修改几个小错。

03函数的导数.tex

@@ -135,19 +135,19 @@ Leibniz记号：记$\Delta f = f(x_0 + \delx) - f(x_0)$，那么
 \end{theorem}
 
 \begin{remark}
-    用Leibniz记号上式可化为
+    用Leibniz记号上式可记为
     \[\frac{\dif y}{\dif t} = \frac{\dif y}{\dif x} \cdot \frac{\dif x}{\dif t}\]
 \end{remark}
 
 \begin{proof}
-    $\deriv{f}(x) = \tolim{\delx}{0}\frac{\Delta y}{\Delta x}$。
+    $\deriv{f}(x) = \tolim{\delx}{0}\dfrac{\Delta y}{\Delta x}$。
     思路：
     \begin{align*}
-        \tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} & = \tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}\\
+        \tolim{\Delta t}{0} \frac{\Delta y}{\Delta t} & = \tolim{\Delta t}{0} \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}\\
         \intertext{注意：这里不能保证$\Delta x \neq 0$}
         & = \tolim{\Delta t}{0} \frac{\Delta x}{\Delta t} \cdot \tolim{\Delta x}{0} \frac{\Delta y}{\Delta x}\\
         & = \deriv{\varphi}(t) \cdot \deriv{f}(x)\\
-        & = \deriv{\varphi}(t) \cdot \deriv{f}(\varphi((t)))
+        & = \deriv{\varphi}(t) \cdot \deriv{f}(\varphi(t))
     \end{align*}
 
     因此证法为：$\Delta x \neq 0$时，引入记号