Question 1200825
<pre>
{{{y(x) = ln(sec^""(x)^""^"" + tan^""(x))}}}

Use the formulas: {{{d/(dx)}}}{{{ln(u)}}}{{{""=""}}}{{{(du/dx)/u^""}}}, 
  {{{d/(dx)}}}{{{sec^""(u)}}}{{{""=""}}}{{{sec^""(u)tan^""(u)expr(du/dx)}}},
and {{{d/(dx)}}}{{{tan^""(u)}}}{{{""=""}}}{{{sec^2(u)expr(du/dx)}}}

we have:

{{{dy/dx}}}{{{""=""}}}{{{(sec^""(x)tan^""(x)+sec^2(x))/(sec^""(x)^""^"" + tan^""(x))}}}

{{{dy/dx}}}{{{""=""}}}{{{(sec^""(x)(tan^""(x)+sec^""(x)^""))/(sec^""(x)^""^"" + tan^""(x))}}}

{{{dy/dx}}}{{{""=""}}}{{{(sec^""(x)(cross(tan^""(x)+sec^""(x)^"")))/(cross(sec^""(x)^""^"" + tan^""(x)))}}}


{{{dy/dx=sec^""(x)}}}

{{{d^2y/dx^2=sec^""(x)tan^""(x)}}}

{{{d^3y/dx^3=sec^""(x)(sec^2(x)^""^"")+tan^""(x)(sec^""(x)tan^""(x)^""^"")}}}

{{{d^3y/dx^3=sec^3(x)+tan^2(x)sec^""(x)}}}

{{{d^3y/dx^2=sec^""(x)(sec^2(x)+tan^2(x)^"")}}}

Use formula {{{1+tan^2(u)=sec^2(u)}}} as {{{tan^2(u)=sec^2(u)-1}}}

{{{d^3y/dx^2=sec^""(x)(sec^2(x)+(sec^2(x)^""-1)^"")}}}

{{{d^3y/dx^2=sec^""(x)(sec^2(x)+sec^2(x)^""-1)}}}

{{{d^3y/dx^2=sec^""(x)(2sec^2(x)^""-1)}}}

Edwin</pre>