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

1 Answer
May 5, 2016

Answer:

#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))#