Can we split a function while doing differentiation? I have a silly question here. Assume a function $$f(x)= \frac{x}{(x-1)^2} \cdot (x+2)$$
Can I write
$$\frac{d}{dx} f(x) = \frac{d}{dx} x \cdot \frac{d}{dx} (x-1)^{-2} \cdot \frac{d}{dx} (x+2)^{-1}$$ 
If not, then why?
 A: It's very easy to see that this doesn't hold:
$$(x\cdot x)'=2x$$
vs.
$$x'\cdot x'=1.$$
By the true rule,
$$(x\cdot x)'=1\cdot x+x\cdot1=2x.$$
A: From the definition of the derivatives,
$$(f(x)g(x))'=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\
=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}\\
=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\lim_{h\to0}g(x+h)+\lim_{h\to0}f(x)\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\\
=f(x)'g(x)+f(x)g'(x).$$

Shortly:
$$f_hg_h-fg=f_hg_h-fg_h+fg_h-fg=(f_h-f)g_h+f(g_h-g),\\(fg)'=f'g+fg'.$$
For addition we just have
$$(f_h+g_h)-(f+g)=(f_h-f)+(g_h-g),\\(f+g)'=f'+g'.$$
Division can be dealt with using
$$f'=\left(\frac fgg\right)'=\left(\frac fg\right)'g+\frac fgg'$$ 
from which we draw
$$\left(\frac fg\right)'=\frac{f'-\dfrac fgg'}g=\frac{f'g-fg'}{g^2}.$$
A: \begin{align}
f(x) &= \frac{x}{(x-1)^2} \cdot (x+2)\\
&= \frac{x^2+2x}{(x-1)^2}\\
&= \frac{x^2}{(x-1)^2} + \frac{2x}{(x-1)^2}\\
\end{align}
Now 
\begin{align}
f^\prime(x) &= \frac{x^2}{(x-1)^2}\,\frac{d}{dx} + \frac{2x}{(x-1)^2}\,\frac{d}{dx}\\
&= -\frac{2x}{(x-1)^3}\ - \frac{2(x+1)}{(x-1)^3}\\
&= -2\left(\frac{x}{(x-1)^3} + \frac{(x+1)}{(x-1)^3}\right)\\
\end{align}

$$f^\prime(x)= -2\frac{2x + 1}{(x-1)^3}$$

A: As others have pointed out, no, you cannot. Dealing with products of functions requires use of the product rule. I'm going to attempt to explain, as intuitively as I can, why you cannot do this.
At the heart of differentiation is linearisation. It's the idea of approximating a function by a linear function $L(x) = mx + b$ around a fixed point $x_0$. What we call the actual derivative is the slope $m$ from this function, and this number and $f(x_0)$ are enough to figure out $L(x)$.
The chain rule can be explained somewhat intuitively like this. If you're looking at $\frac{\mathrm{d}}{\mathrm{d}x} f(g(x))$ at $x = x_0$, then you figure out a linearisation $L_1(x) = m_1 x + b_1$ for $g$ at $x = x_0$, then another linearisation $L_2(x) = m_2 x + b_2$ for $f$ at $x = g(x_0)$. Since $g(x)$ is "approximately" $L_1(x)$ for $x$ around $x_0$, and we are subbing into $f(x)$, which is "approximately" $L_2(x)$ for $x$ around $g(x_0)$, it means that $f(g(x))$ is "approximately" $L_2(L_1(x)) = m_1m_2x + m_2b_1 + b_2$. In particular, look at the slope: it's $m_1m_2$, the product of the derivatives $g'(x_0)$ and $f'(g(x_0))$.
Product rule? No such luck. If we take linear approximations $L_1(x) = m_1 x + b_1$ and $L_2(x) = m_2 x + b_2$ respectively of $f(x)$ and $g(x)$, both around $x = x_0$, then producting $f(x)$ and $g(x)$ together gives us approximately
$$L_1(x)L_2(x) = (m_1 x + b_1)(m_2 x + b_2) = m_1m_2x^2 + (m_1b_2 + m_2b_1)x + b_1b_2.$$
Note that the resulting approximation is no longer in the same form: it's not linear. If you want to recover a linear approximation, you could try differentiating this quadratic, and with enough care, you might be able to recover the product rule formula.
However, I probably wouldn't advise giving it too much thought. This is a hand-wavy explanation. I just hope that it might shed some light, intuitively, why this doesn't work.
A: Yes, you can "split" a function; in fact, that's the standard way for computing derivatives algebraically.
However, you have to know what to do with the pieces.
Some examples (the "addition rule", "product rule", and "chain rule") are
$$ \frac{d}{dx} \big( f(x) + g(x) \big) = f'(x) + g'(x) $$
$$ \frac{d}{dx} \big( f(x) \cdot g(x) \big) = f(x) g'(x) + f'(x) g(x) $$
$$ \frac{d}{dx} \big( f(g(x)) \big) = g'(x) \cdot f'(g(x)) $$
A: You can make life easier using logarithmic differentiation since you just have here product, quotient and powers$$f= \frac{x (x+2)}{(x-1)^2}\implies \log(f)=\log(x)+\log(x+2)-2\log(x-1)   $$ Differentiate both sides
$$\frac {f'}f=\frac 1x+\frac 1{x+2}-\frac 2{x-1}$$ Simplify and, when done, use
$$f'=f \times \left(\frac {f'}f \right)$$ and simplify again.
