Section 3
Second-order Partial Derivatives
The partial derivative of a function of \[n\] variables, is itself a function of \[n\] variables. By taking the partial derivatives of the partial derivatives, we compute the higher-order derivatives. Higher-order derivatives are important to check the concavity of a function, to confirm whether an extreme point of a function is max or min, etc.
Given that a function \[f[x,y]\] is continuously differentiable on an open region, we can derive the following sets of second-order partial derivatives:
Direct second-order partial derivatives:
\[f_{xx} = \frac{\partial f_{x}}{\partial x}\] where \[f_{x}\] is the first-order partial derivative with respect to \[x\].
\[f_{yy} = \frac{\partial f_{y}}{\partial y}\] where \[f_{y}\] is the first-order partial derivative with respect to \[y\].
Cross partial derivatives:
\[f_{xy} = \frac{\partial f_{x}}{\partial y}\] where \[f_{x}\] is the first-order partial derivative with respect to \[x\].
\[f_{yx} = \frac{\partial f_{y}}{\partial x}\] where \[f_{y}\] is the first-order partial derivative with respect to \[y\].
Young's theorem: Corresponding cross partial derivatives are equal. [To read more about Young’s theorem, see Simon & Blume, Mathematics for Economists, p 330.]
Suppose \[\color{red}{y = f[x_1,…,x_n]}\] is a continuously differentiable function of \[n\] variables.
The first order partial derivative with respect to the variable \[\color{red}{x_{i}}\] is \[\color{red}{\partial f / \partial x_{i}}\].
The \[x_{i}x_{j}\] -second order partial derivative is: \[\color{purple}{\frac{\partial }{\partial x_{j}}}[\color{red}{\frac{\partial f}{\partial x_{i}}}] = \frac{\partial^{2}f}{\partial x_{j} \partial x_{i}} = f_{i,j}\]
If \[\color{red}{j = i}\], then \[x_{i}x_{j}\] -second order partial derivative is called \[\frac{\partial^{2}f}{\partial x_{i}^2}\] or second order direct partial derivatives.
If \[\color{red}{j \neq i}\], then \[x_{i}x_{j}\] -second order partial derivative is called the cross partial derivatives.
Example 1: Find the first, second, and cross partial derivatives for the following function: $$f[x,y] = x^{2} + 5xy + 2y^{2}$$ First order partial derivatives: $$f_{x} = 2x + 5y + 0 = 2x + 5y$$ $$f_{y} = 0 + 5x + 4y = 5x + 4y$$ Second order direct partial derivatives: $$f_{xx} = \frac{\partial }{\partial x}[2x + 5y] = 2$$ $$f_{yy} = \frac{\partial }{\partial y}[5x + 4y] = 4$$ Second-order cross partial derivatives: $$f_{xy} = \frac{\partial }{\partial y}[2x + 5y] = 5$$ $$f_{yx} = \frac{\partial }{\partial x}[5x + 4y] = 5$$ See that in this example the cross-partial derivatives are equal.
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This is not the intention of the exercise. They do not want you to indentify the graph. To explain how they want you to do it, I will do 5[a].
It asks to find $f_x[1,2]$, i.e. how $f$ changes as you move in the $x$ direction from the point $[1,2]$. If you go in the positive $x$ direction from the pink spot which represents $[1,2]$, then you can see that the $z$ co-ordinate [i.e. $f$ value] increases. So the sign of $f_x$ is positive at this point.
Similarly you can do 5[b], 6[a] and 6[b].
When it asks to find $f_{xx}[-1,2]$, this means "the rate at which $f_x$ is changing as you advance in the $x$ direction at this point". At that point, $f_x$ is negative, since the graph is going downwards as you advance in the positive $x$ direction. But $f_x$ is increasing, since the "negative slope" is becoming less steep, like going from gradient of $-2$ to $-1$. So $f_{xx}$ is positive.
Similarly you can do 7[b].
When the two letters are different, you need to find "the rate at which $f_x$ changes as you move in the positive $y$ direction". This, combined with the method for question 7, is how to do 8[a] and 8[b].