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

Question

Emily Ammerman · Accepted Answer

#- (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))#

Ezekiel Alexander · Answer

undefined To find the derivative of $ y = \frac{\sin x}{1 - \cos x} $, we use the quotient rule.

Let $ u(x) = \sin x $ and $ v(x) = 1 - \cos x $.

The derivative of $ u(x) $ is $ u'(x) = \cos x $, and the derivative of $ v(x) $ is $ v'(x) = \sin x $.

Now, applying the quotient rule $ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} $, we plug in these derivatives:

$$ y' = \frac{(\cos x)(1 - \cos x) - (\sin x)(-\sin x)}{(1 - \cos x)^2} $$

Simplify the expression if necessary. This yields the derivative of $ y $ with respect to $ x $.