Second Order Derivatives in Becker's Crime and Punishment

Question

I'm trying to understand Becker's seminal paper Crime and Punishment (1968) particularly the parameter of cost of apprehension and conviction and their second order partial derivatives.

The paper states the costs of various activities to apprehend is given by the function $C = C(A)$ where $C' = \frac{dC}{dA}$ and that activities can be approximate by the clearance rate $(p)$ and number of offences $(O)$ by the formula $A ≅ pO$

Now, partially differentiating once with respect to each $(p)$ and $(O)$ gives the following

$$C_p=\frac{∂C(pO)}{∂p} = C'O \quad \textrm{and;} \quad C_O = C'p \quad \textrm{respectively} \quad$$

Becker then says if $pO ≠ 0$ then an increase in either probability of conviction or number of offenses would increase total costs. This is where my confusion begins, when he discusses the marginal cost of increased activity given by second order differentiation where:

$$C_{pp}= C''O^2;\\ C_{OO}= C''p^2\\ C_{pO}=C_{Op}= C''pO + C'$$

Put simply, why is there an added $C'$ for the cross-partial differentiation results? I went through my calculus books and tried to figure out the chain rule application but to no avail.

Answer 1

You have to apply both chain and product rule for derivatives here. In this case:

$$C_p'=\frac{\partial C(pO)}{\partial p}=C_p'(pO)O$$

But note when calculating cross-derivative you take the 2nd order derivative wrt O:

$$\frac{\partial C_p'(pO)O}{\partial O} = \frac{\partial }{\partial O}[C_p'(pO)]\cdot O + C_p'(pO) \cdot\frac{\partial}{\partial O} [O] \\ = C_{pO}''(pO) \cdot pO + C_p'(pO) \cdot 1 $$

Omitting subscripts this is equivalent to: $C'' pO + C'$.

To sum up, $C'$ comes from the product rule for derivatives (i.e. $\frac{d}{dx}[f(x)g(x)]= f'(x)g(x)+ f(x)g'(x)$ - you can learn more about this rule in standard math textbooks such as EMEA by Hammond et al.).