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sipackna
@sipackna
July 2022
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Ставлю 50 баллов. С обьяснениями, пожалуйста. Найти производные
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oganesbagoyan
Verified answer
Найти производные данных функций:
а
)
y =x² +2Lnx ;
y
' = (x² +2Lnx) ' = (
x²) ' +( 2Lnx ) ' =2x
²⁻¹
+2*(Lnx) ' =2x +2/x =2(x+1/x).
===
б)
y =(x
² +1)cos²5x
; * * * (u*v) ' =u'*v+v'*u * * *
y '
= ( (x² +1)cos5x ) ' =(x
² +1) ' *cos5x +
(x² +1)*(cos5x) ' =(2x+0)*cos5x +
(x² +1)*(-sin5x)*(5x) '
= 2xcos5x - 5(x² +1)*sin5x .
====
в)
y = x /
√(3-x²)
;
y '
= ( x /√(3-x²) ) ' =
(x*
(3-x²)^(-1/2) ) ' =
(x) '*(3-x²)^(-1/2) +x *(-1/2)*(3-x²)^(-3/2)*(3-x²)'
=1/√(3-x²) + x/√(3-x²)³
.
* * * иначе (3 - x² + x ) / (3-x²)√
(3-x²) * * *
2 votes
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Answers & Comments
Verified answer
Найти производные данных функций:а)
y =x² +2Lnx ;
y ' = (x² +2Lnx) ' = (x²) ' +( 2Lnx ) ' =2x²⁻¹ +2*(Lnx) ' =2x +2/x =2(x+1/x).
===
б)
y =(x² +1)cos²5x ; * * * (u*v) ' =u'*v+v'*u * * *
y ' = ( (x² +1)cos5x ) ' =(x² +1) ' *cos5x +(x² +1)*(cos5x) ' =(2x+0)*cos5x +
(x² +1)*(-sin5x)*(5x) ' = 2xcos5x - 5(x² +1)*sin5x .
====
в)
y = x /√(3-x²) ;
y ' = ( x /√(3-x²) ) ' = (x*(3-x²)^(-1/2) ) ' =
(x) '*(3-x²)^(-1/2) +x *(-1/2)*(3-x²)^(-3/2)*(3-x²)' =1/√(3-x²) + x/√(3-x²)³ .
* * * иначе (3 - x² + x ) / (3-x²)√(3-x²) * * *