How do you find the derivative of #ln(4x)#?

Question

Isaiah Allen · Accepted Answer

It is #1/x#.

#ln(4x)# is a composite function, composed of the functions #lnx# and #4x#. Because of that, we should use the chain rule: #dy/(dx) = (dy)/(du) (du)/dx# We already know that #(lnx)' = 1/x#. So, we want what's inside of the natural logarithm to be a single variable, and we can do this by setting #u = 4x#. Now we could say that #(lnu)' = 1/u#, with respect to #u#. Essentially, the chain rule states that the derivative of #y# with respect to #x#, is equal to the derivative of #y# with respect to #u#, where #u# is a function of #x#, times the derivative of #u# with respect to #x#. In our case, #y = ln(4x)#. Differentiating #u# with respect to #x# is simple, since #u = 4x#: #u' = 4#, with respect to #x#. So, we see that: #dy/(dx) = 1/u * 4 = 4/u# Now we can change #u# back into #4x#, and get #4/(4x) = 1/x#. Interestingly enough, #[ln(cx)]'# where #c# is a non-zero constant, where it is defined, is equal to #1/x#, just like #(lnx)'#, even though we are using the chain rule.

Emma Aldridge · Answer

undefined To find the derivative of ln(4x), you can use the chain rule. The derivative is 1/(4x) * 4 = 1/x.