what is the derivative of sec x+csc x?

Question

my teacher gave a take home test and never taught me these things

Answer 1

(secx)(tanx) - (cscx)(cotx)

d/dx secx = secxtanx
d/dx cscx = -cscxtanx
These are derived from the quotient rule:
d/dx (f/g) = (gf' - fg')/(g^2)

Answer 2

It's easier to do a problem like this if you convert all the trig functions to sine and cosine, then follow the rules of calculus to find the appropriate derivatives.

sec x = 1/ cos x

csc x = 1 / sin x.

We have to use what's called the quotient rule in calculus to find the derivative of each of these because they are both ratios or fractions, therefore division is implied.

The quotient rule states that for two differentiable functions in x, the derivative of their quotient is found this way:

d(f(x)/g(x))/dx = [g(x) f '(x)] - [f(x) g '(x)]/ [g(x)]^2

In words we would say it like this:

The derivative of the quotient of two differentiable functions in x is the denominator times the derivative of the numerator ( [g(x) f '(x)] ) minus the numerator times the derivative of the denominator ( [f(x) g '(x)] ) divided by the square of the denominator, which of course is [g(x)^2]. Note that g(x) cannot equal zero, since division by zero is an undefined operation in mathematics.

d(sec x)/dx = d(1/cos x)/dx = [(cos x)(0) - (1)(-sin x)]/(cos x)^2 =
(1/cos x)(sin x/ cos x) = sec x tan x


d(csc x)/dx = d(1/sin x)/dx = [(sin x)(0) - (1)(cos x)]/(sin x)^2 =
(-1)(1/sin x)(cos x/ sin x) = -csc x ctn x

So d(sec x + csc x)/dx is:

sec x tan x - csc x ctn x.

Answer 3

secxtanx-cscxcotx