When \(e\) is the base of a logarithm, it becomes straightforward to derive and integration expressions involving the natural logarithm. Calculus of logarithmic functions that do not have a base of \(e\) are possible by transforming the base into \(e\) using log laws.
Here are the go-to formulas for the calculus of the natural log:
\(\frac{d}{dx} \left( \ln{ \left[ f(x) \right]} \right) = \frac{f'(x)}{f(x)} \)
\(\displaystyle \int{\frac{f'(x)}{f(x)}} \ dx = \ln{(|x|)} + c \)
Let \(y = \ln{x} \)
\(\displaystyle \begin{aligned} \frac{dy}{dx} &= \lim_{h \rightarrow 0} {\left[ \frac{\ln{(x+h)} - \ln{x}} {h} \right]} \\[12pt] &= \lim_{h \rightarrow 0} {\left[ \frac{ \ln{( \frac{x+h}{x} )}} {h} \right]} \\[12pt] &= \lim_{h \rightarrow 0} {\left[ \frac{ \ln{( 1+ \frac{h}{x} )}} {h} \right]} \ \ \text{ let } = \frac{h}{x} \\[12pt] &= \lim_{xt \rightarrow 0} {\left[ \frac{ \ln{( 1+ t )}} {tx} \right]} \\[12pt] &= \lim_{xt \rightarrow 0} {\left[ \frac{1}{x} \left( \frac{ \ln{( 1+ t )}} {t} \right) \right]} \\[12pt] &= \frac{1}{x} \times \lim_{t \rightarrow 0} {\left[ \frac{ \ln{( 1+ t )}} {t} \right]} \\[12pt] &= \frac{1}{x} \end{aligned} \)
1) Derive the following
a. \(y = \ln{(2x-4)^3} \)
\(\begin{aligned} \frac{dy}{dx} &= \frac{\left[ (2x-4)^3 \right]'}{(2x-4)^3} \\[5pt] &= \frac{3(2)(2x-4)^2}{(2x-4)^3} \\[5pt] &= \frac{6}{2x-4} \\[5pt] &= \frac{3}{x-2} \end{aligned} \)
b. \(y = \ln{(x+1)(x-1)}\)
\(\begin{aligned} y &= \ln{(x+1)} + \ln{(x-1)} \\[7pt] \frac{dy}{dx} &= \frac{1}{x+1} + \frac{1}{x-1} \\[7pt] &= \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)} \\[7pt] &= \frac{(x-1)+(x+1)}{(x+1)(x-1)} \\[7pt] &= \frac{2x}{(x+1)(x-1)} \end{aligned} \)
c. \(y = \ln \left[{\frac{x+1}{x^2 +1} } \right]\)
\(\begin{aligned} y &= \ln{(x+1)} - \ln{(x^2+1)} \\[7pt] \frac{dy}{dx} &= \frac{1}{x+1} - \frac{2x}{x^2+1} \\[7pt] &= \frac{x^2+1}{(x+1)(x^2+1)} - \frac{2x(x+1)}{(x^2+1)(x+1)} \\[7pt] &= \frac{x^2+1-2x^2-2x}{(x+1)(x^2+1)} \\[7pt] &= -\frac{x^2+2x-1}{(x+1)(x^2+1)} \end{aligned} \)
2) Calculate the following
a. \(f(x) = \int{tan(2x) \ dx} \)
\(\begin{aligned} f(x) &= \int{\frac{\sin{(2x)}}{\cos{(2x)}} \ dx} \\[5pt] &= -\frac{1}{2} \int{ \frac{-2 \sin(2x)}{\cos{(2x)}} \ dx} \\[5pt] &= -\frac{ \ln{| \cos{(2x)}| } }{2} + c \end{aligned} \)
b. \(f(x) = \int{\frac{e^{2x}}{1-e^{2x}} \ dx} \)
\(\begin{aligned} f(x) &= -\frac{1}{2} \int{\frac{-2e^{2x}}{1-e^{2x}} \ dx} \\[5pt] &= - \frac{\ln{|1-e^{2x}|}}{2} + c \end{aligned} \)
\(\begin{aligned} \text{Let } g(x) &= 1-e^{2x} \\[5pt] g'(x) &= -2e^{2x} \end{aligned} \)
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