$f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
Find $f'(0)$
By apply the Quotient Rule $\frac{f'g - g'f}{g^2} $, I can find the answer but it's too long.
Is there any other methods to simplify the steps?
Take $\log$:
\begin{align*} &\log f(x)\\ &=\log(x-3)+\log(x-2)+\log(x-1)+\log x-\log(x+1)-\log(x+2)-\log(x+3)\\ &\dfrac{f'(x)}{f(x)}\\ &=\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\\ &f'(x)\\ &=f(x)\left(\dfrac{1}{x-3}+\dfrac{1}{x-2}+\dfrac{1}{x-1}+\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{x+2}-\dfrac{1}{x+3}\right) \end{align*} and then plugging in $x=0$. But first cancel the same factor $x$ in $f(x)\cdot\dfrac{1}{x}$ and then plugging in $x=0$. For the other factors, they are zero.
Hint:
Use Partial Fraction Decomposition before differentiation
$$f(x)=\dfrac{x(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}=x+\dfrac A{x+1}+\dfrac B{x+2}+\dfrac C{x+3}$$
$$(x+1)(x+2)(x+3)f(x)=?$$
Set $x=-1,-2,-3$ to find $A,B,C$
$$f'(0)=1-\dfrac A{1^2}-\dfrac B{2^2}-\dfrac C{3^2}$$
Put $y = f(x)$. Taking natural logarithm on both sides of the equation,
$$\ln(y) = \ln(x-3) + \ln(x-2) + \ln(x-1) + \ln(x) - \ln(x+1) - \ln(x+2) - \ln(x+3)$$
$$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x-3} + \frac{1}{x-2} + \frac{1}{x-1} + \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} - \frac{1}{x+3}$$
Now, multiplying both sides by $y = f(x)$,
$$f'(x) = f(x)(\frac{1}{x-3} + \frac{1}{x-2} + \frac{1}{x-1} + \frac{1}{x} - \frac{1}{x+1} - \frac{1}{x+2} - \frac{1}{x+3})$$
Then put $x = 0$.
