derivative of trigonometric function?

Question

y=(1+cosx)/(1-cosx)

y'=? derivative

Answer 1

This is a fraction, yes?

The derivative of a fraction is

The derivative of the numerator times the denominator
minus
The derivative of the denominator times the numerator
All divided by the square of the denominator

Let's do it:

[-sin(x) * (1 - cos(x) - sin(x) * (1 + cos(x)] / [1 - cos(x)]^2
[-2sin(x)] / [1 - cos(x)]^2

Notice that your function is discontinuous at x = 0 and x = pi, etc., and so is the derivative.

I LUV screwed up again. The derivative of cos(x) is -sin(x), not sin(x) as she has it.

Steve's solution agrees with mine.

Answer 2

Remember for taking derivatives of fractions: the derivative = bottom times deriv. of top, minus top times deriv. of bottom, all over bottom squared.

f(x) = (1+cosx) ÷ (1-cosx)

f '(x) = [ (1-cosx)(sinx) - (1+cosx)(sinx) ] ÷ (1-cosx) ²

= [sinx - sinxcosx - sinx - sinxcosx ] ÷ (1-cosx) ²

= -2sinxcosx / (1-cosx) ²

= -sin2x / (1-cosx) ²

I used a trig identity for that last step: sin2a = 2sinacosa

~ ♥ ~

Answer 3

ILUV, that's a /, not a * between the parentheses! (I know, you were just testing me...!)

Not seeing any simplification for this expression, just use the normal quotient formula for a derivitive:

((1-cosx)(-sinx) - (1+cosx)(sinx))/(1-cosx)^2

Answer 4

y = (1+cos x)/(1- cos x)
= (1+cos x)^2/(1-cos^2x)
= (1+cosx)^2/sin ^2 x
= (cosec x + tan x)^2

dy/dx = 2 (cosec x + tan x )d/dx(cosec x + tan x)
2(cosec x + tan x) (cosec x cot x +sec^2x)