What is the derivative of #ln(2x)#?

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Answer 1 · 2016-04-03T09:00:58

#(ln(2x))' = 1/(2x) * 2 = 1/x.#

You use the chain rule :

#(f @ g)'(x) = (f(g(x)))' = f'(g(x)) * g'(x)#.

In your case : #(f @ g)(x) = ln(2x), f(x) = ln(x) and g(x) = 2x#.

Since #f'(x) = 1/x and g'(x) = 2#, we have :

#(f @ g)'(x) = (ln(2x))' = 1/(2x) * 2 = 1/x#.

Answer 2 · 2016-04-03T09:04:51

#1/x#

You can also think of it as

#ln(2x) = ln(x) + ln(2)#

#ln(2)# is just a constant so has a derivative of #0#.

#d/dx ln(x) = 1/x#

Which gives you the final answer.