Question

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

Answer 1

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)))`