Answer
(a)$f(x)=g(x)h(x)=3\ln (2+5x).(\ln x)^{-1}$
(b)$f^{'}(x)=\frac{15}{2+5x}(\ln x)^{-1}-3\ln(2+5x)(\ln x)^{-2}\frac{1}{x}$
Work Step by Step
$g(x)=3\ln(2+5x);h(x)=(\ln x)^{-1}$
(a)
Let $f(x)=g(x)h(x)$
$f(x)=g(x)h(x)=3\ln (2+5x).(\ln x)^{-1}$
(b)
$f^{'}(x)=g^{'}(x)h(x)+g(x)h^{'}(x)$
$f^{'}(x)=3\frac{1}{2+5x}(5)(\ln x)^{-1}+3\ln(2+5x)(-1)(\ln x)^{-2}\frac{1}{x}$
$f^{'}(x)=\frac{15}{2+5x}(\ln x)^{-1}-3\ln(2+5x)(\ln x)^{-2}\frac{1}{x}$
