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

1 Answer
Guilherme N. · Stefan V.
Jan 9, 2016

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)#

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