A weird differentiation question.

Question

I was looking at questions on differentiation and came across a weird form that both me and my friend had two different approaches to. This was the question.

Calculate the derivative:

$$ \frac{d(x-\sin(x))}{d(1-\cos(x))} $$

My approach was to factorize out an $x$ in the denominator of the fraction and differentiate as normal like so:

$$ \frac{d(x-\sin(x))}{dx(\frac{1}{x} - \frac{\cos(x)}{x})} = \frac{d}{dx}(\frac{x-\sin(x)}{\frac{1}{x} - \frac{\cos(x)}{x}}) $$

My friend's approach was to let $ u = 1-\cos(x) $ and to then take the approach that way leading to an answer.

Which way would be considered correct and why would this work? Thank you!

Answer 1

Hint:

$$\frac{dy}{dz} = \frac{dy}{dx} \frac{dx}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}}$$

Let $y = x-\sin(x)$ and $z=1-\cos(x)$

Answer 2

Use the "chain rule", $\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}$ with $f(x)= x- \sin(x)$ and $y= 1- \cos(x)$. So we have $\frac{df}{dx}= 1- \cos(x)$ and $\frac{dy}{dx}= \sin(x)$. Putting those into the "chain rule", $1- \cos(x)= \left(\frac{df}{dy}\right)(\sin(x))$ so $\frac{d(x- \sin(x))}{d(1- \cos(x)}= \frac{df}{dy}= \frac{1- \cos(x)}{\sin(x)}$. You can, if you prefer, write that as $\frac{d(x- \sin(x))}{d(1- \cos(x))}= \frac{1}{\sin(x)}- \frac{\cos(x)}{\sin(x)}= \csc(x)- \cot(x)$.