Growth of y with respect to time based on some given assumptions

Question

I would be thankful if someone can help me with the following problem:

Given:

$\frac{\dot{N}(t)}{N(t)} = c_b - \frac{c_d}{y(t)}$

$\frac{\dot{A}(t)}{A(t)} = N(t)g - \delta $

$y(t) = A(t)\cdot(\frac{T}{N(t)})^{1-\alpha}$

$T$ is fixed.

Question:

How is possible to get $\frac{\dot{y}(t)}{y(t)}$ as:

$\frac{\dot{y}(t)}{y(t)} = \frac{\dot{A}(t)}{A(t)} - (1-\alpha)\frac{\dot{N}(t)}{N(t)}$

Edit*

I am trying the following:

$y(t) = A(t)\cdot(\frac{T}{N(t)})^{1-\alpha} \implies$

$\frac{\dot{y}(t)}{y(t)} = \frac{\dot{A}(t)}{A(t)} \cdot (\frac{T}{\frac{\dot{N}(t)}{N(t)}})^{1-\alpha}$, then I take the log:

$e^{log\Big(\frac{\dot{A}(t)}{A(t)} \cdot (\frac{T}{\frac{\dot{N}(t)}{N(t)}})^{1-\alpha}\Big)}$

$e^{log\Big(\frac{\dot{A}(t)}{A(t)}\Big) + log\Big((\frac{T}{\frac{\dot{N}(t)}{N(t)}})^{1-\alpha}\Big)}$

$e^{log\Big(\frac{\dot{A}(t)}{A(t)}\Big) + (1-\alpha)log\Big((\frac{T}{\frac{\dot{N}(t)}{N(t)}})\Big)}$

$e^{log\Big(\frac{\dot{A}(t)}{A(t)}\Big) + (1-\alpha)\Big[log(T) - log({\frac{\dot{N}(t)}{N(t)}})\Big]}$

Now, does this come to:

$\cancel{e}^{\cancel{log}\Big(\frac{\dot{A}(t)}{A(t)}\Big) + (1-\alpha)\Big[\cancel{log}(T) - \cancel{log}({\frac{\dot{N}(t)}{N(t)}})\Big]}$

so I have:

$\frac{\dot{A}(t)}{A(t)} + (1-\alpha)T - (1-\alpha)\frac{\dot{N}(t)}{N(t)}$ ??

I tried substituting, but I got no where - any suggestions are appreciated.

Thanks

Answer 1

You only need the third equation (definition of $y(t)$). Taking $\log$ $$ \log y(t)=\log A(t)+(1-\alpha)\log(T/N(t))=\\ \log y(t)=\log A(t)-(1-\alpha)\log(N(t)/T) $$ then derive wrt $t$.