## Calculus, 10th Edition (Anton)

=$2x^{3}$
Distribute the $\frac{1}{2}$ through the $x^{4}+7$ to get $\frac{1}{2}x^{4}+\frac{7}{2}$. Then, use the power rule for derivatives ($[x^{n}]'=nx^{n-1}$). This gives you $\frac{4}{2}x^{3}+0=2x^{3}$.