请问(1-x)^1⼀2的导数是什么

2025-03-14 10:59:48
推荐回答(3个)
回答(1):

arctanx+C的导数是1/(1+x^2)。C为常数。
解答过程如下:
f(x)=arctanx+C,令y=arctanx;则x=tany
因为
f'(x)=(arctanx)'+0
=1/(tany)'
=1/(siny/cosy)'
=1/[(cos^2y+sin^2y)/cos^2y]
=1/(1+tan^2y)
=1/(1+x^2)
扩展资料:
商的导数公式:
(u/v)'=[u*v^(-1)]'
=u'
*
[v^(-1)]
+[v^(-1)]'
*
u
=
u'
*
[v^(-1)]
+
(-1)v^(-2)*v'
*
u
=u'/v
-
u*v'/(v^2)
通分,易得
(u/v)=(u'v-uv')/v²
常用导数公式:
1.y=c(c为常数)
y'=0
2.y=x^n
y'=nx^(n-1)
3.y=a^x
y'=a^xlna,y=e^x
y'=e^x
4.y=logax
y'=logae/x,y=lnx
y'=1/x
5.y=sinx
y'=cosx
6.y=cosx
y'=-sinx
7.y=tanx
y'=1/cos^2x
8.y=cotx
y'=-1/sin^2x
9.y=arcsinx
y'=1/√1-x^2
10.y=arccosx
y'=-1/√1-x^2

回答(2):

这是个复合函数,外层函数为y^(1/2),内层函数为1-x
所以复合函数的导数=外层函数的导数 * 内层函数的导数
y = (1-x)^(1/2)
y' = [(1-x)^(1/2)]',先对外层求导
y' = (1/2)(1-x)^(1/2-1) * (1-x)',后对内层函数求导,并且与前者相乘
y' = (1/2)(1-x)^(-1/2) * (0-1)
y' = (-1/2)(1-x)^(-1/2)
= (-1/2) * 1/(1-x)^(1/2)
= -1 / [2√(1-x)]

导数公式:(a^n)' = n * a^(n - 1),n是任意常数
复合函数的求导公式:对于y = f[g(x)],y' = f'[g(x)] * g'(x)
或者是dy / dx = df[g(x)] / dg(x) * dg(x) / dx
若设y = f(u),u = g(x),则y = f'(u) * u' = f'(u) * g'(x) = f'[g(x)] * g'(x)
或者是dy / dx = dy / du * du / dx
「y'」表示函数y的导数
「√」表示根号

回答(3):

y = (1-x)^(1/2)
y' = (1/2)(1-x)^(-1/2) . (-1)
= (-1/2)(1-x)^(-1/2)

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