[tex]c(n)=30+5n\\a(n)=\dfrac{c(n)}{n}[/tex]
[tex]\dfrac{d}{dn}(a(n))\quad\text{at n=10}[/tex]
[tex]\dfrac{d}{dn}(a(n))=\dfrac{d}{dn}\left(\dfrac{30+5n}{n}\right)=\dfrac{d}{dn}\left(30n^{-1}+5\right)\\\\=-30n^{-2}\\\\=-30\cdot 10^{-2}\qquad\text{at n=10}\\\\=-0.30[/tex]
The derivative of the average cost function at n=10 is -0.30.