Derivative at A Point

Problem 125: Calculate $f'(1)$ of

(1) $\begin{equation*} f(x) = \ln\left(\frac{e^{2x}}{x(x-2)^2} \right). \end{equation*}$

Solution: We have two different options to find the derivative of (1). First, I will expand the logarithm using properties of logarithm. Then I will find the derivative and evaluate. That is,

(2) $\begin{align*} f(x) & = \ln\left(\frac{e^{2x}}{x(x-2)^2} \right) \\ & = \ln(e^{2x}) - \ln(x(x-2)^2) \qquad (\ln(a/b) = \ln(a) - \ln(b)) \\ & = 2x\ln(e) - \ln(x) - \ln(x-2)^2 \qquad (\ln(a \cdot b) = \ln(a) + \ln(b)) \\ & = 2x - \ln(x) - 2 \ln(x-2) \qquad (\ln(a^b) = b\ln(a)).\end{align*}$

Hence,

(3) $\begin{align*} f(x) & = 2x - \ln(x) - 2\ln(x-2) \\ f'(x) & = 2 - \frac{1}{x} - \frac{2}{x-2} \\ & = 2 - \frac{1}{1} - \frac{2}{1-2} \quad \text{at} \ x = 1 \\ & = 3. \end{align*}$

Therefore, $f'(1) = 3$ .

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