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

1 Answer
Apr 15, 2016

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

Explanation:

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))