How do you differentiate #y=12lnx#?

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Answer 1 · 2018-04-10T06:34:18

The derivative of #12lnx# is #12/x#.

Using the fact that the derivative of #lnx# is #1/x#:

#color(white)=d/dx[12lnx]#

#=12*d/dx[lnx]#

#=12*1/x#

#=12/x#

Here's a short proof for the derivative of #lnx#.

We know that #d/dx[x]=1#, that #d/dx[e^x]=e^x#, also that #e^lnx=x#:

#d/dx[x]=1#

#d/dx[e^lnx]=1#

#e^lnx*d/dx[lnx]=1#

#x*d/dx[lnx]=1#

#d/dx[lnx]=1/x#