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I have the following function

$$f(x_1,x_2)=\frac{ax_1}{ax_1+bx_2}$$

Where the partial derivative of $f$ with respect to $x_2$ is

$$\frac{\partial f}{\partial x_2}=-\frac{abx_1}{(ax_1+bx_2)^2}$$

But why is the absolute value of $\frac{\partial f}{\partial x_2}$ (i.e. the size of the effect of increasing $x_2$) increasing in $a$?

It seems counter-intuitive since clearly as $a \rightarrow \infty$, the effect on $f$ increasing $x_2$ will tend to $0$. What am I overlooking?

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  • $\begingroup$ Do you mean "... the absolute value of $\tfrac{\partial f}{\partial x_2}$..."? If so, this effect is not increasing: in fact, for big "a", the effect decreases and tend to zero as $a \to \infty$. $\endgroup$ – Alejandro Nasif Salum Nov 18 '17 at 17:22
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    $\begingroup$ That is, $$\lim_{a\to \infty} \left | \tfrac{\partial f}{\partial x2}\right| = \lim_{a \to \infty} \frac {|a|}{|a|^2} \frac{|bx_1|}{(x_1+\tfrac b a x_2)^2}= \lim_{a \to \infty} \frac {1}{|a|} \frac{|bx_1|}{(x_1+\tfrac b a x_2)^2}=0$$. $\endgroup$ – Alejandro Nasif Salum Nov 18 '17 at 17:25
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You have $a$ in both numerator and denominator. And, since its power in numerator is equal to $1$ and in denominator this power is equal to $2$, you can see that, indeed, the partial derivative $\partial_{x_2} f$ vanishes as $|a|\to\infty$.

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