derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$  
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$   
I know that the quotient rule is used but I don't know how to do this problem. Would you multiply together all the terms and then differentiate?
 A: Use the following technique,
$$y=f_1(x)f_2(x)...f_n(x)$$
$$ln(y)=ln(f_1(x))+...ln(f_n(x))$$,take the derivative on both side, applying chain rule
$$\frac{y'}{y}=\frac{f_1'(x)}{f_1(x)}+...\frac{f_n'(x)}{f_n(x)}$$
$$y'=y\left( \frac{f_1'(x)}{f_1(x)}+...\frac{f_n'(x)}{f_n(x)} \right)$$
A: $$y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}\\ln (y)=ln(\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5})=\\ln(y)=2lnx +\frac{1}{2}ln(x+1)-ln(x+2)-5ln(x-3)\\\frac{y'}{y}=(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\so\\y'=y*(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\y'= \frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})$$
A: It's a long derivative. First use the quotient rule:
$$\frac{\frac{d}{dx}\left(x^2\sqrt{x+1}\right)(x+2)(x-3)^5- x^2\sqrt{x+1}\frac{d}{dx}\left((x+2)(x-3)^5\right)}{((x+2)(x-3)^5)^2}.$$
Then do the derivatives.
A: $$y'=\dfrac{(x^2\sqrt{x+1})'(x+2)(x-3)^5-x^2\sqrt{x+1}((x+2)(x-3)^5)'}{(x+2)^2(x-3)^{10}}=
\dfrac{((2x\sqrt{x+1}+\frac{x^2}{2\sqrt{x+1}})(x+2)(x-3)^5-x^2\sqrt{x+1}((x-3)^5)+5(x+2)(x-3)^4)}{(x+2)^2(x-3)^{10}}=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}(\dfrac{2}{x}+\dfrac{1}{\sqrt{x+1}\sqrt{x+1}}-(\dfrac{1}{x+2}+\dfrac{5}{x-3})$$
