Derivace-příklady
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x4
= −
2 + x3
x3
We use the quotient rule
u
v
′
=
u′ v
− uv′
v2
.
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
We differentiate (1 − x
3)′ by the sum rule and the power
rule. The expression x2 can be differentiated using the
power rule.
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
We multiply the parentheses.
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
We add the common powers of x.
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
We finished.
Problem 4, y = x
ln
2
x
c
Robert Maˇr´ık, 2004.
Differentiate y = x ln2 x.
y′ = ( x
ln2 x )′ = (x)′ ln2 x + x (ln2 x)′
= 1 ln2 x + x 2 ln x (ln x)′
= ln2 x + x 2 ln x
1
x
= (2 + ln x) ln x
c
Robert Maˇr´ık, 2004.
Differentiate y = x ln2 x.
y′ = ( x
ln2 x )′ = (x)′ ln2 x + x (ln2 x)′
= 1 ln2 x + x 2 ln x (ln x)′
= ln2 x + x 2 ln x
1
x
= (2 + ln x) ln x
We differentiate the product (uv)′ with u = x and