Ответы и объяснения

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  • Minsk00
  • почетный грамотей
2013-07-03T13:01:03+04:00

 

a) y = [(1-xa^(1/2))^2]/x                 y'(0,01)

Найдем производную (дробь но можно и как произведение)

y' =[((1-x^(1/2))^2)' *x-(1-x^(1/2))*x']/x^2 = [2(1-x^(1/2))*(1-x^(1/2))' *x-(1-x^(1/2))^2]/x^2 =

=[(2(1-x^(1/2))*(-1/2)*x^(-1/2)*x-(1-x^(1/2))^2]/x^2 =[ -(1-x^(1/2))*x^(1/2)-(1-x^(1/2))^2]/x^2 =

=(x^(1/2)-1)/x^2

y'(0,01) = ((корень( 0,01)-1)/0,01^2 = -9000

 

 

б) y=2^x *e^(-x)+x    y'(0)

y' = (2^x)' *e^(-x)+2^x *(e^(-x))' +x' = 2^x *ln2 *e^(-x) +2^x *(-e^(-x)) +1 = 2^x *e^(-x)*(ln2-1)+1

y'(0) = 2^0 *e^0*(ln2-1) +1 = ln2-1+1 = ln2 = 0,693

 

 

 в) y=arcsinx/(1-x^2)^(1/2)               y'(0)

y' =(arcsinx' (1-x^2)^(1/2) - arcsinx * [(1-x^2)^(1/2)]')/(1-x^2) =

=((1/(1-x^2)^(1/2))*(1-x^2)^(1/2) -arcsinx * (1/2)*(1-x^2)^(-1/2)*(-2x))/(1-x^2) =

(1+x*arcsinx*(1-x^2)^(-1/2))/(1-x^2)

y'(0) = (1+0*arcsin0*(1-0)^(-1/2))/(1-0) = 1