cleanup (extraneous file)

sections/derivation.tex

@@ -1,43 +0,0 @@
-
-$ z^2 = \sqrt{x^2 + y^2} $
-
-%Logarithm:  $ L(ab) = L(a) + L(b) $
-
-%Exponent:  $ E(a + b) = E(a)E(b) $
-
-Derivative:
-
-\[ f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h} \]
-
-So for the Log function $ L(x) $, then:
-
-\[ L'(x) = \lim_{h\to 0} \frac{L(x + h) - L(x)}{h} \]
-
-Since $ L(a) - L(b) = L(\frac{a}{b})$ then:
-
-\[ \Rightarrow L'(x) = \lim_{h\to 0} \frac{L(\frac{x + h}{x})}{h} = \lim_{h\to 0} \frac{L(1 + \frac{h}{x})}{h} = \lim_{h\to 0} \frac{1}{h}L(1 + \frac{h}{x}) \]
-
-Since $ bL(a) = L(a^b) $ then:
-
-\[ \Rightarrow L'(x) = \lim_{h\to 0} L(1 + \frac{h}{x})^{\frac{1}{h}} \]
-
-Substituting $ j = \frac{h}{x} \Rightarrow h = xj $ then:
-
-\[ \Rightarrow L'(x) = \lim_{xj\to 0} L(1 + j)^{\frac{1}{xj}} = \lim_{j\to 0} L(1 + j)^\frac{1}{xj} \]
-
-or, reverting the limit back to $h$,
-
-\[ = \lim_{h\to 0} L(1 + h)^\frac{1}{xh} \]
-
-Since \[ \lim_{h\to 0} (1 + h)^{\frac{x}{h}} = e^x \Rightarrow \lim_{h\to 0} (1 + h)^{\frac{1}{xh}} = e^\frac{1}{x} \]
-
-then:
-
-\[ \Rightarrow L'(x) = \lim_{h\to 0} L(e^\frac{1}{x}) = \frac{1}{x} \]
-
-for $ L(x) = \log_e(x) $
-
-
-
-
-