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×
| ( x − 1 )′( x
| + 1 ) − ( x + 1 )′( x − 1 )
| =
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| ⋅
| x + 1 − x
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| ( x + 1 )2
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| 3 x − 1
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=
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| x + 1
| .
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| x − 1
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| 3( x + 1 )2
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| x − 1
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| в) y = ( x3 − 2 x + 1 ) ⋅ 3x .
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| Розв’язування. Використавши формули(V)і(XVІІ),маємо
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y′= ( x3 − 2 x + 1 )′ ⋅ 3x + ( x3 − 2 x + 1 )⋅ ( 3x )′= ( 3 x2 − 2 )⋅ 3 x +
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( x3 − 2 x + 1 )⋅ 3 x ln 3 = 3 x ( 3 x2 − 2 + ( x3 − 2 x + 1 )ln 3 ).
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| г) y = ln3( x2 + x + 1 ).
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| Розв’язування. Використавши формули(VIІ)і(XVIa),одер-
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жимо
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y′= 3 ln2 ( x2 + x + 1 )⋅ (ln( x2 + x + 1 )′=
| 3 ln2 ( x2 + x + 1 )
| ×
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| x2 + x + 1
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× ( x2 + x + 1 )′=
| 3 ln2 ( x2 + x + 1 )
| ⋅ ( 2 x + 1 ) =
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| x2 + x + 1
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=
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| 3( 2 x
| + 1 )ln2 ( x2 + x +
| 1 )
| .
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| д)
| y = x arcsin x +
| 1 − x2 .
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| Розв’язування.Використавши формули(IV,V,VIIб,XII),
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одержимо
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y′= ( x )′ arcsin x + x(arcsin x )′+
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| ( 1
| − x2 )′=
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| 1 − x2
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= arcsin x +
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| x
| +
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| (
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| −2 x ) = arcsin x +
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| 1 − x2
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| 1 − x2
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+
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| x
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| 1 − x2
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| 1 − x2
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| е)
| y = ( 1 + x2 )earctgx (Розв’язати самостійно).
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| Відповідь: y′ = ( 2 x + 1 )earctgx .
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