Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 48

$y’=\frac{2x-3}{x(\ln 3)(x-3)}$

$y=\frac{ln(x^2-3x)}{\ln 3}$ $u=ln(x^2-3x)$ $u’=\frac{2x-3}{x^2-3x}$ $v=\ln 3$ $v’= 0$ $y’=\frac{u’v-uv’}{v^2}$ $=\frac{(\frac{2x-3}{x^2-3x})(\ln 3)-(\ln(x^2-3x))(0)}{(\ln 3)^2}$ $y’=\frac{2x-3}{x(\ln 3)(x-3)}$