Задача 65220 Найти производную функции ...
Условие
Решение
f'(x) = 2 + 5*3x^2 = 2 + 15x^2
f'(x0) = f'(2) = 2 + 15*2^2 = 2 + 15*4 = 62
Б) f(x) = 4x^2 - 3x + 2; x0 = 3
f'(x) = 4*2x - 3 = 8x - 3
f'(x0) = f'(3) = 8*3 - 3 = 21
В) f(x) = 3/x^(10) = 3*x^(-10); x0 = 1
f'(x) = 3*(-10)*x^(-11) = -30/x^(11)
f'(x0) = f'(1) = -30/1^(11) = -30
Г) f(x) = 3x^4 + 9x^2 + 5; x0 = -1
f'(x) = 3*4x^3 + 9*2x = 12x^3 + 18x
f'(x0) = f'(-1) = 12(-1)^3 + 18(-1) = -12 - 18 = -30
Д) f(x) = (2x - 5)^3; x0 = -1
f'(x) = 3(2x - 5)^2*(2x - 5)' = 3(2x - 5)^2*2 = 6(2x - 5)^2
f'(x0) = f'(-1) = 6(2(-1) - 5)^2 = 6(-7)^2 = 6*49 = 294
Е) f(x) = e^(x)*sin x; x0 = 0
f'(x) = (e^(x))'*sin x + e^(x)*(sin x)'
f'(x) = e^(x)*sin x + e^(x)*cos x
f'(x0) = f'(0) = e^0*sin 0 + e^0*cos 0 = 1*0 + 1*1 = 1
Ж) f(x) = x^6*ln x; x0 = 1
f'(x) = (x^6)'*ln x + x^6*(ln x)'
f'(x) = 6x^5*ln x + x^6*1/x = 6x^5*ln x + x^5
f'(x0) = f'(1) = 6*1^5*ln 1 + 1^5 = 6*0 + 1 = 1