What is the derivative of #y=tan(ln(2x+1))#?
1 Answer
May 22, 2017
# dy/dx = (2sec^2(ln(2x+1)))/(2x+1) #
Explanation:
Using the standard results:
# d/dx(tanx)=sec^2x # and#d/dxlnx=1/x#
Along with the chain rule then differentiating:
# y = tan(ln(2x+1)) #
gives:
# dy/dx = sec^2(ln(2x+1)) * d/dx ln(2x+1) #
# \ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * d/dx(2x+1) #
# \ \ \ \ \ = sec^2(ln(2x+1)) * 1/(2x+1) * 2 #
# \ \ \ \ \ = (2sec^2(ln(2x+1)))/(2x+1) #
