Question

What are the first and second derivatives of `g(x) =ln(arctanx)-arctan(lnx)`?

Answer 1

We'll need chain rule for both terms.
- Chain rule: `(dy)/(dx)=(dy)/(du)(du)/(dx)`

Explanation:

Also, let's just remember the rule to differentiate `ln` and `arctan`:

• `(ln(u))'=1/u`

• `(arctanu)'=(u')/(1+u^2)`

Therefore, using the chain rule for each term... In the first, we'll rename `ln(u)` with `u=arctanx` and in the second, let's just vary the letter and rename `arctan(v)`, with `v=lnx`.

`(dg(x))/(dx)=1/u(1/(1+x^2))-(v')/(1+v^2)(1/x)`

Substituting `u`, `v` and `v'`:

`(dg(x))/(dx)=1/(arctanx)(1/(1+x^2))-(1/x)/(1+ln^2x)`

`(dg(x))/(dx)=1/(arctanx(1+x^2))-1/(x(1+ln^2x))`

`(dg(x))/(dx)=1/(x^2arctanx+arctanx)-1/(x+xln^2x)`

The second derivative will demand some product rule: `(ab)'=a'b+ab'`. Also, quotient rule, as we have two fractions: `(a/b)'=(a'b-ab')/b^2`

`(dg(x)^2)/(d^2x)=(0-(2xarctanx+x^2(1/(1+x^2))+1/(1+x^2)))/(x^2arctanx+arctanx)^2-(0-(1+ln^2x+cancel(x)(2lnx)/cancel(x)))/(x+xln^2x)^2`

`(dg(x)^2)/(d^2x)=-(2xarctanx+cancel((x^2+1)/(x^2+1)))/(x^2arctanx+arctanx)^2+(1+ln^2x+2lnx)/(x+xln^2x)^2`

Simplifying and rearranging:

`(dg(x)^2)/(d^2x)=(1+lnx)^2/(x^2(ln^2x+1))-(2xarctanx+1)/((x^2+1)arctan^2x)`