Derivace-příklady
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Derivatives
Robert Maˇr´ık
July 20, 2004
1
Contents
Problem 1, y = (x2 +
3x)e−2x
3
Problem 2, y =
x
x2+
1
14
Problem 3, y = 1−x
3
x2
21
Problem 4, y = x
ln2 x
29
Problem 5, y = 3
q
1+x3
1−x3
37
Problem 6, y =
x
−1
x+
1
2
47
Problem 7, y′ = x
ln(x2 − 1)
55
Problem 8, y = 1
4 ln
x
2−1
x2+
1
62
Problem 9, y =
√
x +
1 − ln(1 +
√
x +
1)
70
Problem 10, y =
√
1 − x. arcsin
√
x
76
Problem 1, y = (x2 +
3x)e−2x
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
We differentiate product of the function u = x2 + 3x and
v = e−
2x. We use the product rule.
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
We differentiate the sum. We use the sum rule and the
constant multiple rule.
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e