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(tan(ln(2*x+1)))'The calculation above is a derivative of the function f (x)
(ln(2*x+1))'/((cos(ln(2*x+1)))^2)
((1/(2*x+1))*(2*x+1)')/((cos(ln(2*x+1)))^2)
((1/(2*x+1))*((2*x)'+(1)'))/((cos(ln(2*x+1)))^2)
((1/(2*x+1))*(2*(x)'+(2)'*x+(1)'))/((cos(ln(2*x+1)))^2)
((1/(2*x+1))*(2*(x)'+0*x+(1)'))/((cos(ln(2*x+1)))^2)
((1/(2*x+1))*(0*x+2*1+(1)'))/((cos(ln(2*x+1)))^2)
((0+2)*(1/(2*x+1)))/((cos(ln(2*x+1)))^2)
(2/(2*x+1))/((cos(ln(2*x+1)))^2)
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