Inverse cosine inside floor function derivative

Question

I'm struggling to find the derivative of this function :

$$y=\bigg\lfloor{\arccos\left(\frac{1}{\tan\left(\sqrt{\arcsin x}\right)}\right)}\bigg\rfloor$$

I've been told it should be 0 but how can I find that ?!

Answer 1

It took a while but I reckon the full answer is:

$$ \frac{1}{2 \sqrt{1-x^2} \sqrt{\sin ^{-1}(x)} \left(\sin ^{-1}(x)+1\right) \sqrt{1-\frac{1}{\tan ^{-1}\left(\sqrt{\sin ^{-1}(x)}\right)^2}} \tan ^{-1}\left(\sqrt{\sin ^{-1}(x)}\right)^2}. $$

Stategy:

• Start with letting $u = \frac{1}{\tan(\sqrt{\sin^{-1}x})}$

• Use the chain rule: $\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} y}{\mathrm{d} u}\frac{\mathrm{d} u}{\mathrm{d} x}. $

  • where $\frac{\mathrm{d} y}{\mathrm{d} u}$ is much easier to find.

• Then let $m =\tan(\sqrt{\sin^{-1}x})$ and apply the chain rule again.

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Domain

As the comment rightly stated, tedious maths alone would not suffice in fully describing this function. Because of the finite range of trig functions (e.g. $\cos$), the domain of $\arccos$, $y$ and of its derivative are therefore also finite and limited to determined intervals.

Answer 2

You don't need to compute anything! Equation

$$y=\lfloor f(x) \rfloor$$

with $f$ continuous (as a composition of continuous functions) gives rise to a function that is constant by intervals ; thus its derivative is equal to $0$ on all these intervals, and undefined because of the jumps that occur at the limits of these intervals.

On the figure below, $y=f(x)$ is the red curve, $y=\lfloor f(x) \rfloor$ is the blue curve.