展开全部
z = [cos(6x+7y)]^2
∂z/∂x = -2cos(6x+7y)sin(6x+7y)·6 = -6sin2(6x+7y)
∂z/∂y = -2cos(6x+7y)sin(6x+7y)·7 = -7sin2(6x+7y)
dz = -6sin2(6x+7y)dx - 7sin2(6x+7y)dy = - sin2(6x+7y)(6dx+7dy)
∂z/∂x = -2cos(6x+7y)sin(6x+7y)·6 = -6sin2(6x+7y)
∂z/∂y = -2cos(6x+7y)sin(6x+7y)·7 = -7sin2(6x+7y)
dz = -6sin2(6x+7y)dx - 7sin2(6x+7y)dy = - sin2(6x+7y)(6dx+7dy)
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
1. f(x) = ∫<0, x> (x-t)e^(-t^2)dt = ∫<0, x> xe^(-t^2)dt - ∫<0, x> te^(-t^2)dt
= x∫<0, x> e^(-t^2)dt - ∫<0, x> te^(-t^2)dt (对 t 积分,x相对于常量,可提到积分号外)
f'(x) = ∫<0, x> e^(-t^2)dt + xe^(-x^2) - xe^(-x^2) = ∫<0, x> e^(-t^2)dt
df(x) = f'(x)dx = [∫<0, x> e^(-t^2)dt] dx
2. dy/dx = y'<t>/x'<t> = 3t^2/(2t) = (3/2)t, t = 2 时, 切线斜率 k = (3/2)t = 3,
切点 (5,8), 切线方程 y-8 = 3(x-5), 即 3x-y-7 = 0
= x∫<0, x> e^(-t^2)dt - ∫<0, x> te^(-t^2)dt (对 t 积分,x相对于常量,可提到积分号外)
f'(x) = ∫<0, x> e^(-t^2)dt + xe^(-x^2) - xe^(-x^2) = ∫<0, x> e^(-t^2)dt
df(x) = f'(x)dx = [∫<0, x> e^(-t^2)dt] dx
2. dy/dx = y'<t>/x'<t> = 3t^2/(2t) = (3/2)t, t = 2 时, 切线斜率 k = (3/2)t = 3,
切点 (5,8), 切线方程 y-8 = 3(x-5), 即 3x-y-7 = 0
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询