How do you differentiate #f(x)= sqrt (ln(1/x)/x^2#?

1 Answer
Guilherme N.
Dec 29, 2015

Using chain and quotient rule.

Explanation:

  • Chain rule states that #(dy)/(dx)=(dy)/(du)(du)/(dx)#

  • Quotient rule states that for #y=f(x)/g(x)#, #y'=(f'g-fg')/g^2#

We can go renaming parts of the expression in order to make up the chain.

#f(x)=sqrt(u)#
#u=ln(v)/x^2#
#v=1/x#

#(df(x))=(dx)=1/(2u^(1/2))*((1/v)(-1/x^2)(x^2)-ln(1/x)2x)/x^4#

Substituting #v#, then #u# and aggregating what's possible, so far

#(df(x))/(dx)=1/(2u^(1/2))*((1/(1/x))(-1/cancel(x^2))(cancel(x^2))-ln(1/x)2x)/x^4#

#(df(x))/(dx)=1/(2((ln(1/x))/x^2)^(1/2))*((x)(-1/cancel(x^2))(cancel(x^2))-ln(1/x)2x)/x^4#

#(df(x))/(dx)=(-x-2xln(1/x))/(2x^4sqrt((ln(1/x))/x^2))=color(green)((-1-2ln(1/x))/(2x^3sqrt((ln(1/x))/x^2)))#

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