Derivace-příklady
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−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 differentiate the composite function e−2x. The outside
function is an exponential function which does not
change by differentiation. The inside function is the
linear function −2x.
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
Derivatives of x2 and x can be evaluated by power rule.
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
Derivative of (−2x) can be evaluated by a constant
multiple rule . . .
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
. . . and power rule (x = x1 and hence x′ = 1x0 = 1).
c
Robert Maˇr´ık, 2004.