# Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 8

$$\frac{{dy}}{{dx}} = \frac{3}{{\left( {\ln 5} \right)\left( {3x - 7} \right)}}$$

#### Work Step by Step

\eqalign{ & y = {\log _5}\left( {3x - 7} \right) \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\log }_5}\left( {3x - 7} \right)} \right] \cr & {\text{we can use the formula }}\cr &\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a}} \cdot \frac{1}{u}\frac{{du}}{{dx}}.{\text{ }}\left( {page\,\,390} \right) \cr & {\text{where }}u{\text{ is any differentiable function of }}x \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln 5}}\left( {\frac{1}{{3x - 7}}} \right)\frac{d}{{dx}}\left[ {3x - 7} \right] \cr & {\text{then}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln 5}}\left( {\frac{1}{{3x - 7}}} \right)\left( 3 \right) \cr & \frac{{dy}}{{dx}} = \frac{3}{{\left( {\ln 5} \right)\left( {3x - 7} \right)}} \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.