## Calculus: Early Transcendentals 8th Edition

$r'(t)=\dfrac{(\ln10)10^{2\sqrt{t}}}{\sqrt{t}}$
$r(t)=10^{2\sqrt{t}}$ Differentiate using the chain rule: $r'(t)=(10^{2\sqrt{t}})(\ln10)(2\sqrt{t})'$ Rewriting $2\sqrt{t}$ as $2t^{1/2}$, we proceed with the differentiation process: $r'(t)=(10^{2\sqrt{t}})(\ln10)(2\sqrt{t})'=(10^{2\sqrt{t}})(\ln10)(2t^{1/2})'=...$ $...=(10^{2\sqrt{t}})(\ln10)(2\dfrac{1}{2}t^{-1/2})=(10^{2\sqrt{t}})(\ln10)(t^{-1/2})$ Simplifying: $r'(t)=\dfrac{(\ln10)10^{2\sqrt{t}}}{\sqrt{t}}$