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This is from Simon and Blume's Mathematics for Economists:enter image description here

But, for LHS, applying Chain rule goes:$$\dfrac{\partial f}{\partial(tx_1)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_1)}{\partial x_1}=\dfrac{\partial f}{\partial(tx_1)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial x_1}{\partial x_1}\cdot\dfrac{\partial(tx_1)}{\partial x_1}=\dfrac{\partial f}{\partial x_1}(tx_1,\dots,tx_n),$$and thus derivative is also homogeneous of degree $k$, not $k-1$ .

Where am I wrong?

Silent
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3 Answers3

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As $f$ is homogeneous of degree $k$, we have $f(tx_1, \dots, tx_n) = t^kf(x_1, \dots, x_n)$. Differentiating both sides with respect to $x_i$, we have

\begin{align*} \frac{\partial}{\partial x_i}f(tx_1, \dots, tx_n) &= \frac{\partial}{\partial x_i}(t^kf(x_1, \dots, x_n))\\ \frac{\partial f}{\partial x_i}(tx_1, \dots, tx_n)\frac{\partial}{\partial x_i}(tx_i) &= t^k\frac{\partial}{\partial x_i}f(x_1, \dots, x_n)\\ \frac{\partial f}{\partial x_i}(tx_1, \dots, tx_n)t &= t^k\frac{\partial f}{\partial x_i}(x_1, \dots, x_n)\\ \frac{\partial f}{\partial x_i}(tx_1, \dots, tx_n) &= t^{k-1}\frac{\partial f}{\partial x_i}(x_1, \dots, x_n) \end{align*}

so $\dfrac{\partial f}{\partial x_i}$ is homogeneous of degree $k-1$.

Michael Albanese
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  • Sir, I want to know while using chain rule, isn't this correct: $\dfrac{\partial}{\partial x_i}f(tx_1, \dots, tx_n)=\dfrac{\partial f}{\partial(tx_1)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_1)}{\partial x_1}$ – Silent Dec 28 '13 at 13:01
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    What is $\dfrac{\partial f}{\partial (tx_1)}$? – Michael Albanese Dec 28 '13 at 13:06
  • The chain rule for function $f=g(h(x))$ says $\dfrac{df}{dx}=\dfrac{d(g(h(x)))}{d(h(x))}\cdot\dfrac{d(h(x))}{dx}$. And I have applied just it. Isn't my interpretation correct? – Silent Dec 28 '13 at 13:17
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    Not quite. In your notation the chain rule would read $$\frac{df}{dx} = \frac{dg}{d(h(x))}(h(x))\frac{d(h(x))}{dx},$$ which, when written in Newton notation, becomes $g'(h(x))h'(x)$. When you do the first derivative in the chain rule, you are differentiating the function with respect to one of its arguments, it doesn't matter what is in that argument. In the second derivative in the chain rule, you are differentiating what appears in the relevant argument. – Michael Albanese Dec 28 '13 at 13:47
  • Sir, please do not think I am wasting your time, I am just trying to understand. I learned Chain rule here.I still can't understand how $\dfrac{\partial}{\partial x_i}f(tx_1, \dots, tx_n)=\dfrac{\partial f}{\partial(tx_i)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}=\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}$ – Silent Dec 28 '13 at 15:45
  • Both $\dfrac{\partial f}{\partial(tx_i)}$ and $\dfrac{\partial f}{\partial x_i}$ are being used to denote the $i^{\text{th}}$ partial derivative of the function $f(tx_1, \dots, tx_n)$. – Michael Albanese Dec 28 '13 at 15:50
  • Sir, how is that possible? Doesn't the first one change the scale? – Silent Dec 28 '13 at 15:55
  • No. I'd say they are different notations for the same thing. – Michael Albanese Dec 28 '13 at 16:00
  • Oh Sir, can you please elaborate why differentiating w.r.t. $tx_i$ is the same as w.r.t. $x_i$? – Silent Dec 28 '13 at 16:04
  • In both instances, they denote the function which results from differentiating the function $f$ with respect to the $i^{\text{th}}$ variable. The notation $\frac{\partial f}{\partial x_i}$ denotes this derivative outside of the context of this problem. On the otherhand, $\frac{\partial f}{\partial (tx_i)}$ denotes this derivative within the context of this problem (because $tx_i$ is the expression which appears in the $i^{\text{th}}$ component). – Michael Albanese Dec 28 '13 at 16:26
  • This may help: $$\frac{\partial f}{\partial (tx_i)}(tx_1, \dots, tx_n) = \frac{\partial f}{\partial x_i}(x_1, \dots, x_n)\bigg|_{(tx_1, \dots, tx_n)}.$$ – Michael Albanese Dec 28 '13 at 16:30
  • @Michael Albanese: How is that possible? That identity doesn't always hold, right? (This is with regards to your latest comment). – Rainroad Mar 01 '20 at 15:14
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Hope this help. I find this question because I am also reading a microeconomic book.

\begin{aligned} LHS &= \frac{\partial f(tx)}{\partial x_i} \\ &= \sum_{j} \frac{\partial f(tx)}{\partial t x_j} \frac{\partial t x_j}{\partial x_i} \\ &= \frac{\partial f(tx)}{\partial t x_i} \frac{\partial t x_i}{\partial x_i} \\ &= \frac{\partial f(tx)}{\partial t x_i} t \end{aligned}

dchao
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I guess you already know it. Anyway, I hope this can help others.

  • I think Federica Maggioni's comment, "the guy who has corrected the book is wrong" is inaccurate. I.e., It should be $\partial tx_i$ rather than $\partial x_i$.

  • dchao provides the right answer. By the chain rule: $$ LHS = \frac{\partial f(tx_1,...,tx_n)}{\partial x_i} = \frac{\partial f(tx_1,...,tx_n)}{\partial tx_i} \frac{\partial tx_i}{\partial x_i} = \frac{\partial f(tx_1,...,tx_n)}{\partial tx_i} \cdot t$$ $$ RHS = t^k \frac{\partial f(x_1,...,x_n)}{\partial x_i} $$ These expressions are equivalent due to the definition of "homogeneous," establishing $LHS=RHS$.
    Dividing both sides by $t$, we find it is homogenous of degree k-1.

  • Then it comes to your question, "Where am I wrong?"

    Your equation $LHS=\frac{\partial f(tx_1,...,tx_n)}{\partial x_1}$ is correct, since $\frac{\partial f(tx_1,...,tx_n)}{\partial tx_1}= \frac{1}{t} \frac{\partial f(tx_1,...,tx_n)}{\partial x_1}$ by chain rule.

    However, it is hard to see the equality of both sides if you write the above equation. Since we have $tx_1$ in the numerator and $x_1$ in the denominator.

    To see the equality and take an example of $x_1$, we can write $$LHS=\frac{\partial f(tx_1,...,tx_n)}{\partial \frac{1}{t}tx_1}. \;\;\; \text{ (1) }$$

    Note we want to show $$\frac{\partial f(tx_1,...,tx_n)}{\partial \frac{1}{t}tx_1}=t^{k-1} \frac{\partial f(x_1,...,x_n)}{\partial \frac{1}{t}x_1}, \;\;\;\text{ (2) }$$ because $\frac{\partial f(x_1,...,x_n)}{\partial \frac{1}{t}x_1}$ is a function of $x_1$ for both the numerator and denominator.

    Then, by the definition of homogenous, the numerator: \begin{equation} f(tx_1,...,tx_n)=t^kf(x_1,...,x_n) = t^{2k}f(\frac{1}{t}x_1,...,\frac{1}{t}x_n) \;\;\; \text{ (3) } \end{equation}
    So, since $\frac{\partial f(\frac{1}{t}x_1,...,\frac{1}{t}x_n)} {\partial x_1}=\frac{1}{t}\frac{\partial f(\frac{1}{t}x_1,...,\frac{1}{t}x_n) }{\partial \frac{1}{t} x_1}$ (chain rule) $$ LHS=\frac{\partial f(tx_1,...,tx_n)}{\partial x_1}=\frac{t^{2k}\partial f(\frac{1}{t}x_1,...,\frac{1}{t}x_n) }{\partial x_1} = \frac{t^{2k-1}\partial f(\frac{1}{t}x_1,...,\frac{1}{t}x_n) }{\partial \frac{1}{t} x_1} $$ Reuse (3), $$LHS=t^{k-1}\frac{\partial f(x_1,...,x_n)}{\partial \frac{1}{t} x_1} \;\;\; \text{ (4) }$$ Combine (1) and (4), and the proof is complete.

Ao Liu
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