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

Question

1743 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-23T06:01:20

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

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

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

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

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

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

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