Question

Помогите пожалуйста. ​

Answer 1

Ответ:

Пошаговое объяснение:

a) f'=3x^2-6x+4

f'(2)=3*2^2-6*2+4=3*4-12+4=4

б) f'=1-1/(x-1)^(1/2)

f'(5)=1-1/(5-1)^(1/2)=1-1/2=1/2

в) f'=( x/(2*(x+1)^(1/2))-((x+1)^(1/2)))/x^2

f'(3)= (3/(2*(3+1)^(1/2))-((3+1)^(1/2)))/9=(3/4 - 2)/9=(3-8)/4*9=-5/36

г)f'= 3x^2 +2*e^(2x)

f'(0)=2

д) f'= ((x+1)/(x-1)) * ((x+1)-(x-1))/(x+1)^2 =((x+1)/(x-1))* 2/(x+1)^2

f'($\sqrt{3}$)=$(\sqrt{3}+1)/(\sqrt{3} -1) * 2/(\sqrt{3}+1)^2= 2*(\sqrt{3}+1)/(\sqrt{3} -1)*(\sqrt{3}+1)^2=\\=2/(\sqrt{3} -1)*(\sqrt{3}+1)=2/(3-1)=2/2=1$

е)

$f'=(sinx*sinx )'=cosx*sinx+sinx*cosx=2sinx*cosx=sin(2x)\\f'(pi/4)=sin(2*Pi/4)=sin(Pi/2)=1$