Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 226: 53

$\frac{-4c^2+40c+49900}{100}$

We are given $D(p)=\frac{-p^2}{100}+500$ and $p(c)=2c-10$ The demand in terms of the cost is: $D(p(c))=D(2c-10)$ $=\frac{-(2c-10)^2}{100}+500$ $=\frac{-(4c^2-40c+100)+50000}{100}$ $=\frac{-4c^2+40c+49900}{100}$