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

1 Answer
Feb 28, 2018

#g'(x)=cot(x)+tan(x)-(sec^2(ln(x)))/x#
#g''(x)=sec^2(x)-csc^2(x)-(2sec^2(lnx)tan(lnx)-sec^2(lnx))/x^2#

Explanation:

Recall that #d/dxln(x)=1/x#

Furthermore, if we have a natural logarithm whose argument is another function, #ln(f(x)),# we can say

#d/dxln(f(x))=1/f(x)*d/dxf(x)# As per the Chain Rule.

Let's take a look at #ln(tan(x)):#

#d/dxln(tan(x))=1/tan(x) * d/dxtan(x)#

#d/dxln(tan(x))=sec^2(x)/tan(x)#

Let's simplify this a bit, using the fact that #sec^2(x)=1+tan^2(x)#:

#d/dxln(tan(x))=(1+tan^2(x))/tan(x)=1/tan(x)+(tan^(cancel(2)1)(x))/canceltan(x)=1/tan(x)+tan(x)=cot(x)+tan(x)#

Recall that #d/dxtan(x)=sec^2(x)#.

If we have a tangent function whose argument is another function, #tan(f(x)),# we can say

#d/dxtan(f(x))=sec^2(f(x))*d/dxf(x)# As per the Chain Rule.

Looking at #tan(ln(x)):#

#d/dxtan(ln(x))=sec^2(ln(x))*1/x=sec^2(ln(x))/x#

Combining our two derivatives together, we get

#g'(x)=d/dx(ln(tan(x))-tan(ln(x))#

#g'(x)=d/dxln(tan(x))-d/dxtan(ln(x))#

#g'(x)=cot(x)+tan(x)-(sec^2(ln(x)))/x#

For #g''(x):#

#d/dxcot(x)=-csc^2(x)#
#d/dx(tan(x))=sec^2(x)#

#d/dx(sec^2(ln(x)))/x=(xd/dxsec^2(ln(x))-sec^2ln(x)d/dxx)/x^2#

This gives us:

#((2cancelx2sec(lnx)sec(lnx)tan(lnx))/cancelx-sec^2(lnx))/x^2#

#(sec^2(lnx)tan(lnx)-sec^2(lnx))/x^2#

#g''(x)=sec^2(x)-csc^2(x)-(2sec^2(lnx)tan(lnx)-sec^2(lnx))/x^2#