vault backup: 2025-02-16 18:47:21

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

@@ -1,27 +0,0 @@
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education/math/MATH1210 (calc 1)/Derivatives.md

@@ -115,8 +115,13 @@ $$ \dfrac{d}{dx} f(g(x)) = f'(g(x))*g'(x) $$
 
 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 ($f(x) = x^4$):
-$$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 $$
-2. 
+$$ \dfrac{d}{dx} (x^2 +3)^4 = 4(g(x))^3 ...$$
+2. Multiply that by the derivative of the inside function, $g(x)$, or $x^2 + 3$. 
+$$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$ 
+> Apply the chain rule to $x^4$
+
+If we treat the above as a function along the lines of $f(x) = (x)^4$, and $g(x) = x$, then the chain rule can be used like so:
+$$ 4(x)^3 * x $$
 # Trig Functions
 $$ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 $$
 $$ \lim_{x \to 0} \dfrac{\cos x - 1}{x} = 0 $$
@@ -149,8 +154,8 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 
 > Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$ 
 
-3. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
-4. $= 4x^\frac{1}{3} - x^{-6}$  
-5. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$ 
-6. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
-7. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
+2. $f(x) = 4\sqrt[3]{x} = \dfrac{1}{x^6}$
+3. $= 4x^\frac{1}{3} - x^{-6}$  
+4. $f'(x) = \dfrac{1}{3} * 4x^{-\frac{2}{3}} -(-6)(x^{-6-1})$ 
+5. $= 4x^{-2-\frac{2}{3}} + 6x^{-7}$
+6. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$