How do you find the derivative of #2sec((3x+1)^(1/2))#?

1 Answer
Mar 23, 2018

Answer:

#3*{sec[sqrt(3x+1)]*tan[sqrt(3x+1)]}/[sqrt(3x+1)]#

Explanation:

#2 d/dx sec[sqrt(3x+1)]#

#=2 {[secsqrt(3x+1)]*[tansqrt(3x+1)]}*d/dx [sqrt(3x+1)]}#

#=2*{[sec sqrt(3x+1)]*[tan sqrt(3x+1)]}*{1/[2(sqrt (3x+1)]}*d/dx(3x+1)}#

#=2*{[sec sqrt(3x+1)]*[tan sqrt(3x+1)]}*[1/[2(sqrt (3x+1)]]*(3(1)+0)}#

#=2*{[sec sqrt(3x+1)]*[tan sqrt(3x+1)]}*{1/[2*sqrt (3x+1)]}*3#