# How do you find the second derivative of #y=ln((2x)/(x+3))# ?

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To find the second derivative of ( y = \ln\left(\frac{2x}{x+3}\right) ), follow these steps:

- Find the first derivative ( y' ) using the chain rule and quotient rule.
- Once you have ( y' ), differentiate it again to find the second derivative ( y'' ).

The first derivative ( y' ) is: [ y' = \frac{d}{dx}\left[\ln\left(\frac{2x}{x+3}\right)\right] ] [ = \frac{1}{\frac{2x}{x+3}} \cdot \frac{d}{dx}\left[\frac{2x}{x+3}\right] ] [ = \frac{x+3}{2x} \cdot \left(\frac{2(x+3) - 2x}{(x+3)^2}\right) ] [ = \frac{x+3}{2x} \cdot \left(\frac{6}{(x+3)^2}\right) ] [ = \frac{3(x+3)}{x(x+3)} ] [ = \frac{3}{x} ]

Now, to find the second derivative ( y'' ): [ y'' = \frac{d}{dx}\left(\frac{3}{x}\right) ] [ = -\frac{3}{x^2} ]

So, the second derivative of ( y = \ln\left(\frac{2x}{x+3}\right) ) is ( y'' = -\frac{3}{x^2} ).

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