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If I'm asked to differentiate a function that is simply of the form: $ f(tx_1, tx_2, ... tx_n) $ (i.e., the function is not "defined" to be particularly anything), how should I go about this? I want to differentiate with respect to $ t$.

Context: Let's say we have a homogeneous function of with degree $ k $.

Thus,

$$ f( tx_1, tx_2, ... tx_n) = t^k f(x_1, x_2, ... x_n) $$

Differentiating on both sides with respect to $ t$, we get,

$$ x_1 f_1 (tx_1, tx_2, ... tx_n) + ... + x_n f_n (tx_1, tx_2, ... tx_n) = k \cdot t^{k-1} \cdot f(x_1, .. x_n) $$

  1. On the RHS, why we differnetiate with respect to to get $ k \cdot t^{k-1} $. Because the function on the RHS is a constant (i.e., it doesn't depend on t, we write it as it is)

  2. On the LHS, why do we have: $ x_n f_n(tx_1, tx_2,\dots, tx_n)$ for each term?

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  • $\begingroup$ Shouldn't your RHS be $k \cdot t^{k-1} \cdot f(x_1, \dots, x_n)$? (without the $t$ inside) $\endgroup$ – BigbearZzz Jan 17 at 11:22
  • $\begingroup$ Yes, sorry. That was a mistake on my part. $\endgroup$ – WorldGov Jan 17 at 11:23
  • $\begingroup$ You need to assume that $k>1$. For $k>1$, you may fact out the variable $x_1$ first, then factor out the variable $x_2$ from the remainder and so on. Try to write down an example and you'll see it is simple. $\endgroup$ – James Jan 17 at 11:27
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On the RHS, you just apply the usual rules of single variable calculus. The factor $t^k$ becomes $k t^{k-1}$ and $f(x_1,\dots,x_n)$ is constant with respect to $t$, so you can just keep it.

For the LHS you are looking for the chain rule for multivariable functions. If $f(x_1,\dots,x_n)$ is a function of $n$ variables and you have $n$ functions $z_1(t),\dots,z_n(t)$ each of one variable $t$, then

$$ \frac{\mathrm d}{\mathrm dt} f(z_1(t),\dots,z_n(t)) = \frac{\partial f}{\partial x_1}(z_1(t),\dots,z_n(t)) \cdot \frac{\mathrm dz_1}{\mathrm dt}(t) + \dots +\frac{\partial f}{\partial x_n}(z_1(t),\dots,z_n(t)) \cdot \frac{\mathrm dz_n}{\mathrm dt}(t) $$

In your example you have $z_i(t) = tx_i$ so that $\frac{\mathrm dz_i}{\mathrm dt}(t) = x_i$.

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