How do you use the chain rule to differentiate #sqrt(ln4x)#?

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Answer 1 · 2018-01-25T11:58:50

#1/(2xsqrt(ln4x))#

Apply the chain rule:
#d/dxsqrt(ln4x)#

So the chain rule states that #(df(u))/dx=(df)/(du)*(du)/dx#.
Here, #f=sqrt(u)#, and #u=ln4x#.

According to the power rule #d/dxx^a=ax^(a-1)#.

So #d/(du)sqrt(u)#
#=d/(du)u^(1/2)#
#=1/2(1/sqrt(u))#
#=1/(2sqrt(u))#

Now that we've solved #(df)/(du)#, we have to solve #(du)/dx#.

But we can't. We cannot differentiate #ln4x# directly.

Use the chain rule again: Here, #f=lnu#, and #u=4x#
We know that #d/(du)lnu=1/u#, and that #d/dx4x=4#

So now we have #d/(dx)ln4x=1/u*4#

But #u=4x,# so input:

#1/(4x)*4=1/x#

So now we have #1/(2sqrt(u))*1/x#

#=1/(2sqrt(ln4x))*1/x#

#=1/(2xsqrt(ln4x))#