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Find the derivative

veronica1999

Member
Jun 4, 2012
63
I don't understand why my answer is wrong...
 

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Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Re: find the derivative

Hi Veronica,

It would be a good idea to take some time to learn Latex. We have a Latex forum with some tips here to get you started.

This is what I think you wrote out.
$$y = \csc ^{-1} \left[ \sec(x) \right]$$
Find $y'$. Is that correct? :)
 

veronica1999

Member
Jun 4, 2012
63
Re: find the derivative

Hi Veronica,

It would be a good idea to take some time to learn Latex. We have a Latex forum with some tips here to get you started.

This is what I think you wrote out.
$$y = \csc ^{-1} \left[ \sec(x) \right]$$
Find $y'$. Is that correct? :)
Yes!
I promise i will learn it soon.
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Re: find the derivative

Yes!
I promise i will learn it soon.
It's ok, Veronica :)

I really need to head to bed now but I believe your error is in the cancellation. Your second line of in the PDF looks good to me, so if there's an error it seems like it should be with the cancellation. $\tan(x)$ can be negative but $\sqrt{ \left[ \tan(x) \right] ^2}$ cannot.

Someone will be along to help you soon.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Re: find the derivative

I don't understand why my answer is wrong...
Operating as follows...

$\displaystyle y= \csc^{-1} (\sec x) \implies \csc y = \frac{1}{\cos x} \implies \sin y = \cos x \implies y = \sin^{-1} (\cos x)$

... You have to operate on more comfortable inverse trigonometric functions...

Kind regards

$\chi$ $\sigma$
 

soroban

Well-known member
Feb 2, 2012
409
Re: find the derivative

Hello, veronica1999!

I can't open your file, but I'll give it a try.


[tex]\text{Differentiate: }\:y \:=\:\csc^{-1}(\sec x)[/tex]
[tex]\text{Formula: }\:\text{If }y \:=\:\csc^{-1}u,\,\text{then: }\:y' \:=\:\frac{-u'}{u\sqrt{u^2-1}}[/tex]

[tex]\text{We have: }\:y \:=\:\csc^{-1}(\sec x)[/tex]

[tex]\text{Then: }\:y' \;=\;\frac{-\sec x\tan x}{\sec x\sqrt{\sec^2x-1}} \;=\;\frac{-\sec x\tan x}{\sec x\tan x} \;=\;-1[/tex]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: find the derivative

This is a surprisingly tricky problem. The first thing to notice is that \(\displaystyle \frac d{dx}(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}\). (See here for example, though there are several web sites that carelessly give that formula without the mod signs on the $x$ in the denominator.) Therefore $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{-\sec x\tan x}{|\sec x|\sqrt{\sec^2 x-1}} = \frac{-\sec x\tan x}{|\sec x\tan x|}.$$ Next, $\sec x\tan x = \sin x\sec^2x$, and $\sec^2x$ is always positive. So we can cancel a factor $\sec^2x$ from the top and bottom of that last displayed fraction, to get $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{-\sin x}{|\sin x|} = \begin{cases}-1 & \text{if }\sin x>0,\\ +1 & \text{if }\sin x<0.\end{cases}$$
 

veronica1999

Member
Jun 4, 2012
63
Re: find the derivative

This is a surprisingly tricky problem. The first thing to notice is that \(\displaystyle \frac d{dx}(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}\). (See here for example, though there are several web sites that carelessly give that formula without the mod signs on the $x$ in the denominator.) Therefore $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{-\sec x\tan x}{|\sec x|\sqrt{\sec^2 x-1}} = \frac{-\sec x\tan x}{|\sec x\tan x|}.$$ Next, $\sec x\tan x = \sin x\sec^2x$, and $\sec^2x$ is always positive. So we can cancel a factor $\sec^2x$ from the top and bottom of that last displayed fraction, to get $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{-\sin x}{|\sin x|} = \begin{cases}-1 & \text{if }\sin x>0,\\ +1 & \text{if }\sin x<0.\end{cases}$$

Thank you!!!
Now it is clear.:D
 

soroban

Well-known member
Feb 2, 2012
409
Re: find the derivative

Hello, veronica1999!

There is a reason for the constant answer.

Differentiate: .[tex]y \:=\:\csc^{-1}(\sec x)[/tex]
Assume angle [tex]x[/tex] is acute.

[tex]x[/tex] is an angle in a right triangle.
Let [tex]z[/tex] be the other acute angle.

Code:
                        *
                     * z*
             c    *     *
               *        * a
            *           *
         * x            *
      *  *  *  *  *  *  *
               b
We have: .[tex]\sec x \,=\,\tfrac{c}{b}[/tex]

Then we want: .[tex]\csc^{-1}\left(\tfrac{c}{b}\right)[/tex]

The angle whose cosecant is [tex]\tfrac{c}{b}[/tex] is angle [tex]z.[/tex]

Hence: .[tex]y \:=\:\csc^{-1}(\sec x) \:=\:z [/tex]

. . . . . . [tex]y \:=\:\tfrac{\pi}{2} - x[/tex]

Therefore: .[tex]y' \;=\;-1[/tex]
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
\(\displaystyle \displaystyle \begin{align*} y &= \csc^{-1} { \left[ \sec{(x)} \right] } \\ \csc{(y)} &= \sec{(x)} \\ \frac{1}{\sin{(y)}} &= \frac{1}{\cos{(x)}} \\ \cos{(x)} &= \sin{(y)} \\ \frac{d}{dx} \left[ \cos{(x)} \right] &= \frac{d}{dx} \left[ \sin{(y)} \right] \\ -\sin{(x)} &= \cos{(y)}\,\frac{dy}{dx} \\ -\frac{\sin{(x)}}{\cos{(y)}} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{ \sqrt{ 1 - \frac{1}{\csc^2{(y)}} } } &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{ \sqrt{ 1 - \frac{1}{ \left( \csc{ \left\{ \csc^{-1}{ \left[ \sec{(x)} \right] } \right\} } \right) ^2 } } } &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sqrt{ 1 - \frac{1}{\sec^2{(x)}} }} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sqrt{1 - \cos^2{(x)}}} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sin{(x)}} &= \frac{dy}{dx} \\ \frac{dy}{dx} &= -1 \end{align*}\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
\(\displaystyle \displaystyle \begin{align*} y &= \csc^{-1} { \left[ \sec{(x)} \right] } \\ \csc{(y)} &= \sec{(x)} \\ \frac{1}{\sin{(y)}} &= \frac{1}{\cos{(x)}} \\ \cos{(x)} &= \sin{(y)} \\ \frac{d}{dx} \left[ \cos{(x)} \right] &= \frac{d}{dx} \left[ \sin{(y)} \right] \\ -\sin{(x)} &= \cos{(y)}\,\frac{dy}{dx} \\ -\frac{\sin{(x)}}{\cos{(y)}} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{ \sqrt{ 1 - \frac{1}{\csc^2{(y)}} } } &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{ \sqrt{ 1 - \frac{1}{ \left( \csc{ \left\{ \csc^{-1}{ \left[ \sec{(x)} \right] } \right\} } \right) ^2 } } } &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sqrt{ 1 - \frac{1}{\sec^2{(x)}} }} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sqrt{1 - \cos^2{(x)}}} &= \frac{dy}{dx} \\ -\frac{\sin{(x)}}{\sin{(x)}} &= \frac{dy}{dx}\quad \color{red}{\text{The denominator here should be }|\sin x|} \\ \frac{dy}{dx} &= -1\quad \color{red}{\text{ if }\sin x>0, \text{ but +1 if }\sin x<0.}\end{align*}\)
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