Derivation of $\frac{d}{dx} \arccos{\frac{1}{x}}$

calculustrigonometry

I am trying to derive $\frac{d}{dx} \arccos{\frac{1}{x}}$ and found a (rather longer than necessary) method below. I realise there are more concise ways, but I wanted to check the method below is technically correct, and interpretation of the $\frac{d}{dx}$ and $dx$ variables.

Starting with

$\cos{y} = \frac{1}{x}$

$y = \arccos{\frac{1}{x}}$

taking the derivative of both sides and expanding $y$ to $\arccos{\cos{y}}$

$\frac{d}{dx} \arccos{(\cos{y})} = \frac{d}{dx} \arccos{\frac{1}{x}}$

$\frac{d}{\cos{y}}\arccos{(\cos{y})} \cdot (-\sin{y}) \cdot \frac{dy}{dx} = \frac{dy}{dx} $

$\frac{d}{\cos{y}} y = \frac{-1}{\sin{y}}$

$\frac{d}{\cos{y}}y = \frac{-1}{\sqrt{1 – \cos^2{y}}}$

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

$\frac{d}{\frac{d(\frac{1}{x})}{dx}dx}y = \frac{-1}{\sqrt{1 – \cos^2{y}}}$

$\frac{d}{dx}y \frac{dx}{d(\frac{1}{x})} = \frac{-1}{\sqrt{1 – \cos^2{y}}}$

$\frac{d}{dx}\arccos{\frac{1}{x}} = \frac{-1}{\sqrt{1 – \frac{1}{x^2}}} \frac{d(\frac{1}{x})}{dx}$

Which is the correct answer. My questions are:

  1. Is this valid? For some reason I thought we could not manipulate $\frac{dy}{dx}$ like that, but have since seen it done in my text book.
  2. Interpreting $dx$, this is the arbitary choice of $\delta{x}$, and when we differentiate we take the limit to 0. But before we differentiate we can read this as some arbitary segment along x. Is this correct?
  3. Similar, $\frac{dy}{dx}$ is the change in y w.r.t change in x when we make a small $\delta{x}$. As such $\frac{dy}{dx}dx$ can be interpreted as $dy$ because we are multiplying $\frac{dy}{dx}$ with are arbitary change in x, giving an arbitary change in y. Is this correct, is there anything further to consider?

Thank for your help

Best Answer

$1.$ Your answer is valid, if longer. From a purely technical standpoint, if you know the derivative of $\arccos x$, then the most concise way of getting this derivative is through the chain rule, rather than implicit differentiation.

$$y = \arccos \bigg(\frac {1}{x}\bigg) \\ y' = \bigg({\frac {-1}{\sqrt{1-\frac{1}{x^2}}}}\bigg)\bigg({\frac {-1}{x^2}}\bigg) \\ y' = \dfrac {1}{x^2 \sqrt {1-\frac{1}{x^2}}}$$

$2$ and $3.$: You are correct. If you use the definition of the derivative $$\frac {dy}{dx} = \lim_{\Delta x \rightarrow 0} \frac {f(x+\Delta x)-f(x)}{\Delta x}$$ we can take $\Delta x$ smaller and smaller as $x$ approaches zero, and then we can call it $dx$ when it does approach $0$.

Thence we have $${dy} = \lim_{\Delta x \rightarrow 0} \frac {f(x+\Delta x)-f(x)}{\Delta x}\ dx$$

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