Derivace-příklady
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Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
The common factor e−2x can be taken out.
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
We simplify inside the parentheses.
c
Robert Maˇr´ık, 2004.
Differentiate y = (x2 + 3x)e−2x
y
′ =
x
2 + 3x
′
e
−2x + (x2 + 3x)
e
−2x
′
=
(x2)′ + 3(x)′
e
−2x + (x2 + 3x)e−2x(−2x)′
=
2x + 3 · 1
e
−2x + (x2 + 3x)e−2x(−2)(x)′
=
2x + 3
e
−2x + (x2 + 3x)e−2x(−2)1
=
2x + 3 + (−2)(x
2 + 3x)
e−
2x
=
−2x
2 − 4x + 3
e−
2x = −
2x2 + 4x − 3
e−
2x
We take out the minus sign. Finished!
Problem 2, y =
x
x2 +
1
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
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 function is a quotient.
c
Robert Maˇr´ık, 2004.
Differentiate y =
x
x2 +
1
.
y′ =
x
x2 +
1
′
=
(x)′ · (x2 + 1) − x · (x2 + 1)′