Question 1024913
Let's do it in sections.
{{{y[1]=ln(x^2)}}}
{{{dy/dx=(1/x^2)(2x)=2/x}}}
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{{{y[2]=1}}}
{{{dy/dx=0}}}
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{{{y[3]=e^(2x)/(sec(x)+1))}}}
{{{dy/dx=(2e^(2x)(sec(x)+1)-e^(2x)*tan(x)sec(x))/(sec(x)+1)^2}}}
{{{dy/dx=(e^(2x)/(sec(x)+1)^2)*(2sec(x)+2-tan(x)sec(x))}}}
{{{dy/dx=(e^(2x)/(sec(x)+1)^2)*(sec(x)(2-tan(x))+2))}}}
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Putting it all together,
{{{dy/dx=2/x+(e^(2x)/(sec(x)+1)^2)*(sec(x)(2-tan(x))+2))}}}