vault backup: 2025-02-16 19:02:21

.obsidian/plugins/obsidian-git/data.json

@@ -1,27 +0,0 @@
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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. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
+## Examples