diff --git "a/_posts/2024-03-01-\346\225\260\345\255\246\345\255\246\344\271\240.md" "b/_posts/2024-03-01-\346\225\260\345\255\246\345\255\246\344\271\240.md"
index 3d3bfc1..bc37112 100644
--- "a/_posts/2024-03-01-\346\225\260\345\255\246\345\255\246\344\271\240.md"
+++ "b/_posts/2024-03-01-\346\225\260\345\255\246\345\255\246\344\271\240.md"
@@ -203,9 +203,13 @@ ln 答案 = $ln(\lim_{n \to \infty} (1 + \frac{1}{n})^n)$ = 1
 
 ---
 
-$f(x) \approx f(x) + f(x)' f(x - x0)$
+$f(x) \approx f(x_0) + f'(x_0)*(x - x_0)$
 
-$\frac{\Delta f} {\Delta x} \approx f'(x_0)$ ($\Delta x = x - x_0$)
+$\frac{\Delta f} {\Delta x} \approx f'(x_0)$ ($\Delta x = x - x_0$) 当x=x0时，和上面公式一样
 
+ln 1.1 = ?
 
+x0 在 0 时，ln(1+x) ~= f(0) + f'(0)f(x-x0) ~= ln(1) + 1/1+0 * (x - 0) ~= x
+
+x0 在 0 时，(1+x)^r ~= 1 + r(1+0)^r-1 * (x - 0) ~= 1 + rx