How do you differentiate # y= log (6x-2)#?

Question

1743 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-15T01:53:22

#dy/dx=(3)/(ln10*(3x-1))#

Assuming your #log# is an understood #log_10#...

You can rewrite it using #ln#, which is more familiar to us.

#log_10(6x-2) = ln(6x-2)/ln(10)=1/ln10*ln(6x-2)#

Now differentiation will be relatively easy.

we know that differentiating #ln# goes like this:
#d/dx[ln(u)]=1/u*u'#

so let's differentiate #y# now.
#y=1/ln10*ln(6x-2)#

#dy/dx=1/ln10*1/(6x-2)*(6x-2)'#

#dy/dx=1/ln10*1/(6x-2)*6#

#dy/dx=6/(ln10*(6x-2))#

#dy/dx=(cancel2*3)/(ln10*cancel2*(3x-1))#

#dy/dx=(3)/(ln10*(3x-1))#