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Why is $$\dfrac{\partial}{\partial x_i}f(tx_1, \dots, tx_n)=\dfrac{\partial f}{\partial(tx_i)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}=\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}$$ here ? I think it is NOT equal as, I have this answer, and according to it, I should go this way: $$\begin{align}\dfrac{\partial f}{\partial(tx_i)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}&=\dfrac{\partial f}{\partial(tx_i)}(tx_1,\dots,tx_n)\cdot\dfrac{\partial x_i}{\partial x_i}\cdot\dfrac{\partial(tx_i)}{\partial x_i}\\&=\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\cdot\dfrac{\partial x_i}{\partial(tx_i)}\cdot\dfrac{\partial(tx_i)}{\partial x_i}\\&=\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\cdot\dfrac{t}{t}\\&=\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\\& \neq\dfrac{\partial f}{\partial x_i}(tx_1,\dots,tx_n)\cdot\dfrac{\partial(tx_i)}{\partial x_i}\end{align}$$

Please help! I am stuck in my Microeconomics course.

Community
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Silent
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1 Answers1

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This has nothing to do with partial derivatives, but instead it is confusion about the chain rule.

Consider $f(g(x))$. We have $$\frac{d}{dx}[f(g(x))] = \frac{df}{dx}(g(x))\frac{dg}{dx}(x) = \frac{d}{dg}[f(g(x))]\frac{dg}{dx}(x).$$

The key part is that $$\frac{df}{dx}(g(x)) = \frac{d}{dg}[f(g(x))].$$

In your situation $$\frac{df}{dx}(tx) = \frac{d}{d(tx)}[f(tx)].$$

Sorry if I have led you astray with some of my previous comments.

Michael Albanese
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  • But, again, $ \frac{d}{d(tx)}[f(tx)]=\frac{d}{d(tx)}[f(tx)]\cdot \frac{dx}{dx}=\frac{d}{dx}[f(tx)]\cdot\frac{dx}{d(tx)}=\frac{d}{dx}[f(tx)]\cdot \frac{1}{t}\neq\frac{df}{dx}(tx) $, isn't it? – Silent Dec 29 '13 at 05:03
  • How did you get from $\frac{d}{d(tx)}[f(tx)]\cdot\frac{dx}{dx}$ to $\frac{d}{dx}[f(tx)]\cdot\frac{dx}{d(tx)}$? – Michael Albanese Dec 29 '13 at 05:51