Can we differentiate Equation 9 and get Expression 10?

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beaf123
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Homework Statement


Text reading: A division of Equation (8) by L yields the labor productivity of the overall economy. Y/L = y = e^(rt)(a + by)/(1 + ye^((r-s)t)) (9). Thus, productivity growth of the economy, ŷ, is equal to: ŷ = (r + s y e^((r-s)t))/(1 + y e^((r-s)t)) (10).


Homework Equations


[/B]
I came over these Equations when I was Reading an article. I tried to replicate the results by differentiating Equation 9 w.r.t time. However I could not do it.
Just to be sure. The growth of a variable is Equal to the time derivative of the same variable? What is most important for me is to know that, that is what they have done.

The Attempt at a Solution



[/B]
I tried to use:
Quotient Rule f/g (f' g − g' f )/g2

And I think I did it right, but maybe I could not simplify it to get Expression 10.

I really hope someone can help me with this.
 
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  • Typeset equations: Y/L = y = e^(rt)(a + by) / (1 + ye^((r-s)t)) labelled (9); then ŷ = (r + sye^((r-s)t)) / (1 + ye^((r-s)t)) labelled (10).
    upload_2017-10-15_19-43-51.png
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beaf123 said:

Homework Statement


View attachment 213096

Homework Equations


[/B]
I came over these Equations when I was Reading an article. I tried to replicate the results by differentiating Equation 9 w.r.t time. However I could not do it.
Just to be sure. The growth of a variable is Equal to the time derivative of the same variable? What is most important for me is to know that, that is what they have done.

The Attempt at a Solution



[/B]
I tried to use:
Quotient Rule f/g (f' g − g' f )/g2

And I think I did it right, but maybe I could not simplify it to get Expression 10.

I really hope someone can help me with this.
How is ##\hat y## defined? You seem to be assuming that ##\hat y = \frac{dy}{dt}##, which might not be what is meant by that symbol.
 
It is defined as "labor Productivity Growth", but it is not defined any further. Are you (or someone else) able to see if we can differentiate (9) and get (10)?
 
beaf123 said:
It is defined as "labor Productivity Growth", but it is not defined any further. Are you (or someone else) able to see if we can differentiate (9) and get (10)?
I don't see how (9) could possibly be differentiated (with respect to t) to yield (10). Equation (10) would have to have a factor of ##(a + b\gamma)##, which isn't present. Besides this, I differentiated (9) w.r.t. t, and didn't get anything remotely close to (10).
 
beaf123 said:
It is defined as "labor Productivity Growth", but it is not defined any further. Are you (or someone else) able to see if we can differentiate (9) and get (10)?

Nobody can get it because it is false. For example, if ##a = 1.5, b = 2, \gamma = 2.5, r = .1, s = 3## we have
$$y = \frac{6.5 \exp(.1\,t)}{1+2.5 \exp(-2.9 \,t)} $$
and
$$Dy_1 \equiv \frac{dy}{dt} = \frac{6.5(7.5 \exp(-2.8\,t)+0.1 \exp(0.1\,t)}{(1+2.5 \exp(-2.9\,t))^2} $$
The expression you wrote is
$$Dy_2 = \frac{.1+7.5 \exp(-2.9\,t)}{1+2.5 \exp(-2.9\,t)}$$
These last two are definitely unequal; here is a plot:
Plot of two curves against t from 0 to 2: a red curve starting near 4, rising to a peak near 5.2 at t about 0.4, then falling to about 1; a blue curve starting near 2.2 and decreasing toward 0.


The top curve is ##Dy_1## (the true derivative), while the bottom curve is ##Dy_2## (the false derivative in equation (10)). I did all the algebra and the plot using the computer algebra package Maple, so there should not be any errors. (However, I copied your expressions from hard-to-read and slightly out-of-focus images, so maybe there is something wrong. You should avoid posting images, and just type out everything, the way most helpers do.)
 
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