## Calculus: Early Transcendentals 8th Edition

$G'(x)=-\dfrac{C(\ln4)4^{C/x}}{x^{2}}$
$G(x)=4^{C/x}$ (Here, $C$ is a constant) Differentiate using the chain rule: $G'(x)=4^{C/x}(\ln4)(\dfrac{C}{x})'=...$ Rewrite $\dfrac{C}{x}$ as $Cx^{-1}$ and continue with the differentiation process: $...=4^{C/x}(\ln4)(\dfrac{C}{x})'=4^{C/x}(\ln4)(Cx^{-1})'=4^{C/x}(\ln4)(-Cx^{-2})$ $G'(x)=-\dfrac{C(\ln4)4^{C/x}}{x^{2}}$