Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 175: 73

$$y'=\frac{1}{\theta\ln 2}$$

$$y=\log_2(5\theta)$$ Recall Theorem 7: $$\frac{d}{d\theta}\log_au=\frac{1}{u\ln a}\frac{du}{d\theta}$$ Apply here to find $y'$: $$y'=(\log_2(5\theta))'=\frac{1}{5\theta\ln 2}(5\theta)'=\frac{5}{5\theta\ln 2}$$ $$y'=\frac{1}{\theta\ln 2}$$