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

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

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Answer 1 · 2018-03-23T06:01:20

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

Explanation:

$2 \frac{d}{d x} sec [\sqrt{3 x + 1}]$ $2 d/dx sec[sqrt(3x+1)]$

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

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

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

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

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

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