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

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Answer 1 · 2015-04-11T13:39:38

You could immediately use: #d/(dx)( ln u) = 1/u (du)/(dx)#

which is also written #d/(dx)( ln g(x)) = 1/g(x) g'(x) = (g'(x))/g(x)#.

That will work, another way is to rewrite before differentiating.

#y=ln((2x)/(x+3)) = ln(2) +ln(x) - ln(x+3)#

So #y' = = 1/(x) - 1/(x+3)#

Is you want to rewrite this as a single fraction, do so.