Taking the derivative of capital per effective unit of labor in the Solow model I'm taking an intro macro course, and my calculus is pretty rusty.  I'm going through some lectures notes right now where they derive the growth rate of capital in the Solow model like so:



I'm wondering what the process is for getting to the very first step where they take the derivative with respect to time.  Generally, are there good resources for learning how to do this, as I imagine I'll be doing it a lot?
I've tried to "fill the blanks", hope this helps.

$$\tilde{k} = \frac{K}{AL}$$
$$\frac{d[\tilde{k}]}{dt} = \frac{d}{dt}\left[\frac{K}{AL} \right] \quad (1)$$ 
Now we use the quocient rule: to differentiate $f(x) = \frac{g(x)}{h{x}}$ (for $h(x) \neq 0$) we can apply the formula $f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{[h(x)]^2}$. In your problem, $f(x) = \dot{k}$, $g(x) = K$ and $h(x) = AL$, but instead $x$, you have $t$. So, applying the rule in (1), we have:
$$\frac{d[\tilde{k}]}{dt} = \frac{AL\left(\frac{d[K]}{dt}\right) - K\left(\frac{d[AL]}{dt}\right)}{(AL)^2} \quad (2)$$

The second term in the numerator has the derivative of a product ($AL$), so we need again to use the appropriate rule: $(f(x)\cdot g(x))' = f'(x)\cdot g(x) + f(x)\cdot g'(x)$

So, using the rule in (2) we have:

$$\frac{d[\tilde{k}]}{dt} = \frac{AL\left(\frac{d[K]}{dt}\right) - K\left(L\frac{d[A]}{dt}+A\frac{d[L]}{dt}\right)}{(AL)^2}$$

Now we re-write using dot notation - $\dot{x} \equiv \frac{d[x]}{dt}$:

$$\dot{\tilde{k}} = \frac{AL\dot{K} - K(L\dot{A}+A\dot{L})}{(AL)^2} \quad = \frac{AL\dot{K}}{(AL)^2}-\frac{KL\dot{A}}{(AL)^2}-\frac{KA\dot{L}}{(AL)^2}$$

Since $\frac{\dot{A}}{A} = g$ and $\frac{\dot{L}}{L} = n$, we have

$$\dot{\tilde{k}} = \frac{\dot{K}}{AL} - Kg -  Kn = \frac{\dot{K}}{AL} -K(n+g)$$

ps: I'm not sure why your textbook is using dot notation over $\tilde{k}$ before taking the derivative.

EDIT:

I'm adding an alternative way to achieve the same result. 

Starting with $\tilde{k} = \frac{K}{AL}$, you can rewrite as $K = \tilde{k}AL$.

Now, take log on both sides and remember that $log(ab) = log(a) + log(b)$:

$log(K) = log(\tilde{k}AL) = log(\tilde{k}) + log(A) + log(L) \quad (1)$

Now we are going to take the derivative (in t), using the log rule: if $f(x) = log(x)$, then $f'(x) = \frac{1}{x}$. But now we have to use the chain rule, because all variables are in reality functions of $t$. The chain rule is: if z depends on y and y depends on t, then z depends on t: $\frac{dz}{dt} = \frac{dz}{dy}\frac{dy}{dt}$. Using $log(K)$ as example, we have: $\frac{d}{dt}[log(K)] = \frac{d}{dK}[log(K)]\cdot \frac{d}{dt}[K]$.

(1) then became:
$$\frac{d}{dt}[log(K)] =  \frac{d}{dt}[log(\tilde{k})] + \frac{d}{dt}[log(A)] + \frac{d}{dt}[log(L)]$$

Which is...
$$\frac{d}{dK}[log(K)]\cdot \frac{d}{dt}[K] =  \frac{d}{d\tilde{k}}[log(\tilde{k})]\cdot\frac{d}{dt}[\tilde{k}] + \frac{d}{dA}[log(A)]\cdot\frac{d}{dt}[A] + \frac{d}{dL}[log(L)]\cdot\frac{d}{dt}[L]$$

Now we substitute all derivatives of logs and use dot notation:

$$\frac{1}{K}\dot{K} = \frac{1}{\tilde{k}}\dot{\tilde{k}} + \frac{1}{A}\dot{A} + \frac{1}{L}\dot{L} = \frac{\dot{\tilde{k}}}{\tilde{k}} + g + n$$

In the SS we have $\frac{\dot{\tilde{k}}}{\tilde{k}} = 0$, so:

$$\frac{\dot{K}}{K} = g + n$$