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lululilovo
@lululilovo
July 2022
2
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С 3 по 5, решите пжл срочно. Заранее выражаю огромную благодарность
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amin07am
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Ответ ответ ответ ответ ответ
1 votes
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oganesbagoyan
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3)
f(x) = 1/(2x+7)⁴ -(1-x)³ ; x₀ = -3.
----
f ' x
₀) -?
f ' (x) =( (1/(2x+7)⁴ -(1-x)³ )' = ( (1/(2x+7)⁴ )' -((1-x)³ ) ' = ( (2x+7)⁻⁴ )' -((1-x)³ )' =
- 4*(2x+7)⁻⁵ *(2x+7) ' - 3(1-x)
² *(1-x)' =
- 8/(2x+7)⁵+ 3(1-x)² .
f ' (x₀)
= f '( -3) = -8/(2*(-3) +7)⁵ +3(1 -(-3))
² = -8 +48 = 40.
--------------------------
4)
f(x) =x² - 4x ; g(x) =√x .
----
f ' (g(x) ) -?
f (g(x))=(√x)² - 4√x = x - 4√x ⇒ f '(g(x))=( x - 4√x ) ' =(x)' - 4*(√x)' =1 - 2/√x.
--------------------------
5)
Докажите тождество :
g ' (x)= (g(x) /cosx) ² ,если g(x) = -ctqx +ctqπ/2.
----
g(x)
= -ctqx +ctqπ/2 = -ctqx +0 =
-
ctqx .
( g(x) ) ' = (-ctqx) ' =1 /sin² x = (cos²x /sin²x)* (1 / cos²x) =(ctqx)²*(1/cos²x) =
(-ctqx)² / (1/cosx)² = (g(x))
²*(1/cosx)²
=(g(x) /cosx)²
* * * a² =(-a)² * *
0 votes
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Verified answer
Ответ ответ ответ ответ ответVerified answer
3) f(x) = 1/(2x+7)⁴ -(1-x)³ ; x₀ = -3.----
f ' x₀) -?
f ' (x) =( (1/(2x+7)⁴ -(1-x)³ )' = ( (1/(2x+7)⁴ )' -((1-x)³ ) ' = ( (2x+7)⁻⁴ )' -((1-x)³ )' =
- 4*(2x+7)⁻⁵ *(2x+7) ' - 3(1-x)² *(1-x)' =
- 8/(2x+7)⁵+ 3(1-x)² .
f ' (x₀) = f '( -3) = -8/(2*(-3) +7)⁵ +3(1 -(-3))² = -8 +48 = 40.
--------------------------
4) f(x) =x² - 4x ; g(x) =√x .
----
f ' (g(x) ) -?
f (g(x))=(√x)² - 4√x = x - 4√x ⇒ f '(g(x))=( x - 4√x ) ' =(x)' - 4*(√x)' =1 - 2/√x.
--------------------------
5) Докажите тождество :
g ' (x)= (g(x) /cosx) ² ,если g(x) = -ctqx +ctqπ/2.
----
g(x) = -ctqx +ctqπ/2 = -ctqx +0 = -ctqx .
( g(x) ) ' = (-ctqx) ' =1 /sin² x = (cos²x /sin²x)* (1 / cos²x) =(ctqx)²*(1/cos²x) =
(-ctqx)² / (1/cosx)² = (g(x))²*(1/cosx)² =(g(x) /cosx)² * * * a² =(-a)² * *