Derivace-příklady
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=
3
s
1 + x3
1 − x3
2x2
1 − x6
The problem is solved.
Problem 6, y =
x
− 1
x +
1
2
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
− 1
x +
1
2
.
y′ =
2
x
− 1
x +
1
x
− 1
x +
1
′
= 2
x
− 1
x +
1
·
(x − 1)′(x + 1) − (x − 1)(x + 1)′
(x + 1)2
= 2
x
− 1
x +
1
·
1.(x + 1) − (x − 1).1
(x + 1)2
= 2
x
− 1
x +
1
·
2
(x + 1)2
= 4
x
− 1
(x + 1)3
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
− 1
x +
1
2
.
y′ =
2
x
− 1
x +
1
x
− 1
x +
1
′
= 2
x
− 1
x +
1
·
(x − 1)′(x + 1) − (x − 1)(x + 1)′
(x + 1)2
= 2
x
− 1
x +
1
·
1.(x + 1) − (x − 1).1
(x + 1)2
= 2
x
− 1
x +
1
·
2
(x + 1)2
= 4
x
− 1
(x + 1)3
The function is a second power of fraction. The power
function is the outside function and is differentiated as
the first, using the rule (x2)′ = 2x.
The derivative of the inside function follows.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
− 1
x +
1
2
.
y′ =
2
x
− 1
x +
1
x
− 1
x +
1
′
= 2
x
− 1
x +
1
·
(x − 1)′(x + 1) − (x − 1)(x + 1)′
(x + 1)2
= 2
x
− 1
x +
1
·
1.(x + 1) − (x − 1).1
(x + 1)2
= 2
x
− 1
x +
1
·
2
(x + 1)2
= 4
x
− 1
(x + 1)3
The fraction is differentiated by the rule
u
v
′
=
u′v
− uv′
v2
.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
− 1
x +
1
2
.
y′ =
2
x
− 1
x +
1
x
− 1
x +
1
′
= 2
x
− 1
x +
1
·
(x − 1)′(x + 1) − (x − 1)(x + 1)′
(x + 1)2
= 2
x
− 1
x +
1
·
1.(x + 1) − (x − 1).1
(x + 1)2
= 2
x
− 1
x +
1
·
2
(x + 1)2
= 4
x
− 1
(x + 1)3
It is easy to differentiate the expressions in the numerator.
We use the sum rule and the rules (x)′ = 1 and (1)′ = 0.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
− 1
x +