diff --git a/.obsidian/plugins/obsidian-git/data.json b/.obsidian/plugins/obsidian-git/data.json
index bef4c6e..e69de29 100644
--- a/.obsidian/plugins/obsidian-git/data.json
+++ b/.obsidian/plugins/obsidian-git/data.json
@@ -1,27 +0,0 @@
-{
-  "commitMessage": "vault backup: {{date}}",
-  "autoCommitMessage": "vault backup: {{date}}",
-  "commitDateFormat": "YYYY-MM-DD HH:mm:ss",
-  "autoSaveInterval": 5,
-  "autoPushInterval": 0,
-  "autoPullInterval": 5,
-  "autoPullOnBoot": true,
-  "disablePush": false,
-  "pullBeforePush": true,
-  "disablePopups": false,
-  "listChangedFilesInMessageBody": false,
-  "showStatusBar": true,
-  "updateSubmodules": false,
-  "syncMethod": "merge",
-  "customMessageOnAutoBackup": false,
-  "autoBackupAfterFileChange": false,
-  "treeStructure": false,
-  "refreshSourceControl": true,
-  "basePath": "",
-  "differentIntervalCommitAndPush": false,
-  "changedFilesInStatusBar": false,
-  "showedMobileNotice": true,
-  "refreshSourceControlTimer": 7000,
-  "showBranchStatusBar": true,
-  "setLastSaveToLastCommit": false
-}
\ No newline at end of file
diff --git a/education/math/MATH1210 (calc 1)/Derivatives.md b/education/math/MATH1210 (calc 1)/Derivatives.md
index 90fa3d5..03f5542 100644
--- a/education/math/MATH1210 (calc 1)/Derivatives.md	
+++ b/education/math/MATH1210 (calc 1)/Derivatives.md	
@@ -15,6 +15,15 @@ If we have the coordinate pair $(a, f(a))$, and the value $h$ is the distance be
 $$\lim_{h \to 0}\dfrac{f(a + h) - f(a)}{h}$$
 The above formula can be used to find the *derivative*. This may also be referred to as the *instantaneous velocity*, or the *instantaneous rate of change*.
 
+## Examples
+
+> Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$ 
+
+1. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
+2. $= 4x^\frac{1}{3} - x^{-6}$  
+3. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$ 
+4. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
+5. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
 # Point Slope Formula (Review)
 $$  y - y_1 = m(x-x_1) $$
 Given that $m = f'(a)$ and that $(x_1, y_1) = (a, f(a))$, you get the equation:
@@ -114,9 +123,9 @@ $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
 > Given the function $(x^2+3)^4$, find the derivative.
 
 Using the chain rule, the above function might be described as $f(g(x))$, where $f(x) = x^4$, and $g(x) = x^2 + 3)$.
-1. First find the derivative of the outside function function ($f(x) = x^4$):
+6. First find the derivative of the outside function function ($f(x) = x^4$):
 $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
-2. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
+7. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
 $$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$ 
 > Apply the chain rule to $x^4$
 
@@ -152,18 +161,8 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$.
 
 Given these facts:
-1. Let $y$ be some function of $x$
-2. $\dfrac{d}{dx} x = 1$
-3. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
-What's the derivative of $y^2$?
-$\dfrac{d}{dx} y^2 = 2(y)^1 *\dfrac{dy}{dx}$ 
+8. Let $y$ be some function of $x$
+9. $\dfrac{d}{dx} x = 1$
+10. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
 
-# Examples
-
-> Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$ 
-
-4. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
-5. $= 4x^\frac{1}{3} - x^{-6}$  
-6. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$ 
-7. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
-8. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
\ No newline at end of file
+## Examples