Answer
$f'(x)=\frac{6+5\sqrt{x}}{6\sqrt[3]{x^2}}$
Work Step by Step
Product Rule $f'(x)=((u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=\sqrt[3]{x} ;u’(x)=\frac{1}{3\sqrt[3]{x^2}} $
$v(x)=3+\sqrt{x} ;v’(x)=\frac{1}{2\sqrt{x}} $
$f'(x)=(\frac{1}{3\sqrt[3]{x^2}})(3+\sqrt{x})+(\frac{1}{2\sqrt{x}})(\sqrt[3]{x})$
$=\frac{3+\sqrt{x}}{3\sqrt[3]{x^2}}+\frac{1}{2\sqrt[6]{x}}=\frac{6+5\sqrt{x}}{6\sqrt[3]{x^2}}$
