How do you find the derivative of #y=ln(5x)#?

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Jim H Share
Mar 14, 2015

A student comfortable with the natural logarithm function and its properties might think of this:

One could reason as follows:
#y=ln(5x)=ln(5)+ln(x)#.

But #ln(5)# is a constant, so its derivative is #0#.

Therefore, #(dy)/(dx) = d/(dx)(ln5+lnx)=d/(dx)(lnx)=1/x#.

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Gaurav Share
Jul 29, 2014

#y'=1/x#

Explanation

Suppose, #y=ln(b(x))#

then using Chain Rule,

#y'=1/(b(x))*(b(x))'#

Similarly following for the above function yields,

#y'=1/(5x)*(5x)'#

#y'=1/(5x)*5#

#y'=1/x#

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Gió Share
Mar 24, 2015

Using the Chain Rule you'll get:

#y'=1/(5x)*5=1/x#

(Derive #ln# as it is and then multiply by the derivative of the argument, #5x#).

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