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

1 Answer
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)))