Calculus: Early Transcendentals 8th Edition

$f'(x)=\dfrac{5}{2\sqrt{5x+1}}$
$f(x)=\sqrt{5x+1}$ Let's write the function like this: $f(x)=(5x+1)^{1/2}$ Differentiate using the chain rule: $f'(x)=\dfrac{1}{2}(5x+1)^{-1/2}(5x+1)'=...$ $...=\dfrac{1}{2}(5x+1)^{-1/2}(5)=\dfrac{5}{2}(5x+1)^{-1/2}=\dfrac{5}{2(5x+1)^{1/2}}=...$ $...=\dfrac{5}{2\sqrt{5x+1}}$