z=ln(x+e^(-y))
dz/dx=1/(x+e^(-y))*(x+e^(-y))'=1/(x+e^(-y))
d2z/dx2=((x+e^(-y))^(-1))'=-(x+e^(-y))^(-2)*(x+e^(-y))'=-1/(x+e^(-y))^2
d3z/dx2dy=(-(x+e^(-y))^(-2))'=-(-2(x+e^(-y)))^(-3)*(x+e^(-y))'=2(x+e^(-y))^(-3)*(-e^(-y))=-2e^(-y)/(x+e^(-y))^3
dz/dy=1/(x+e^(-y))*(x+e^(-y))'=1/(x+e^(-y))*(-e^(-y))=-e^(-y)/(x+e^(-y))
d2z/dydx=(-e^(-y)*(x+e^(-y))^(-1))'=-e^(-y)*((x+e^(-y))^(-1))'=
-e^(-y)*(-((x+e^(-y))^(-2)))*(x+e^(-y))'=e^(-y)/(x+e^(-y))^2
d3z/dydx2=(e^(-y)/(x+e^(-y))^2)'=e^(-y)((x+e^(-y))^(-2))'=
e^(-y)*(-2((x+e^(-y))^(-3)))*(x+e^(-y))'=-2e^(-y)/(x+e^(-y))^3
и все
-2e^(-y)/(x+e^(-y))^3-(-2e^(-y)/(x+e^(-y))^3)=-2e^(-y)/(x+e^(-y))^3+2e^(-y)/(x+e^(-y))^3=0
Объяснение:
9.
log₁₄ 7 = m найдем log₁₇₅ 56 - ?
log₁₄ 5 = n
Используем формулу перехода другому основанию:
log₁₇₅ 56 = log₁₄ 56/log₁₄ 175 = log₁₄ (8×7)/log₁₄ (25×7) = log₁₄ (2³×7)/log₁₄ (5²×7) = log₁₄ 2³ × log₁₄ 7/log₁₄ 5² × log₁₄ 7 = 3log₁₄ 2 × log₁₄ 7/2log₁₄ 5 × log₁₄ 7
Нам нужно найти log₁₄ 2:
log₁₄ 2 = log₁₄ 14/7 = log₁₄ 14 - log₁₄ 7 = 1 - m
Получаем:
log₁₇₅ 56 = 3×(1 - m) + m/2n + m = 3 - 3m + m/2n + m = 3 - 2m/2n + m
ответ: log₁₇₅ 56 = 3 - 2m/2n + m
10.
log₅ 5 = 1
log₁₁ 15 = log₁₀ 15/log₁₀ 11 ≈ 1,17609/1,04139 ≈ 1,12934
Следовательно:
1 < 1,12934
log₅ 5 < log₁₁ 15
ответ: log₅ 5 < log₁₁ 15
