Ответы
task/29490961 Найти производную функции в точке x₀
1) f(x) = 4x³+6x +3 , x₀ = 1 .
* * * правила: (u+v) ' = u ' + v ' ; [C*f(x )] ' =C*f ' (x) ; (xⁿ )' nxⁿ ⁻ ¹ * * *
f '(x) =( 4x³+6x +3 )' =( 4x³) ' +(6x)' +3' = 4*(x³) ' +6*(x)' +0=4*3x²+6*1=12x²+6.
f '(x₀) = f '( 1) = 12*1²+6 = 18 .
2) f(x) = x²- 4√x , x₀ = 4 .
f '(x) =( x²- 4√x ) ' =( x²)' - 4*(√x ) ' = 2x - 4*(1 / 2√x ) = 2x -2 /√x =2(x - 1/√x)
f '(x₀) = f '( 4) = 2(4 - 1 /√4 ) = 2(4 -0,5)=2*3,5 = 7 .
3) f(x) = (4x - 7) / (x² +4) , x₀ = 0 . * * * правила: (u/v) ' = (u '*v - u*v ' ) / v² * * *
f '(x) = [ ( 4x-7) / x² +4)] ' =[ ( 4x-7)' *( x² +4 ) - ( 4x-7)*( x² +4) '] / (x² +4)² =
=[ 4( x² +4) -(4x-7)2x ] / (x² +4)² = -2(2x² -7x - 8) / ( x² +4)²
f '(x₀) = f '( 0) = -2(2*0² +7*0 - 8) / ( 0² +4)² = -2*(-8)8/16 = 1.
4) f(x) =x*sinx , x₀ = π/2 . * * * правила: (u*v) ' = u ' *v + u* v ' * * *
f '(x) = (x*sinx) ' =(x) '*sinx +x*(sinx)' = 1*sinx +x*cosx = sinx +xcosx .
f '(x₀) = f '( π/2) =sin(π/2) +(π/2)*cos(π/2) =1 +(π/2)*0 = 1 .
5) f(x) =2x*cosx , x₀ = 0 .
f '(x) = (2x*cosx) ' =2*[(x) '*cos +x*(cos)' ] = 2(cosx- xsinx ) .
f '(x₀) = f '( 0) =2(cos0 - 0*sin0) =2(1 -0) = 2 .