# Show that the derivative of #(sinx)/(1 - cosx)# is #-cscx#?

and not

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We can't, because it's false...

Let's use the product rule.

Therefore:

Clearly, we have found that this identity is false. More definitively, we can plot this equation to get:

graph{sinx + cosx [-10, 10, -5, 5]}

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To find the derivative of (\frac{\sin(x)}{1 - \cos(x)}), use the quotient rule: [ \frac{d}{dx}\left(\frac{\sin(x)}{1 - \cos(x)}\right) = \frac{(1 - \cos(x))\cos(x) - (-\sin(x))(\sin(x))}{(1 - \cos(x))^2} ] [ = \frac{\cos(x) - \cos^2(x) + \sin^2(x)}{(1 - \cos(x))^2} ] [ = \frac{\cos(x) - \cos^2(x) + (1 - \cos^2(x))}{(1 - \cos(x))^2} ] [ = \frac{\cos(x) - \cos^2(x) + 1 - \cos^2(x)}{(1 - \cos(x))^2} ] [ = \frac{1 - \cos^2(x) - \cos^2(x) + \cos(x) + 1}{(1 - \cos(x))^2} ] [ = \frac{2 - 2\cos^2(x) + \cos(x)}{(1 - \cos(x))^2} ] [ = \frac{2(1 - \cos^2(x)) + \cos(x)}{(1 - \cos(x))^2} ] [ = \frac{2\sin^2(x) + \cos(x)}{(1 - \cos(x))^2} ] [ = \frac{2\sin^2(x) + \cos(x)}{\sin^2(x)} ] [ = \frac{2\sin^2(x)}{\sin^2(x)} + \frac{\cos(x)}{\sin^2(x)} ] [ = 2 + \frac{\cos(x)}{\sin^2(x)} ] [ = 2 + \csc^2(x) ] [ = -\csc^2(x) ] Therefore, the derivative of (\frac{\sin(x)}{1 - \cos(x)}) is (-\csc^2(x)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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