<p>You have to apply both chain and <a href="https://en.wikipedia.org/wiki/Product_rule" rel="nofollow noreferrer">product rule</a> for derivatives here. In this case:</p>
<p><span class="math-container">$$C_p'=\frac{\partial C(pO)}{\partial p}=C_p'(pO)O$$</span></p>
<p>But note when calculating cross-derivative <strong>you take the 2nd order derivative wrt O</strong>:</p>
<p><span class="math-container">$$\frac{\partial C_p'(pO)O}{\partial O} =  \frac{\partial }{\partial O}[C_p'(pO)]\cdot O +  C_p'(pO)  \cdot\frac{\partial}{\partial O} [O] \\
= C_{pO}''(pO) \cdot pO + C_p'(pO)  \cdot 1 $$</span></p>
<p>Omitting subscripts this is equivalent to: <span class="math-container">$C'' pO + C'$</span>.</p>
<p>To sum up, <span class="math-container">$C'$</span> comes from the product rule for derivatives (i.e. <span class="math-container">$\frac{d}{dx}[f(x)g(x)]= f'(x)g(x)+ f(x)g'(x)$</span> - you can learn more about this rule in standard math textbooks such as EMEA by Hammond et al.).</p>

CC BY-SA 4.0
2021-02-28T12:25:45.153Z
Thank you so very much sir! I finally get it now :)
Ardhi
CC BY-SA 4.0
2021-02-28T12:26:40.177Z
@Ardhi you are welcome, if you think the answer above answered your question consider accepting it
1muflon1
CC BY-SA 4.0
2021-02-28T11:50:32.463Z
2021-02-28T11:58:01.490Z
2021-02-28T11:58:01.490Z
1muflon1
<p>I'm trying to understand Becker's seminal paper <em>Crime and Punishment (1968)</em> particularly the parameter of cost of apprehension and conviction and their second order partial derivatives.</p>
<p>The paper states the costs of various activities to apprehend is given by the function <span class="math-container">$C = C(A)$</span> where <span class="math-container">$C' = \frac{dC}{dA}$</span> and that activities can be approximate by the clearance rate <span class="math-container">$(p)$</span> and number of offences <span class="math-container">$(O)$</span> by the formula <span class="math-container">$A ≅ pO$</span></p>
<p>Now, partially differentiating once with respect to each <span class="math-container">$(p)$</span> and <span class="math-container">$(O)$</span> gives the following</p>
<p><span class="math-container">$$C_p=\frac{∂C(pO)}{∂p} = C'O \quad \textrm{and;} \quad C_O = C'p \quad \textrm{respectively} \quad$$</span></p>
<p>Becker then says if <span class="math-container">$pO ≠ 0$</span> 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:</p>
<p><span class="math-container">$$C_{pp}= C''O^2;\\
C_{OO}= C''p^2\\
C_{pO}=C_{Op}= C''pO + C'$$</span></p>
<p>Put simply, why is there an added <span class="math-container">$C'$</span> 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.</p>

<cost-functions>
Second Order Derivatives in Becker's Crime and Punishment
Ardhi
# Second Order Derivatives in Becker's Crime and Punishment

(Asked by: username_0 on 2021-02-28)

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.

**Tags:** cost-functions


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## 1 answers

### Answer by username_1 on 2021-02-28. Score: 0 | Accepted answer

You have to apply both chain and [product rule](https://en.wikipedia.org/wiki/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.).

> **Comments:**
> **username_0**: Thank you so very much sir! I finally get it now :)  
> **username_1**: @username_0 you are welcome, if you think the answer above answered your question consider accepting it  

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