y 2 = 3 at the point (1, 1).
Solution: Differentiation gives
2x + y + xy 0 + 2yy 0 = 0.
Setting m = y 0 (1, 1) and substituting (x, y) = (1, 1), this becomes
2 + 1 + m + 2m = 0.
So m = −1 and the equation of the tangent line at (1, 1) reads
y = 1 + m(x − 1) = −x + 2.
6. [5] Differentiate the function y = log5 (xex ).
Solution:
(x + 1)ex
y =
=
ln(5)xex
0
1
1+
x
ln
1
5
Page 5 of 7
page 6
7. [5] Use logarithmic differentiation to differentiate y = xsin x .
Solution: Applying ln gives
ln y = sin(x) ln(x).
Then differentiate to obtain
y0
sin x
= cos(x) ln(x) +
y
x
which yields
0
y =x
sin(x)
sin x
cos(x) ln(x) +
x
.
8. [5] If a snow ball melts so that its surface area decreases at a rate of 1 cm2 /min, find
the rate at which the diameter decreases when the diameter is 10 cm.
Solution: The surface area of the snow ball is
A = 4πr2 = πd2
where r = r(t) and d = d(t) denote the radius and diameter, respectively. So, at
the time t = t0 when d(t0 ) = 10, we have
−1 = A0 (t0 ) = 2πd(t0 )d0 (t0 ) = 20πd0 (t0 )
1
.
and hence d0 (t0 ) = − 20π
Page 6 of 7
page 7
9. [5] The radius of a circular disc is given as 10 cm with a maximum error in measurement
of 0.1 cm. Use differentials to estimate the relative error in the calculated area of the
disc.
Solution: From A = πr2 , we compute dA = 2πrdr, and then the relative error is
∆A
dA
dr
≈
= 2 = 0.02 = 2%.
A
A
r
10. [5] Find the absolute maximum and minimum values of the function f (x) =
where −4 ≤ x ≤ 4.
Solution: Since
f 0 (x) =
16x
,
(x2 + 4)2
the critical numbers are −4, 0, 4. Therefore,
f (±4) =
3
and f (0) = −1
5
are the maximum and minimum values.
End of examination
Total pages: 7
Total marks: 50
x2 −4
,
x2 +4