Question

How do you find the derivative of `ln(ln x^2)`?

Answer 1

`1/(xlnx)`

Explanation:

We have:

`d/dx(ln(lnx^2))`

According to the chain rule, `d/dxf(g(x))=d/dx(f)*d/dxg(x)`

Here, `f(u)=lnu` where `u=lnx^2`

Since `d/(du)lnu=1/u`, we now have:

`1/u*d/dxlnx^2`

Here, `f(v)=lnv` where `v=x^2`, so we have:

`1/u*1/v*2x`

Since `u=lnx^2` and `v=x^2`, we have:

`(2x)/(ln(x^2)x^2)`

Since `lnx^n=nlnx`, we get:

`(2x)/(2x^2lnx)`

`1/(xlnx)`

Answer 2

`d/dxln(lnx^2)=2/(xlnx^2)=1/(xlnx)`

Explanation:

We can use chain rule here. We can write `f(x)=ln(lnx^2)` as

`f(x)=ln(g(x))`, `g(x)=ln(h(x))` and `h(x)=x^2`

then `(df)/(dg)=1/(g(x))`, `(dg)/(dh)=1/(h(x))` and `(dg)/(dh)=2x`

and using chain rule as `(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)`

= `1/(g(x))xx1/(h(x))xx2x`

= `1/lnx^2*1/x^2*2x`

= `2/(xlnx^2)`

Hence `d/dxln(lnx^2)=2/(xlnx^2)=2/(x2lnx)=1/(xlnx)`