Answer
$f^{'}(x)= [\frac{12}{x} (17-3\ln 4x)+(19+12\ln x)(-\frac{3}{x})]$
Work Step by Step
$f(x)=(19+12\ln x)(17-3\ln 4x)$
$f^{'}(x)= [\frac{d(19+12\ln x) }{dx}] (17-3\ln 4x)+(19+12\ln x)\frac{d(17-3 \ln 4x)}{dx}]$
$f^{'}(x)= [\frac{12}{x} (17-3\ln 4x)+(19+12\ln x)(-3\frac{1}{4x}(4))]$
$f^{'}(x)= [\frac{12}{x} (17-3\ln 4x)+(19+12\ln x)(-\frac{3}{x})]$
