Logarithmic Differentiation

Problem 116: Find $dy/dx = y'$ using the method of logarithmic differentiation when $y = x^{9\tan x}$ .

Solution: We need to use property of logarithm to solve this problem. That is,

(1) $\begin{align*} y & = x^{9 \tan x} \\ \ln(y) & = \ln\left(x^{9\tan x}\right) \\ \ln(y) & = 9 \cdot \tan(x) \cdot \ln(x) \quad \text{(Property of logarithm)} \\ \frac{1}{y} \cdot y' & = \left( 9\sec^2(x) \ln(x) + \frac{9\tan(x)}{x}\right) \\ y' & = x^{9\tan(x)} \cdot \left( 9\sec^2(x) \ln(x) + \frac{9\tan(x)}{x}\right) \\ & = 9x^{9\tan(x)} \cdot \left( \sec^2(x) \ln(x) + \frac{\tan(x)}{x}\right). \end{align*}$

Hence,

(2) $\begin{equation*} \boxed{y' = 9x^{9\tan(x)} \cdot \left( \sec^2(x) \ln(x) + \frac{\tan(x)}{x}\right)}.\end{equation*}$

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