MHB Calc Help: Find Derivative & Understand Why Answer Wrong

[veronica1999]

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

 

[Jameson]

Re: find the derivative

Hi Veronica,

It would be a good idea to take some time to learn Latex. We have http://www.mathhelpboards.com/f26/ 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]

Re: find the derivative

Hi Veronica,

It would be a good idea to take some time to learn Latex. We have http://www.mathhelpboards.com/f26/ 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.

 

[Jameson]

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]

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]

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]

Re: find the derivative

This is a surprisingly tricky problem. The first thing to notice is that $$\frac d{dx}(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}$$. (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]

Re: find the derivative

This is a surprisingly tricky problem. The first thing to notice is that $$\frac d{dx}(\operatorname{arccsc} x) = \frac{-1}{|x|\sqrt{x^2-1}}$$. (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]

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.

                        *
                     * 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&#039; \;=\;-1

 

[Prove It]

[math]\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*}[/math]

 

[Opalg]

[math]\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*}[/math]

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