Exam (elaborations)
Single Variable Calculus
- Module
- Mathematics
- Institution
- Massachusetts Institute Of Technology
Solutions to exam 1 for the single variable calculus
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18.01 Solutions to Exam 1Problem 1(15 points) Use the definition of the derivative as a limit of difference quotients to
compute the derivative of y = x + x1 for all points x > 0. Show all work.
Solution to Problem 1 Denote by f (x) the function x + x1 . By definition, the derivative of f (x)
at x = a is,
f (a + h) − f (a)
f � (a) = lim .
h→0 h
The increment f (a + h) − f (a) equals,
� � � � � �
1 1 1 1
(a + h) + − a+ =h+ − .
a+h a a+h a
To compute the second term, clear denominators,
1 1 1 a 1a+h a − (a + h) −h
− = − = = .
a+h a a+ha aa+h a(a + h) a(a + h)
Thus the increment f (a + h) − f (a) equals,
h
h− .
a(a + h)
Factoring h from each term, the difference quotient equals,
f (a + h) − f (a) 1
=1− .
h a(a + h)
Thus the derivative of f (x) at x = a equals,
� �
� 1 1 1
f (a) = lim 1 − =1− = 1 − 2.
h→0 a(a + h) a(a + 0) a
Therefore the derivative function of f (x) equals,
f � (x) = 1 − 1
x2
.
1
, 2 /2
Problem 2(10 points) For the function f (x) = e−x , compute the first, second and third deriva
tives of f (x).
Solution to Problem 2 Set u equals −x2 /2 and set v equals eu . So v equals f (x). By the chain
rule,
dv dv du
= .
dx du dx
Since v equals eu , dv/du equals (eu )� = eu . Since u equals −x2 /2, du/dx equals −(2x)/2 = −x.
Thus, backsubstituting,
dv 2 2
f � (x) = = (eu )(−x) = e−x /2 (−x) = −xe−x /2 .
dx
For the second derivative, let u and v be as defined above, and set w equals −xv. So w equals
f � (x). By the product rule,
dw dv
= (−x)� v + (−x)v � = −v − x .
dx dx
By the last paragraph,
dv 2
= −xe−x /2 .
dx
Substituting in,
dw 2 2 2 2 2
f �� (x) = = −e−x /2 − x(−xe−x /2 ) = −e−x /2 + x2 e−x /2 = (x2 − 1)e−x /2 .
dx
For the third derivative, take u and v as above, and set z equals (x2 − 1)v. So z equals f �� (x). By
the product rule,
dz dv
= (x2 − 1)� v + (x2 − 1)v � = 2xv + (x2 − 1) .
dx dx
By the first paragraph,
dv 2
= −xe−x /2 .
dx
Substituting in,
dz 2 2 2 2 2
f ��� (x) = = 2xe−x /2 + (x2 − 1)(−xe−x /2 ) = 2xe−x /2 + (−x3 + x)e−x /2 = (−x3 + 3x)e−x /2 .
dx
2
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