Find the derivative of 3cos^-1(sqrt x)

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mario99 said:
Hi skeeter.

I am a professional calculus student, but I don't understand this:

Why do we need to put an absolute symbol around [imath]f(x)[/imath] for the derivative of secant and cosecant inverse?🤔
Click to expand...
Please show us your work and either you'll see why there are absolute value bars or someone here will point it out. Or is skeeter wrong? ( I doubt that!)
 
Shown is graph of the basic secant function in black with domain restricted to [imath]\left[0,\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\pi\right][/imath] so that it passes the horizontal line test for the existence of an inverse function.

Reflection over the line y = x is the inverse secant function in red ... note its domain is [math](-\infty,-1] \cup [1, \infty)[/math] and range [math]\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},1\right][/math].

What do you notice about the slope of the inverse secant function?

sec&inv_sec.jpg
 
mario99 said:
Hi skeeter.

I am a professional calculus student, but I don't understand this:

Why do we need to put an absolute symbol around [imath]f(x)[/imath] for the derivative of secant and cosecant inverse?🤔
Click to expand...
We need to do that because that's what it works out to be!

On the other hand ...

My understanding is that some people define arcsec with a different range, namely [0,π/2)∪(π,3π/2). in order to avoid the need for the absolute value in the derivative. I see no other reason to prefer that definition, and several reasons not to, which probably explains what it is not popular.

I mentioned this here; I think this is the place where I learned about it.

It is also mentioned in Wikipedia, with no citation:

Note: Some authors[citation needed] define the range of arcsecant to be
{\textstyle (0\leq y<{\frac {\pi }{2}}{\text{ or }}\pi \leq y<{\frac {3\pi }{2}})}
, because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range,
{\displaystyle \tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}},}
whereas with the range
{\textstyle (0\leq y<{\frac {\pi }{2}}{\text{ or }}{\frac {\pi }{2}}<y\leq \pi )}
, we would have to write
{\displaystyle \tan(\operatorname {arcsec}(x))=\pm {\sqrt {x^{2}-1}},}
since tangent is nonnegative on
{\textstyle 0\leq y<{\frac {\pi }{2}},}
but nonpositive on
{\textstyle {\frac {\pi }{2}}<y\leq \pi .}

I don't think I've ever found an actual use of the alternative range, myself.

What range does it have in your experience?

On the other hand, MathWorld presents a version of the derivative (using the standard definition) that avoids the absolute value:

The derivative of
sec^(-1)z
is​
​
​
d/(dz)sec^(-1)z=1/(z^2sqrt(1-1/(z^2))),
(2)​
​
which simplifies to​
​
​
d/(dx)sec^(-1)x=1/(xsqrt(x^2-1))
(3)​
​
for
x>0
.​
 
Steven G said:
Please show us your work and either you'll see why there are absolute value bars or someone here will point it out. Or is skeeter wrong? ( I doubt that!)
Click to expand...
skeeter said:
Shown is graph of the basic secant function in black with domain restricted to [imath]\left[0,\frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},\pi\right][/imath] so that it passes the horizontal line test for the existence of an inverse function.

Reflection over the line y = x is the inverse secant function in red ... note its domain is [math](-\infty,-1] \cup [1, \infty)[/math] and range [math]\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2},1\right][/math].

What do you notice about the slope of the inverse secant function?

View attachment 37178
Click to expand...
Dr.Peterson said:
We need to do that because that's what it works out to be!

On the other hand ...

My understanding is that some people define arcsec with a different range, namely [0,π/2)∪(π,3π/2). in order to avoid the need for the absolute value in the derivative. I see no other reason to prefer that definition, and several reasons not to, which probably explains what it is not popular.

I mentioned this here; I think this is the place where I learned about it.

It is also mentioned in Wikipedia, with no citation:

Note: Some authors[citation needed] define the range of arcsecant to be
{\textstyle (0\leq y<{\frac {\pi }{2}}{\text{ or }}\pi \leq y<{\frac {3\pi }{2}})}
, because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range,
{\displaystyle \tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}},}
whereas with the range
{\textstyle (0\leq y<{\frac {\pi }{2}}{\text{ or }}{\frac {\pi }{2}}<y\leq \pi )}
, we would have to write
{\displaystyle \tan(\operatorname {arcsec}(x))=\pm {\sqrt {x^{2}-1}},}
since tangent is nonnegative on
{\textstyle 0\leq y<{\frac {\pi }{2}},}
but nonpositive on
{\textstyle {\frac {\pi }{2}}<y\leq \pi .}

I don't think I've ever found an actual use of the alternative range, myself.

What range does it have in your experience?

On the other hand, MathWorld presents a version of the derivative (using the standard definition) that avoids the absolute value:

The derivative of
sec^(-1)z
is​
​
​
​
​
d/(dz)sec^(-1)z=1/(z^2sqrt(1-1/(z^2))),
(2)​

​
which simplifies to​
​
​
​
​
d/(dx)sec^(-1)x=1/(xsqrt(x^2-1))
(3)​

​
for
x>0
.​
Click to expand...
So, basically, the derivative of the secant inverse is always positive. Omitting the absolute symbol will result in negative slope when taking [imath]x < -1[/imath].

In my opinion, changing the range to get rid of the absolute symbol is just making the function more complicated as well as we need to define that new range every time we encounter the secant inverse function.

I think that the best solution is just to stick with the absolute symbol or to write its derivative in this clever way:

[imath]\displaystyle \frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}[/imath]

A few years ago, when I wrote the derivative of the secant inverse as [imath]\displaystyle \frac{1}{x\sqrt{x^2-1}}[/imath] did not know that this only true for [imath]x > 1[/imath].

No one told me that I have to include the absolute symbol in my derivative. Thank god I have seen this post and got the correct full picture of the secant and cosecant inverse derivatives.

Thank you guys.
🥺
 
mario99 said:
A few years ago, when I wrote the derivative of the secant inverse as [imath]\displaystyle \frac{1}{x\sqrt{x^2-1}}[/imath] did not know that this only true for [imath]x > 1[/imath].

No one told me that I have to include the absolute symbol in my derivative. Thank god I have seen this post and got the correct full picture of the secant and cosecant inverse derivatives.
Click to expand...
Where had you found that expression for the derivative? That's what my ultimate question for you was: Did it come from a source with a different definition for the range, or was it just a mistake of some sort (yours or theirs)?

Looking around, I do find at least one site that leaves out the absolute value, apparently out of neglect; their proof omits consideration of sign. So there are definitely mistakes out there.
 
Dr.Peterson said:
Where had you found that expression for the derivative? That's what my ultimate question for you was: Did it come from a source with a different definition for the range, or was it just a mistake of some sort (yours or theirs)?

Looking around, I do find at least one site that leaves out the absolute value, apparently out of neglect; their proof omits consideration of sign. So there are definitely mistakes out there.
Click to expand...
It's the mistake of the instructor who has derived the derivative of the secant inverse without mentioning the absolute value or the domain at all. She derived it the same way of this website you are giving and we have taken that method as a standard way of deriving other inverse trigonometric functions. I now remember the book of the course that I was reading was using the absolute value, but that did not bother me because I thought that it might be a typo or something not important. When skeeter wrote the derivatives of the inverses, I was wondering if that absolute value was a typo too. I was curious to understand its story and did not come in my mind at all that it was because of the positive slope of the secant inverse function. I just knew that now. Everyday, we learn something new. Dr.Peterson said that.​
 
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