# Differentiating a function that is defined generally (for eg, $f (x_1, x_2, … x_n)$)

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?

• Shouldn't your RHS be $k \cdot t^{k-1} \cdot f(x_1, \dots, x_n)$? (without the $t$ inside) – BigbearZzz Jan 17 at 11:22
• Yes, sorry. That was a mistake on my part. – WorldGov Jan 17 at 11:23
• 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. – James Jan 17 at 11:27

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 $$\frac{\mathrm dz_i}{\mathrm dt}(t) = x_i$$.