Exponential growth and decay calculus. I'm reading this text about exponential growth functions and derivatives and I'm a bit confused by this (questions in bold):


Is the above text true because the derivative of $e^x$ is $e^x$? and also the derivative of $5e^x$ is $5e^x$ and by chain rule the derivative of $5e^{2x}$ is $2 \cdot 5e^{2x}$. Is that why functions that fit the form "constant multiple of itself" are in the form $Ce^{kt}$?
In Newton's law of cooling example, I don't see how we arrived at $\frac{dy}{dt}$. I don't get the section shaded in blue. On a high level, what is going on?

 A: The fact that $\dfrac d {dt} e^{kt} = k e^{kt}$ explains why the function $f(t) = e^{kt}$ satisfies the differential equation $f'(t) = kf(t),$ and why $t\mapsto Ce^{kt}$ also satisfies that differential equation, but it does not explain why only functions of that form satisfy the equation.
To go from $df = kf\, dt$ to $\dfrac{df} f = k\,dt$ can be made logically rigorous by saying some things about various theorems, provided we assume $f$ is nowhere $0$ (since otherwise we'd be dividing by $0$) and would raise questions about whether some functions that are $0$ at some points could satisfy the equation. And the function that is $0$ everywhere is also a solution although that division of both sides by $f$ is unsound in that case.
But here's another way. Suppose some function $g$ is a solution, i.e. we have $g'=kg.$ Consider the function
$$
h(t) = \frac{g(t)}{e^{kt}}.
$$
Then by the quotient rule
\begin{align}
h'(t) & = \frac{e^{kt} g'(t) - ke^{kt} g(t)}{e^{2kt}} \\[10pt]
& = \frac{g'(t) - kg(t)}{e^{kt}} \\[10pt]
& = \frac{0}{e^{kt}} \text{ where we see that this is $0$ precisely because $g'(t) = kg(t)$} \\[10pt]
& = 0.
\end{align}
Since $h'(t) = 0$ for all values of $t,$ it follows that $h$ is constant. So $g(t) = (\text{constant} \cdot e^{kt}).$ And note that the denominator in what is done above is never $0.$
The proof that a function whose derivative is everywhere $0$ is constant uses the mean value theorem.
Note that among solutions are functions like $g(t) = 5^t,$ since $5^t$ is the same as $e^{kt}$ if $k=\log_e 5.$
A: A function $y(t) = Ce^{kt}$ does satisfy $y' = ky$, because:
$$
   y'(t) = Ce^{kt}(k) = kCe^{kt} = ky(t)
$$
I think you asked why it's true that any function that satisfies $y'=ky$ must be in this form.  There's a different answer to that.  Given such a $y$, let $z(t) = y(t)e^{-kt}$.  Then
$$
    z'(t) = y'(t) e^{-kt} + y(t)e^{-kt}(-k)
          = ky(t) e^{-kt} -ky(t) e^{-kt} = 0
$$
So $z(t)$ is a constant $C$, which means $y(t) = Ce^{kt}$.
Your second question is more straightforward.  The equation 
$$\frac{dT}{dt} = k(T-T_s)\tag{$*$}$$
is a mathematical formulation of Newton's Law of Cooling: An object cools in proportion to the difference in temperature between the object and its surroundings.  
Make sure that makes sense to you.  If an object is much hotter than its surroundings, it will cool rapidly.  If an object is close to “room temperature”, it won't change its temperature much at all.  It works for warming, too.
Given such a function $T(t)$, let $y(t) = T(t) - T_s$.  Then $y'(t) = T'(t)$, because the derivative of a constant is zero.  On the other hand, by ($*$) and the definition of $y(t)$, $T'(t) = k y(t)$.  So $y'(t) = ky(t)$, implying $y=Ce^{kt}$.
A: First question:
Is the above text true because the derivative of $e^x$ is $e^x$? and also the derivative of $5e^x$ is $5e^x$ and by chain rule the derivative of $5e^{2x}$ is $2 \cdot 5e^{2x}$. Is that why functions that fit the form "constant multiple of itself" are in the form $Ce^{kt}$?
--Yes, you are absolutely correct.
Second question:
In Newton's law of cooling example, I don't see how we arrived at $\frac{dy}{dt}$. I don't get the section shaded in blue. On a high level, what is going on?
Long Answer. We have the function T(t) or simply T, which shows temperature of the cooling object as a function of time (t). It doesn’t matter what letter we use f(x) or g(t) etc. So I'm gonna keep everything the way it is but I'm gonna use 70F instead of $T_s$ for simplicity. It is a constant here. So we have:
$$T’(t)=k(T(t)-70)$$
That’s the law of physics we are dealing with here. Now the derivative by definition is $dy/dx$ or in our case $dT/dt$. So we have (I'm using T instead of T(t) for simplicity):
$$\frac{dT}{dt}=k(T-70)$$
Now we multiply the both parts of this equation by $dt$ and divide by $(T-70)$:
$$\frac{dT}{T-70}=kdt$$
Now we integrate the both parts of this equation:
$$\int\frac{1}{T-70}dT=\int kdt$$
We get 
$$ln\mid T-70\mid=kt+C$$ 
where C is some constant which, if you reverse this step by differentiating, will turn to zero.
Because our object is cooling to reach the room temperature, the temperature of the object (T) is greater than $70F$ (our room/surroundings temperature). Therefore no need for absolute value symbols with “ln”. Now by definition of logarithm our previous expression is the same as:
$$T-70=e^{kt+C}= e^{kt}\cdot e^{C}\qquad(1)$$
Let’s denote the initial temperature $T(0)=T_0$ -- it’s when the process of cooling starts that is when we set time to zero: $t=0$
$$T(0)-70=T_0-70 =e^{k\cdot0}\cdot e^{C}\;$$ Therefore $\;T_0=70+e^{C}.\;$ So, $\;e^{C}=T_0-70\;$ which we now can plug into formula $(1)$:
$$T-70=e^{kt}\cdot e^{C}=e^{kt}\cdot(T_0-70)$$
$$T-70=e^{kt}\cdot(T_0-70)$$
It may be denoted slightly differently in physics but it's the same. Of course the temperature of surroundings doesn't have to be 70F. It's just a constant $T_s$ in this equation (and you can denote it by any letter you like).
Short answer
$$\frac{dT}{dt}=k(T(t)-T_s)$$
Now the author denotes $y(t)=T(t)-T_s$ so we have:
$$\frac{dT}{dt}=ky(t)$$
Now $\;dy/dt=y'(t)=(T(t)-T_s)'=T'(t)-0=T'(t)=dT/dt\;$ or simply $\;dy/dt=dT/dt\;$ And that's why the author wrote what you have highlighted in blue:
$$\frac{dy}{dt}=ky(t)$$
$$* * *$$
I advise you to have an alternative book to fall back on in order to cover these topics from a slightly different angle, perhaps a bit deeper. You might (advisable) need to cover a few sections on integrating techniques and several pages on differential equations. A good alternative book with detailed explanations (and perhaps a bit more depth) will be a nice way to supplement your current book. For example, Calculus by Larson or Stewart are simple and comprehensive books that explain things well. You may also try Thomas' Calculus or Swokowski for more challenging/interesting derivations and explanations. Any editions of these books will do. Please note your book suggested a simple change of variables in the equation, and it's a must when studying calculus. Such changes occur frequently when you integrate. Studying them thoroughly will make the page you posted here look a whole lot it easier.
