What is the derivative of #ln(8x)^(1//2)#?

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Answer 1 · 2017-07-10T20:51:21

#d/(dx) [ln(8x)^(1"/"2)] = color(blue)(1/(2xsqrt(ln(8x)))#

If you mean #d/(dx) [ln(sqrt(8x))]#, see Truong-Son N.'s answer.

If you mean #d/(dx) [sqrt(ln(8x))]#, see my explanation here:

We're asked to find the derivative

#d/(dx) [ln(8x)^(1/2)]#

This can also be written

#d/(dx) [sqrt(ln(8x))]#

Use the chain rule:

#d/(dx) [sqrt(ln(8x))] = d/(du) [sqrtu] (du)/(dx)#

where

#u = ln(8x)#

#d/(du) [sqrtu] = 1/(2sqrtu)#:

#= (d/(dx) [ln(8x)])/(2sqrt(ln(8x)))#

Use the chain rule again:

#d/(dx) [ln(8x)] = d/(du) [lnu] (du)/(dx)#

where

#u = 8x#

#d/(du) [lnu] = 1/u#:

#= (d/(dx) [8x])/((8x)(2sqrt(ln(8x)))#

#= 8/(16xsqrt(ln(8x)))#

#= color(blue)(1/(2xsqrt(ln(8x)))#

Answer 2 · 2017-07-10T23:49:57

Nathan has a good approach, but I figured I'd add an alternative answer. I got

#(dy)/(dx) = 1/(2x)#

I assume you mean #ln(8x)^"1/2"# as-written (rather than #(ln(8x))^"1/2"# or #ln^"1/2"(8x)#).

You can rewrite this as:

#y = ln(8x)^"1/2"#

#= 1/2 ln(8x)#

#= 1/2 ln8 + 1/2 lnx#

So, the derivative is just

#color(blue)((dy)/(dx) = 1/(2x))#