Derivace-příklady
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(x2 + 1)2
=
1 · (x2 + 1) − x · (2x + 0)
(x2 + 1)2
=
1 − x2
(1 + x2)2
We use the quotient rule
u
v
′
=
u′ v
− uv′
v2
.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
x2 +
1
.
y′ =
x
x2 +
1
′
=
(x)′ · (x2 + 1) − x · (x2 + 1)′
(x2 + 1)2
=
1 · (x2 + 1) − x · (2x + 0)
(x2 + 1)2
=
1 − x2
(1 + x2)2
x
′ = 1 by the power rule.
(x2 + 1)′ = (x2)′ + (1)′ = 2x + 0 = 2x by the sum rule
and the power rule.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
x2 +
1
.
y′ =
x
x2 +
1
′
=
(x)′ · (x2 + 1) − x · (x2 + 1)′
(x2 + 1)2
=
1 · (x2 + 1) − x · (2x + 0)
(x2 + 1)2
=
1 − x2
(1 + x2)2
We multiply the parentheses and simplify the numerator.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
x2 +
1
.
y′ =
x
x2 +
1
′
=
(x)′ · (x2 + 1) − x · (x2 + 1)′
(x2 + 1)2
=
1 · (x2 + 1) − x · (2x + 0)
(x2 + 1)2
=
1 − x2
(1 + x2)2
The problem is finished.
Problem 3, y =
1 − x
3
x2
c
Robert Maˇr´ık, 2004.
Differentiate y =
1 − x3
x2
y′ =
1 − x3
x2
′
=
(1 − x3)′ · x2 − (1 − x3) · (x2)′
(x2)2
=
(0 − 3x2) · x2 − (1 − x3) · 2x
(x2)2
= −
3x4 − 2x + 2x4
x4
= −
2 + x3
x3
c
Robert Maˇr´ık, 2004.
Differentiate y =
1 − x3
x2
y′ =
1 − x3
x2
′
=
(1 − x3)′ · x2 − (1 − x3) · (x2)′
(x2)2
=
(0 − 3x2) · x2 − (1 − x3) · 2x
(x2)2
= −
3x4 − 2x + 2x4
x4
= −
2 + x3
x3
The function is a quotient.
c
Robert Maˇr´ık, 2004.
Differentiate y =
1 − x3
x2
y′ =
1 − x3
x2
′
=
(1 − x3)′ · x2 − (1 − x3) · (x2)′
(x2)2
=
(0 − 3x2) · x2 − (1 − x3) · 2x
(x2)2
= −
3x4 − 2x + 2x4