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We have the following function:

$$f(\boldsymbol{x}) = \left(\frac{x_1}{x_2}\right)^\sigma + x_1$$

We would like to derivate it by the fraction of $x_1$ and $x_2$:

$$\frac{\partial \left[ \left( \frac{x_1}{x_2} \right)^\sigma + x_1 \right]}{\partial \frac{x_1}{x_2}}$$

The derivation of the first term seems easy to accomplish:

$$\frac{\partial \left( \frac{x_1}{x_2} \right)^\sigma}{\partial \frac{x_1}{x_2}} = \sigma \left( \frac{x_1}{x_2} \right)^{\sigma-1}$$

But what would be the following equal to:

$$\frac{\partial x_1}{\partial \frac{x_1}{x_2}} = \boldsymbol{?}$$

Athaeneus
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1 Answers1

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HINT: Let $z=\frac{x_1}{x_2}$. You can then define $x_1(z)=zx_2$ and differentiate $f(z,x_1(z))$ with respect to $z$, right?

Weierstraß Ramirez
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  • Thank you very much! Do I understand it correctly that the result would be: $\sigma \left( \frac{x_1}{x_2} \right)^{\sigma-1} + x_2$? Can I use this to skip that $x_2$ is also a function of $z$ like $x_2(z) = \frac{x_1}{z}$? – Athaeneus Mar 09 '23 at 10:10
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    Indeed, it is an application of the Chain Rule. You might find this useful: https://en.wikipedia.org/wiki/Total_derivative#Example:_Differentiation_with_direct_dependencies – Weierstraß Ramirez Mar 09 '23 at 10:15