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In a linear regression of the form Y=bX we often have ln transformed Y and X. ie., lnY=b*lnX This is interpreted as a 1% change in X resulting in a b% change in Y (approximately)

The derivation of this interpretation comes from taking the derivatives of both sides: d (ln Y) = d (b*lnX) because of the fact that the derivative of ln is it's inverse gives 1/Y dy = b*1/X dx ie., dy/Y = b* dx/X or %y=b*%x or b= %y/%x ie., a 1% change in X results in a b % change in Y

is this true that you can take the derivative of both sides as in the above? for instance does it mean that:

for y^2=x^2 (note:this simplifies to y=+-x and taking the derivative of this is dy/dx=+-1) however taking the derivative of both sides of the original equation leads to: 2y dy = 2x dx or y dy = x dx or y dy/dx = x and dy/dx= x/y which is different from dy/dx=+-1

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  • $\begingroup$ Take care : you go from $Y=b*X$ to $ln(Y)=ln(b) + ln(X)$. This how you can make the analysis of the impact of a fractional change of $x$. Now, take the derivative of both sides and change $\delta Y$ by $\Delta Y$ and same for $X$. Is this better or do you want me to elaborate ? $\endgroup$ – Claude Leibovici Feb 17 '14 at 17:09
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Two functions, $f(x),g(x)$ have the same determinant iff they differ by a constant factor. This is just from the fundamental theorem of calculus. Thus if you know that two functions are equal, their derivatives are equal (since they differ by the constant factor 0).

$y^2=x^2\Rightarrow 2ydy=2xdx\Rightarrow y\frac{dy}{dx}=x\Rightarrow \pm y=x$. Notice that the solution set of the first and the last (and every equation I wrote along the way) are the same. That's what the theorem is saying. To use this to compute the derivative, you wind up at $\frac{dy}{dx}=\frac{x}{y}$ and then have to go back to the original equation and notice that if $y^2=x^2$, then $y=\pm x$ and then go plug that into the previous formula to get $\pm 1$

In a more abstract sense, $d(f+g)=df+dg$. So if $f=g$, then $(f-g)=0\Rightarrow d(f-g)=d(0)=0\Rightarrow df-dg=0\Rightarrow df=dg$

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  • $\begingroup$ excellent I hadn't thought of taking the derivative with respect to y of the left side and with respect to x of the right side would preserve the equality very useful! $\endgroup$ – user129467 Feb 18 '14 at 20:15
  • $\begingroup$ You want to be very careful saying that. I'm not taking a derivative with respect to anything per say. Notice I have a $dx$ and a $dy$ term on each side from the chain rule. You are only applying the linear operator $d$ to both sides, and then doing algebra, and noting that the ratio of $\frac{dy}{dx}$ is something important here. $\endgroup$ – Stella Biderman Feb 19 '14 at 7:18
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It's easier if, rather than thinking about "taking the derivative of both sides," you instead gather everything on one side of the equation.

So, for example, if you have $y=x^2$, then you can instead write $y-x^2 = 0$, and then you differentiate with respect to $x$. But, since the right-hand side is just zero, we don't have to worry about it:

$$\frac{d}{dx}\left[y-x^2\right] = \frac{d}{dx} 0,\\ \frac{dy}{dx}-\frac{d}{dx}\left[x^2\right] = 0, \\ \frac{dy}{dx}=\frac{d}{dx}\left[x^2\right], \\ \frac{dy}{dx} = 2x.$$

This is a trivial example, but it is sometimes useful to use to make sure that we're differentiating everything with respect to the same variable. And we can do this because, in general, $\frac{d}{dx}\left(f(x)+g(x)\right) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$. More generally, differentiation is a linear operator, so in a sense you can think of it as "distributing" over addition.

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