# Find the derivative

#### veronica1999

##### Member
I don't understand why my answer is wrong...

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#### Jameson

##### Administrator
Staff member
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
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
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
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
Re: find the derivative

Hello, veronica1999!

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

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

$$\text{We have: }\:y \:=\:\csc^{-1}(\sec x)$$

$$\text{Then: }\:y' \;=\;\frac{-\sec x\tan x}{\sec x\sqrt{\sec^2x-1}} \;=\;\frac{-\sec x\tan x}{\sec x\tan x} \;=\;-1$$

#### Opalg

##### MHB Oldtimer
Staff member
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
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.

#### soroban

##### Well-known member
Re: find the derivative

Hello, veronica1999!

There is a reason for the constant answer.

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

$$x$$ is an angle in a right triangle.
Let $$z$$ be the other acute angle.

Code:
                        *
* z*
c    *     *
*        * a
*           *
* x            *
*  *  *  *  *  *  *
b
We have: .$$\sec x \,=\,\tfrac{c}{b}$$

Then we want: .$$\csc^{-1}\left(\tfrac{c}{b}\right)$$

The angle whose cosecant is $$\tfrac{c}{b}$$ is angle $$z.$$

Hence: .$$y \:=\:\csc^{-1}(\sec x) \:=\:z$$

. . . . . . $$y \:=\:\tfrac{\pi}{2} - x$$

Therefore: .$$y' \;=\;-1$$

#### Prove It

##### Well-known member
MHB Math Helper
\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
\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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