Question

How do you differentiate `f(x)=sqrt(ln2x)` using the chain rule?

Answer 1

`1/(2xsqrt(ln2x))`

Explanation:

using the`color(blue)" chain rule"`

If f(x)=[f(g(x))] then f'(x) = f'(g(x)).g'(x)
`"--------------------------------------------"`
Rewrite `sqrt(ln2x)=(ln2x)^(1/2)`

f(g(x))=`(ln2x)^(1/2)rArrf'(g(x))=1/2(ln2x)^(-1/2)`

and g(x)`=ln2xrArrg'(x)=1/(2x).d/dx(2x)=1/(2x)xx2=1/x`
`"-------------------------------------------------------------------------"`
Substituting these values into f'(x)

`f'(x)=1/2(ln2x)^(-1/2)xx1/x=1/(2x(ln2x)^(1/2))=1/(2xsqrt(ln2x))`