Answer
(a)$f(x)=g(x)h(x)=(6e^{-x}+\ln x) (4x^{2.1})$
(b)$f^{'}(x)=(4x^{2.1})[-6e^{-x} +\frac{1}{x}]+8.4(6e^{-x}+\ln x)(x^{1.1})$
Work Step by Step
$g(x)=6e^{-x}+\ln x ; h(x)=4x^{2.1}$
(a)
Let $f(x)=g(x)h(x)$
Then
$f(x)=g(x)h(x)=(6e^{-x}+\ln x) (4x^{2.1})$
(b)
Taking derivative with respect to $x$ of $f(x)$
$f^{'}(x)=\frac{d[( (6e^{-x}+\ln x) (4x^{2.1}) )]}{dx}$
Using the product rule
$f^{'}(x)=(4x^{2.1})\frac{d(6e^{-x}+\ln x )}{dx} +(6e^{-x}+ \ln x)\frac{d(4x^{2.1})}{dx}$
$f^{'}(x)=(4x^{2.1})[\frac{d(6e^{-x})}{dx}+\frac{d(\ln x)}{dx}]+4(6e^{-x}+\ln x)\frac{dx^{2.1}}{dx}$
$f^{'}(x)=(4x^{2.1})[6\frac{d(e^{-x})}{dx}+\frac{1}{x}]+4(6e^{-x}+\ln x)(2.1)x^{2.1-1}$
$f^{'}(x)=(4x^{2.1})[6e^{-x} \frac{d(-x)}{dx}+\frac{1}{x}]+8.4(6e^{-x}+\ln x)(x^{1.1})$
$f^{'}(x)=(4x^{2.1})[-6e^{-x} +\frac{1}{x}]+8.4(6e^{-x}+\ln x)(x^{1.1})$
