1

For example:

Find the following derivatives for the given values of $x$.

$$\frac{d}{dx}\arccos (x^2), x \in (−1, 1).$$

Now finding the derivative is easy: $$=-\frac{2x}{\sqrt{1-x^4}}.$$

Am I suppose to find the derivatives at every point in that interval? If so, do I graph it? Or do I just find the derivative of the function and that's it? I am confused why the $x \in (−1, 1)$, is it because it is undefined?

amWhy
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Jon Wick
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  • Yes, because the domain of the derivative is $(-1,1)$. – Vishu Apr 02 '21 at 15:08
  • If $\lvert x\rvert > 1$ we get a negative number inside the square root and if $\lvert x\rvert =1$ we divide by zero. – vitamin d Apr 02 '21 at 15:14
  • In my opinion, the answer that you have given is what the problem composer is looking for, as is, with nothing else needed. Also, the previous comments explain why the problem composer included the specification of $x \in (-1,1)$. It was so that you wouldn't have to worry whether your answer was valid throughout the domain of the function. – user2661923 Apr 02 '21 at 15:28
  • Note also that although the function is valid for $x \in {-1,1}$, because the range of the cosine function is $[-1,1]$, the formula for the derivative is not valid for $x \in {-1,1}.$ – user2661923 Apr 02 '21 at 15:31

1 Answers1

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Why $x\in (-1,1)$ only? It is because $\cos t$ has values between $-1$ and $+1$ only when $t$ is real so $x$ such that $|x|\gt 1$ is not considered. Now the problem is why $1$ and $-1$ are not considered (Refer Note).

Also note that it is not always so direct to find derivative of inverse trig. functions so be careful while differentiating these and make sure that you consider only that domain where inverse exists (that is, trig. functions are bijective).

In your case, you may proceed as below also: $\frac{d}{dx}\arccos (x^2)$ where $x\in (−1, 1)$
$x^2\in [0,1)$

Let $\theta=\arccos (x^2)$, clearly $\theta \in (0,\pi/2]$ (considering principal domain of arccos as $[0,\pi]$) so $\cos\theta =x^2$

$\frac{d}{dx}\cos\theta=2x\implies -\sin\theta \frac{d}{dx}(\theta)=2x\implies \frac{d}{dx}(\theta)=-\frac{2x}{\sqrt{1-\cos^2\theta}}=-\frac{2x}{\sqrt{1-x^4}}\tag 1$

Note: Let $f(x)=\arccos x^2$,where $x\in (-1,1)$ then by definition , derivative of $f$ at $x=1$$\begin{align}&=\lim_{h\to 0^+}\frac{\arccos (1)-\arccos(1-h)^2}{h}\\&=\lim_{h\to 0^+}\frac{0-\arccos(1-h)^2}{h}\\&=\lim_{h\to 0^+}-\frac{2(1-h)}{\sqrt{1-(1-h)^4}}\\&=\lim_{h\to 0^+}-\frac{2(1-h)}{\sqrt{(1-(1-h)^2)(1+(1-h)^4)}}\to -\infty\end {align}$.

In 3rd line above, L'Hopital's rule has been applied using result obtained in $(1)$.

Similarly can you show that $f'(-1)$ is also not finite?

Koro
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