# What is the derivative of #g(u) = ln(sqrt((3u+6)/(3u-6)))#?

First, notice that you can simplify the function o get

You can differentiate this function by using the chain rule twice and the quotient rule once.

More specifically, you'll need to use the chain rule for

So, start with the target derivative

This is where the quotient rule comes in handy

This means that the target derivative will be equal to

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To find the derivative of ( g(u) = \ln\left(\sqrt{\frac{3u+6}{3u-6}}\right) ), we use the chain rule and the properties of logarithmic and square root functions. The derivative is:

[ g'(u) = \frac{d}{du} \ln\left(\sqrt{\frac{3u+6}{3u-6}}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right)^{\frac{1}{2}} ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right)^\frac{1}{2} ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\left(\frac{3u+6}{3u-6}\right)^\frac{1}{2}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\left(\frac{3u+6}{3u-6}\right)^\frac{1}{2}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{(3u+6)^{\frac{1}{2}}}{(3u-6)^{\frac{1}{2}}}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \left(\frac{\frac{1}{2}(3u+6)^{-\frac{1}{2}}}{(3u-6)^{\frac{1}{2}}} - \frac{\frac{1}{2}(3u-6)^{-\frac{1}{2}}}{(3u+6)^{\frac{1}{2}}}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \left(\frac{3u+6}{2(3u-6)} - \frac{3u-6}{2(3u+6)}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{3u+6 - (3u-6)}{2(3u-6)(3u+6)} ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{12}{(3u-6)(3u+6)} ] [ = \frac{6}{(3u-6)(3u+6)\sqrt{\frac{3u+6}{3u-6}}} ]

So, the derivative of ( g(u) ) is ( \frac{6}{(3u-6)(3u+6)\sqrt{\frac{3u+6}{3u-6}}} ).

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