Derivace-příklady
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2x
= ln(x2 − 1) +
2x2
x2
− 1
Derivative of u = x is easy. The function ln(x2 − 1) is a
composite function with the outside function ln(·) and
the inside function x2 − 1.
c
Robert Maˇr´ık, 2004.
Differentiate y′ = x ln(x2 − 1).
y
′ = x′ ln(x2 − 1) + x
ln(x2 − 1)
′
= 1 ln(x2 − 1) + x
1
x2
− 1
(x2 − 1)′
= ln(x2 − 1) + x
1
x2
− 1
2x
= ln(x2 − 1) +
2x2
x2
− 1
(x2 − 1)′ = 2x − 0 = 2x
c
Robert Maˇr´ık, 2004.
Differentiate y′ = x ln(x2 − 1).
y
′ = x′ ln(x2 − 1) + x
ln(x2 − 1)
′
= 1 ln(x2 − 1) + x
1
x2
− 1
(x2 − 1)′
= ln(x2 − 1) + x
1
x2
− 1
2x
= ln(x2 − 1) +
2x2
x2
− 1
We simplify.
c
Robert Maˇr´ık, 2004.
Differentiate y′ = x ln(x2 − 1).
y
′ = x′ ln(x2 − 1) + x
ln(x2 − 1)
′
= 1 ln(x2 − 1) + x
1
x2
− 1
(x2 − 1)′
= ln(x2 − 1) + x
1
x2
− 1
2x
= ln(x2 − 1) +
2x2
x2
− 1
Finished!
Problem 8, y =
1
4
ln
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)
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)
The function is a constant multiple of a logarithm. We
use the multiple 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)
The logarithm contains the fraction as its inside function.
We use the rule (ln(x))′ =