<s>[INST] 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?
 [/INST] 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:
$$y' = \frac{dy}{dx} = \frac{x-2}{x(x+1)} \frac{(x+1)^3}{x^2} = \frac{(x-2)(x+1)^2}{x^3}$$
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