Помогите с 2-5, срочно срочно срочно. Created by lululilovo. algebra-ru

Помогите с 2-5, срочно срочно срочно...

lululilovo @lululilovo

July 2022 1 17 Report

Помогите с 2-5, срочно срочно срочно

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oganesbagoyan

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2)Решите уравнение f'(x) =0 : f(x) = 4cos(x/8)*cos(x/8)  .
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f(x) =4cos(x/8)*cos(x/8) = 4cos²(x/8) =4*(1+cos2*(x/8) ) / 2 =2+2cos(x/4).
f ' (x) =(2+2cos(x/4) ) ' = (2)' +(2cos(x/4) ) '= 0 -2*sin(x/4) *(x/4) ') = -0,5sin(x/4).
f ' (x) =0 ⇔  -0,5sin(x/4) = 0 ⇔sin(x/4)=0 ⇒x/4 =π*n , n∈Z.
x =4π*n , n∈Z.
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3) f(x) = 1/(2x+7)⁴ -(1-x)³ ;   x₀ = -3.
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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.
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4)  f(x) =x² - 4x ; g(x) =√x .
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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.
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5) Докажите тождество :
g ' (x)= (g(x) /cosx) ² ,если   g(x) = -ctqx +ctqπ/2.
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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 * * *

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