Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 51

$y’=\frac{x}{(x^2-1)(\ln 5)}$

$y=\frac{(\frac{1}{2})\ln(x^2-1)}{\ln 5}$ $u=\frac{1}{2} \ln(x^2-1)$ $u’=\frac{x}{x^2-1}$ $v=\ln 5$ $v’=0$ $y’=\frac{u’v-uv’}{v^2}$ $=\frac{\left(\frac{x}{x^2-1}\right)(\ln 5)-(\frac{1}{2} \ln(x^2-1))(0)}{(\ln 5)^2}$ $y’=\frac{x}{(x^2-1)(\ln 5)}$