
需要知识:
-
(
x
n
)
′
=
n
x
n
−
1
(x^n)'=nx^{n-1}
(xn)′=nxn−1
-
(
s
i
n
x
)
′
=
c
o
s
x
(sinx)'=cosx
(sinx)′=cosx
-
[
f
(
g
(
x
)
)
]
′
=
f
′
(
g
(
x
)
)
×
g
′
(
x
)
[f(g(x))]'=f'(g(x))\times g'(x)
[f(g(x))]′=f′(g(x))×g′(x)
推完之后,考虑导函数与x轴的交点。可以设
t
=
c
o
s
x
t=cosx
t=cosx,解二元一次方式。然后用
a
c
o
s
(
t
)
acos(t)
acos(t) 反求
x
x
x