如题，

供参考：
设曲线方程 y=f(x) 求过定点(x0,y0)的切线。
设切线为： y=k*(x-x0)+y0
设切点为：(x1,y1)
则 y1=f(x1)=k*(x1-x0)+y0
又根据切线的性质，有
f'(x1)==k 于是方程
f(x) = f'(x)*(x-x0)+y0
的根就是我们要求的x1。

你打算怎么求f'(x)呢

Private Sub Command2_Click()
Dim e As Single, k As Single, y2 As Single, y As Single, a As Single
yd = 0.8386
xd = 0.8386
For i = 0.8382 To 0 Step -0.01
x = i
If x <= 0.3 Then
a = 1.1213 * (x + 0.2) ^ (-1.5235)
Else
a = 0.8938 * x ^ (-1.0632)
End If
Next i
y = a * x / (1 + (a - 1) * x)
While e >= 10 ^ (-4)
k = (yd - y) / (xd - x)
y2 = (x * (1 + (a - 1) * x) - a * x * (a - 1)) / (1 + (a - 1) * x) ^ 2
e = Abs(k - y2)
Wend
Print "x="; x, "y="; y
End Sub
帮我看看哪里错了！谢谢

你怎么知道错了的呢？还是，这里的数都是哪来的？看不明白，好歹你把什么方程发来看看哈，要不打个包发上来好了