Question

how to find derivative using first principal in trignometry?
f(x)= sec^2 (x)

Answer 1

any one here to answer me>?

Answer 2

i do not know what you mean by first principal in trigonometry, but teh derivative of sec^2(x) is 2tan(x)sec^2(x)

Answer 3

yeh but how?

Answer 4

i think its a rule sorry I do not really remember the rule

Answer 5

f(x)=f(x+h)-f(x)/h this the formula

Answer 6

I think I would use the quotient rule, and re-write sec^2(x) as 1/cos^2(x). If I have the quotient rule correct, it is something like d/dx of f(x)/g(x) is g(x)f'(x)-g'(x)f(x) all over g(x)^2. So if we just do the denominator real quick to get it out of the way, we have (cos^2(x))^2 which becomes cos^4(x). Now for the numerator, we already know f(x) = 1, and d/dx of that = 0. For finding g'(x), I went ahead and just did the product rule (f'g+fg') on cos^2(x). Since we know f' is 0, that makes part of the work on the numerator easy, it will just be 0 - (something). That some is g'(x)f(x), which when I did the product rule on cos^2(x) came out to be -2sin(x)cos(x). Since there was already a minus before the (something) in the numerator, our problem is now looking like [2sin(x)cos(x)]/[cos^4(x)]. Canx out a cos(x) from top and bottom, and that leaves us with the 2 up top, a sin(x) on top and one of the three cos(x) in the denom to make the tan(x), and cos^(x) left over in the denom which we can just go ahead and rewrite as sec^2(x).

Answer 7

sorry, typo: in that last sentence I meant to say "that leaves us with a cos^2(x) in the denominator, which we can go ahead and write as sec^2(x)."