diff --git a/education/math/MATH1210 (calc 1)/Derivatives.md b/education/math/MATH1210 (calc 1)/Derivatives.md
index dc85cdf..1d3fa69 100644
--- a/education/math/MATH1210 (calc 1)/Derivatives.md	
+++ b/education/math/MATH1210 (calc 1)/Derivatives.md	
@@ -28,7 +28,13 @@ The above formula can be used to find the *derivative*. This may also be referre
 $$  y - y_1 = m(x-x_1) $$
 Given that $m = f'(a)$ and that $(x_1, y_1) = (a, f(a))$, you get the equation:
  $$ y - f(a) = f'(a)(x - a) $$
-As a more practical example, given an equation with a slope of $6$, 
+As a more practical example, given an equation with a slope of $6$ at the point $(-2, -4)$:
+$$ y - (-4) = 6(x - -2)$$
+Solving for $y$ looks like this:
+1. $y + 4 = 6(x + 2)$
+2. $y = 6(x + 2) - 4$ 
+3. $y = 6x + 12 - 4$
+4. $y = 6x + 8$ 
 # Line Types
 ## Secant Line
 A **Secant Line** connects two points on a graph.
@@ -123,9 +129,9 @@ $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
 > 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 $g(x) = x^2 + 3)$.
-6. First find the derivative of the outside function function ($f(x) = x^4$):
+5. First find the derivative of the outside function function ($f(x) = x^4$):
 $$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
-7. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
+6. 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 $x^4$
 
@@ -161,8 +167,8 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 - Given the equation $y = x^2$, $\dfrac{d}{dx} y = \dfrac{dy}{dx} = 2x$.
 
 Given these facts:
-8. Let $y$ be some function of $x$
-9. $\dfrac{d}{dx} x = 1$
-10. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
+7. Let $y$ be some function of $x$
+8. $\dfrac{d}{dx} x = 1$
+9. $\dfrac{d}{dx} y = \dfrac{dy}{dx}$\
 
 ## Examples