Derivace-příklady
Níže je uveden pouze náhled materiálu. Kliknutím na tlačítko 'Stáhnout soubor' stáhnete kompletní formátovaný materiál ve formátu PDF.
1
x
and the chain rule.
Remember that
1
x2
−1
x2+
1
=
x
2 + 1
x2
− 1
.
c
Robert Maˇr´ık, 2004.
Differentiate y =
1
4
ln
x
2 − 1
x2 +
1
.
y
′ =
1
4
x
2 + 1
x2
− 1
2x(x2 + 1) − (x2 − 1)2x
(x2 + 1)2
=
1
4
x
2 + 1
x2
− 1
4x
(x2 + 1)2
=
x
(x2 − 1)(x2 + 1)
We continue in using the chain rule. We evaluate the
derivative of the inside function by the quotient rule.
c
Robert Maˇr´ık, 2004.
Differentiate y =
1
4
ln
x
2 − 1
x2 +
1
.
y
′ =
1
4
x
2 + 1
x2
− 1
2x(x2 + 1) − (x2 − 1)2x
(x2 + 1)2
=
1
4
x
2 + 1
x2
− 1
4x
(x2 + 1)2
=
x
(x2 − 1)(x2 + 1)
We simplify the numerator of the last fraction. Terms
with x3 cancel and 4x remains.
c
Robert Maˇr´ık, 2004.
Differentiate y =
1
4
ln
x
2 − 1
x2 +
1
.
y
′ =
1
4
x
2 + 1
x2
− 1
2x(x2 + 1) − (x2 − 1)2x
(x2 + 1)2
=
1
4
x
2 + 1
x2
− 1
4x
(x2 + 1)2
=
x
(x2 − 1)(x2 + 1)
We multiply the fraction.
c
Robert Maˇr´ık, 2004.
Differentiate y =
1
4
ln
x
2 − 1
x2 +
1
.
y
′ =
1
4
x
2 + 1
x2
− 1
2x(x2 + 1) − (x2 − 1)2x
(x2 + 1)2
=
1
4
x
2 + 1
x2
− 1
4x
(x2 + 1)2
=
x
(x2 − 1)(x2 + 1)
Finished!
Problem 9,
y =
√
x +
1 − ln(1 +
√
x +
1)
c
Robert Maˇr´ık, 2004.
Differentiate y =
√
x +
1 − ln(1 +
√
x +
1).
y
′ =
1
2
√
x +
1
−
1
1 +
√
x +
1
0 +
1
2
√
x +
1
=
1
2
√
x +
1
1 −
1
1 +
√
x +
1
=
1
2
√
x +
1
√
x +
1
1 +
√
x +
1
=
1
2(1 +
√
x +
1)
c
Robert Maˇr´ık, 2004.
Differentiate y =
√
x +
1 − ln(1 +
√
x +
1).
y
′ =
1
2
√
x +
1
−
1
1 +
√
x +
1
0 +
1
2
√
x +
1
=
1
2
√
x +
1
1 −
1
1 +
√
x +
1
=
1
2
√
x +
1
√
x +
1
1 +
√
x +
1
=
1
2(1 +
√
x +
1)
(
√
x)′ =
x
1
2
′
=
1
2
x
1
2 −1
=
1
2
x−
1
2
=
1
2
√
x
by the power rule. We combine this rule by the chain rule
and hence
(
√
x +
1)′ =
1
2
√
x +
1
· 1 =
1
2
√
x +
1
c
Robert Maˇr´ık, 2004.
Differentiate y =
√