Can anyone explain this process of solving? (Differentiation) I'm at differentiation of algebraic functions. There's an example in the module that I couldn't quite get how it led to that.
$y=\frac {(x+1)^3}{x^2}$
It's solved by using a combination of quotient and power rules. I'll enumerate how it's solved.
$(1) y′= \frac{(x^2(3(x+1)^2))–(x+1)^32x}{x^4}$
$(2) y′= \frac{x(x+1)^2 (3x–2(x+1))}{x^4}$
$(3) y′= \frac {(x+1)^2 (3x-2x-2)}{x^3}$
$(4) y′= \frac {(x+1)^2 (x-2)}{x^3}$
How did $(2)$ came to be? Why was it done like that? I just can't wrap my head around it. It just looked like it skipped a couple steps (at least to me). Can anyone help me with this?

If it gets too messy, you can always use logarithmic differentiation. Take the natural logarithm of both sides to get:
$$\ln y = \ln((x+1)^3) - \ln(x^2)$$
$$\ln y = 3 \ln(x+1) - 2 \ln x$$
and so using the chain rule, we have that $\frac{d}{dx} (\ln y)$, where $y$ is a function of $x$), is $\frac{1}{y} \cdot y' = \frac{y'}{y}$:
$$\frac{y'}{y} = \frac{3}{x+1} - \frac{2}{x} = \frac{3x - 2(x+1)}{x(x+1)} = \frac{x-2}{x(x+1)}$$
so using what $y$ is, we have that:
$$y' = \frac{dy}{dx} = \frac{x-2}{x(x+1)} \frac{(x+1)^3}{x^2} = \frac{(x-2)(x+1)^2}{x^3}$$
