Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 49

$h’(t)=-\frac{2}{(4-t)(\ln 5)}$

$h(t)=\frac{2\ln(4-t)}{\ln 5}$ $u=2\ln(4-t)$ $u’=\frac{-2}{4-t}$ $v=\ln 5$ $v’=0$ $h’(t)=\frac{u’v-uv’}{v^2}$ $=\frac{\left(-\frac{2}{4-t}\right)(\ln5)-(2)(\ln(4-t))(0)}{(\ln 5)^2}$ $h’(t)=-\frac{2}{(4-t)(\ln 5)}$