What is the derivative of #ln(4x)#?

1 Answer
Jim H
Jun 29, 2015

1/x

Explanation:

#y = ln4x#

We have a choice. We can either use the chain rule in the form:
#d/dx(ln(u)) = 1/u * (du)/dx# OR we can use properties of logarithms to rewrite the function.

Chain Rule Solution

#d/dx(ln4x) = 1/(4x) * d/dx(4x) = 1/(4x) * 4 = 1/x#

Rewrite Solution

Use #lnab = lna + lnb#, to get:

#d/dx(ln4x) = d/dx(ln4+lnx) = d/dx(ln4) + d/dx(lnx) = 0+(1/x) = 1/x#

(Note that #ln4# is some constant, hence its derivative is #0#.)

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