Confusion in finding derivative of $\sqrt{\frac{1-\cos(2x)}{1 + \cos(2x)}}$

Question

Find $f'(x)$ where $f(x) = \sqrt{\dfrac{1-\cos(2x)}{1 + \cos(2x)}}$.

This question is given in my textbook but I don't agree with the solution given in my book and various sites on the internet.

The book shows the following method:

$$f(x) = \sqrt{\dfrac{1-\cos(2x)}{1 + \cos(2x)}} =\sqrt{\dfrac{2\sin^2(x)}{2\cos^2(x)}}=\sqrt{\tan^2x}= \tan(x)$$ So the derivative of $\tan(x)$ is $\sec^2(x).$

But my confusion is that, $\sqrt{\tan^2(x)}$ should be $|\tan(x)|$ and so, it's derivative cannot be equal to $\sec^2x$.

Derivative of $|x|$ is $\dfrac{x}{|x|}$.

So, the derivative of $|\tan(x)|$ should be $\dfrac{\tan(x) \cdot \sec^2(x)}{|\tan(x)|}.\bf\qquad\qquad\qquad\qquad....(1)$

Or we can also say that derivative of $|\tan(x)|= \begin{cases}\sec^2{x},\rm If\, tan(x)\ge0\\-\sec^2x, \rm If \tan(x) < 0 \end{cases}.\bf\qquad....(2)$

Am, I right in (1) and (2)?

Although I agree with you, one should be careful about the domain of definition of functions - when you write the derivative of $\vert x\vert$, it is not define if $x=0$ — Sora, Feb 26 '22 at 09:28
More concisely, $f^\prime(x)=\operatorname{sgn}(\tan x)\sec^2x$. — J.G., Feb 26 '22 at 09:33
Using the function floor the derivative can be written as $f'(x)= (-1)^{\lfloor 2 x/\pi\rfloor }\frac{1}{\cos(x)^2}$ — Dr. Wolfgang Hintze, Mar 04 '22 at 14:51
The function is even, hence its derivative (for $x\neq 0$) is an odd function. Therefore the textbook answer is certainly wrong. — Ryszard Szwarc, May 02 '22 at 09:18
f(x) is not defined where cos 2x = -1. For any other x, f(x) is the square root of a number >= 0, divided by a number > 0, so well defined. But clearly for example f(-0.1) ≠ tan (-0.1). — gnasher729, Jun 08 '22 at 07:30
$$f’(x) = \sqrt{\frac{1+\cos 2x }{1 - \cos 2x}} \tan x\sec^2x$$ — Quanto, Jun 28 '22 at 14:07

score 5 · Accepted Answer · edited Jun 08 '22 at 07:03

5

You are absolutely correct. The function is equal to $|\tan(x)|$. You can also plot the given function in desmos. There is a sharp edge at $x=0$ and the function is not differentiable at $x=0$. $$ f'(x)= \begin{cases}\hspace{10 pt} \sec^2{x},& \text{if }\, \tan(x)>0,\\-\sec^2x,& \text{if }\, \tan(x) < 0. \end{cases}\bf\qquad $$

edited Jun 08 '22 at 07:03

Gary

31,845

answered Feb 26 '22 at 09:31

Shivam M

322

2

Continuity does not implies differentiability. However for a function to be differentiable a point continuity is necessary. Differentiability implies continuity and smoothness. A function having sharp edge at a point implies left hand derivative and right hand derivative are not equal hence the function is not differentiable at that point. – Shivam M Feb 26 '22 at 11:56
2

If a function is not continuous at a point then It cannot be differentiable at that point. But If it is continuous then It may or may not be differentiable provided If it has a sharp edge at that point or not. – Shivam M Feb 26 '22 at 12:08

devam_04 · Answer 2 · 2022-02-26T09:39:41.337

Both your formulations in (1) and (2) are correct, and we do indeed require the absolute value for this.

The answer of $f'(x)=\sec^2x$ is clearly incorrect as $f(x)≠\tan x,$ but rather $$f(x)=\frac{\mid\sin x\mid}{\mid \cos x\mid}= {\mid\tan x\mid}$$

If we differentiate this then we see that you are definitely correct with (1)

Your answer in (2) is also correct and we can see $f(x)$ [red] and $f'(x)$ [purple] graphically

This makes the "piecewise" nature of $f'(x)$ clear as it is equal to $+\sec^2 x$ only for $\tan x>0$

score 0 · Answer 3 · answered Feb 26 '22 at 10:08

0

There is another to prove that you are correct.

Use the logarithmic differentiation $$f(x) = \sqrt{\dfrac{1-\cos(2x)}{1 + \cos(2x)}} \implies \frac{f'(x)}{f(x)}=\csc (x) \sec (x)$$

answered Feb 26 '22 at 10:08

Claude Leibovici

260,315

@Anonymous. Don't you see the discontinuity at $x=0$ because of $\csc(x)$ ? – Claude Leibovici Feb 26 '22 at 10:13
@Anonymous Take logarithm of both sides: $\ln(f(x))=\frac12(\ln(1-\cos(2x))-\ln(1+\cos(2x)))$. Now take derivative. – Etemon Feb 26 '22 at 13:13

Confusion in finding derivative of $\sqrt{\frac{1-\cos(2x)}{1 + \cos(2x)}}$

3 Answers3

Linked