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

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23
Apr 3, 2016

Answer:

#1/x#

Explanation:

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.

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Shura Share
Apr 3, 2016

Answer:

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

Explanation:

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#.

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