<s>[INST] How to solve it in simple ways, Find $f'(0)$ $f(x) = \dfrac{\left(x-3\right)\left(x-2\right)\left(x-1\right)x}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$
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?
 [/INST] 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 $x = 0$. 
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