I don't know if this answer to a log derative is right please check for me?

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Answer 1 · 2018-02-01T01:06:43

#-1/(xln2)#

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We need to use the Log rule of differentiation:

#Log_bu=lnu/lnb#

#d/dxLog_8(2/x^3)=d/dx(ln(2/x^3)/ln8)#

Let #u=2/x^3# and #b=8#

Let's take the constant out and take the derivative:

#=1/ln8*d/dxln(2/x^3)=1/ln8*d/(du)lnu=D#

#du=-6/x^4dx#

#D=1/ln8*1/udu#

Now, we can substitute back:

#D=1/ln8(1/(2/x^3))(-6/x^4)=1/ln8(x^3/2)(-6/x^4)#

#D=1/ln8(-3/x)=(1/ln2^3)(-3/x)=(1/(3ln2))(-3/x)#

#D=(1/(cancelcolor(red)3ln2))(-cancelcolor(red)3/x)=-1/(xln2)#