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Expert: Sherman D. - 7/23/2007


Question
i know it says you dont do calculus but since you are the only one available i need it by tomarrow i thought i would try...

prove that (d/dx)sec(x)=Sec(x)Tan(x)

Answer
sec(x)' =
(1/cos(x))' =
(1'cos(x) - 1cos(x)')/(cos(x)^2) =
(0 - (-sin(x)))/(cos(x)^2) =
(sin(x))/(cos(x)^2) =
(sin(x)/cos(x)) * (1/cos(x)) =
tan(x)sec(x)

so

(d/dx)sec(x) = sec(x)tan(x)

once i figured out that you weren't asking for the antiderivative, then it was easy to figure out.

this website will also show you step by step of how to solve it.

http://archives.math.utk.edu/visual.calculus/2/trig.1/7.html

for others like this, go to http://archives.math.utk.edu/visual.calculus/2/trig.1/index.html

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Sherman D.

Expertise

I can answer questions dealing in mathematics of all kinds except for Physics and Calculus, but i can answer questions in Pre-Calculus and Chemistry. I can also answer questions in Recipes of all kinds. I can find games cheats/walkthroughs, but i can`t find a specific game online or offline. I can also do history and recipes for alcoholic beverages.

Experience

Mathematics, Recipes, History, and Games.

Education/Credentials
High School graduated. I graduated with honors, and i was in Beta Club for a year and a half.

Awards and Honors
Principle's list and A and B honor roll in high school only.

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