利用导数定义,求y=√2x+1 的导函数
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/04 18:07:18
利用导数定义,求y=√2x+1 的导函数
是用导数定义
是用导数定义
![利用导数定义,求y=√2x+1 的导函数](/uploads/image/z/8103250-10-0.jpg?t=%E5%88%A9%E7%94%A8%E5%AF%BC%E6%95%B0%E5%AE%9A%E4%B9%89%2C%E6%B1%82y%3D%E2%88%9A2x%2B1+%E7%9A%84%E5%AF%BC%E5%87%BD%E6%95%B0)
看的不是很清楚,给三个可能性
y=(√2)x+1
y'=lim[h→0] [√2(x+h)+1-√2x-1]/h
=lim[h→0] (√2x+√2h-√2x)/h
=lim[h→0] √2h/h
=lim[h→0] √2
=√2
y=√(2x)+1
y'=lim[h→0] {√[2(x+h)]+1-√(2x)-1}/h
=lim[h→0] [√(2x+2h)-√(2x)]/h
=lim[h→0] 1/h*[√(2x+2h)-√(2x)][√(2x+2h)+√(2x)]/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*(2x+2h-2x)/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*2h/[√(2x+2h)+√(2x)]
=lim[h→0] 2/[√(2x+2h)+√(2x)]
=2/[√(2x)+√(2x)]
=1/√(2x)
y=√(2x+1)
y'=lim[h→0] {√[2(x+h)+1]-√(2x+1)}/h
=lim[h→0] [√(2x+2h+1)-√(2x+1)]/h
=lim[h→0] 1/h*[√(2x+2h+1)-√(2x+1)][√(2x+2h+1)+√(2x+1)]/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*(2x+2h+1-2x-1)/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*2h/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 2/[√(2x+2h+1)+√(2x+1)]
=2/[√(2x+1)+√(2x+1)]
=1/√(2x+1)
y=(√2)x+1
y'=lim[h→0] [√2(x+h)+1-√2x-1]/h
=lim[h→0] (√2x+√2h-√2x)/h
=lim[h→0] √2h/h
=lim[h→0] √2
=√2
y=√(2x)+1
y'=lim[h→0] {√[2(x+h)]+1-√(2x)-1}/h
=lim[h→0] [√(2x+2h)-√(2x)]/h
=lim[h→0] 1/h*[√(2x+2h)-√(2x)][√(2x+2h)+√(2x)]/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*(2x+2h-2x)/[√(2x+2h)+√(2x)]
=lim[h→0] 1/h*2h/[√(2x+2h)+√(2x)]
=lim[h→0] 2/[√(2x+2h)+√(2x)]
=2/[√(2x)+√(2x)]
=1/√(2x)
y=√(2x+1)
y'=lim[h→0] {√[2(x+h)+1]-√(2x+1)}/h
=lim[h→0] [√(2x+2h+1)-√(2x+1)]/h
=lim[h→0] 1/h*[√(2x+2h+1)-√(2x+1)][√(2x+2h+1)+√(2x+1)]/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*(2x+2h+1-2x-1)/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 1/h*2h/[√(2x+2h+1)+√(2x+1)]
=lim[h→0] 2/[√(2x+2h+1)+√(2x+1)]
=2/[√(2x+1)+√(2x+1)]
=1/√(2x+1)