How do you find the derivative of # y =sinx/(1-cosx)#?

Question

10210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-04T08:40:39

#- (1) / (1 - cos(x))#

We have: #y = (sin(x)) / (1 - cos(x))#

This function can be differentiated using the "quotient rule":

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = ((1 - cos(x)) cdot (d) / (dx) (sin(x)) - (sin(x)) cdot (d) / (dx) (1 - cos(x))) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = ((1 - cos(x)) cdot cos(x) - sin(x) cdot - (- sin(x))) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = (cos(x) - cos^(2)(x) - sin^(2)(x)) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = (cos(x) -1 (cos^(2)(x) + sin^(2)(x))) / ((1 - cos(x))^(2))#

Let's apply the Pythagorean identity #cos^(2)(x) + sin^(2)(x) = 1#:

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = (cos(x) - 1 cdot 1) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = (cos(x) - 1) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = ((-1)(1 - cos(x))) / ((1 - cos(x))^(2))#

#=> (d) / (dx) ((sin(x)) / (1 - cos(x))) = - (1) / (1 - cos(x))#