vault backup: 2025-02-16 18:52:21

education/math/MATH1210 (calc 1)/Derivatives.md

@@ -121,7 +121,7 @@ $$ \dfrac{d}{dx} (x^2 + 3)^4 = 4(x^2 + 3)^3 * (2x)$$
 > 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:
-$$ 4(x)^3 * x $$
+$$ 4(x)^3 * (1) $$
 # Trig Functions
 $$ \lim_{x \to 0} \dfrac{\sin x}{x} = 1 $$
 $$ \lim_{x \to 0} \dfrac{\cos x - 1}{x} = 0 $$
@@ -150,12 +150,20 @@ $$ \dfrac{d}{dx} \cot x = -\csc^2 x $$
 - $\dfrac{d}{dx} x = \dfrac{dx}{dx} = 1$ : The derivative of $x$ with respect to $x$ is one
 - $\dfrac{d}{dx} y = \dfrac{dy}{dx} = y'$ 
 - 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 $y^2$?
+$\dfrac{d}{dx} y^2 = 2(y)^1 *\dfrac{dy}{dx}$ 
+
 # Examples
 
 > Differentiate $f(x) = 4\sqrt[3]{x} - \dfrac{1}{x^6}$ 
 
-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. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$
+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. $= \dfrac{4}{3\sqrt[3]{x^2}} + \dfrac{6}{x^7}$