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veronica1999
Member
 Jun 4, 2012
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I don't understand why my answer is wrong...
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Yes!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?
It's ok, VeronicaYes!
I promise i will learn it soon.
Operating as follows...I don't understand why my answer is wrong...
[tex]\text{Formula: }\:\text{If }y \:=\:\csc^{1}u,\,\text{then: }\:y' \:=\:\frac{u'}{u\sqrt{u^21}}[/tex][tex]\text{Differentiate: }\:y \:=\:\csc^{1}(\sec x)[/tex]
This is a surprisingly tricky problem. The first thing to notice is that \(\displaystyle \frac d{dx}(\operatorname{arccsc} x) = \frac{1}{x\sqrt{x^21}}\). (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 $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{\sec x\tan x}{\sec x\sqrt{\sec^2 x1}} = \frac{\sec x\tan x}{\sec x\tan x}.$$ 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 $$\frac d{dx}\bigl(\operatorname{arccsc}(\sec x)\bigr) = \frac{\sin x}{\sin x} = \begin{cases}1 & \text{if }\sin x>0,\\ +1 & \text{if }\sin x<0.\end{cases}$$
Assume angle [tex]x[/tex] is acute.Differentiate: .[tex]y \:=\:\csc^{1}(\sec x)[/tex]
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...\(\displaystyle \displaystyle \begin{align*} y &= \csc^{1} { \left[ \sec{(x)} \right] } \\ \csc{(y)} &= \sec{(x)} \\ \frac{1}{\sin{(y)}} &= \frac{1}{\cos{(x)}} \\ \cos{(x)} &= \sin{(y)} \\ \frac{d}{dx} \left[ \cos{(x)} \right] &= \frac{d}{dx} \left[ \sin{(y)} \right] \\ \sin{(x)} &= \cos{(y)}\,\frac{dy}{dx} \\ \frac{\sin{(x)}}{\cos{(y)}} &= \frac{dy}{dx} \\ \frac{\sin{(x)}}{ \sqrt{ 1  \frac{1}{\csc^2{(y)}} } } &= \frac{dy}{dx} \\ \frac{\sin{(x)}}{ \sqrt{ 1  \frac{1}{ \left( \csc{ \left\{ \csc^{1}{ \left[ \sec{(x)} \right] } \right\} } \right) ^2 } } } &= \frac{dy}{dx} \\ \frac{\sin{(x)}}{\sqrt{ 1  \frac{1}{\sec^2{(x)}} }} &= \frac{dy}{dx} \\ \frac{\sin{(x)}}{\sqrt{1  \cos^2{(x)}}} &= \frac{dy}{dx} \\ \frac{\sin{(x)}}{\sin{(x)}} &= \frac{dy}{dx}\quad \color{red}{\text{The denominator here should be }\sin x} \\ \frac{dy}{dx} &= 1\quad \color{red}{\text{ if }\sin x>0, \text{ but +1 if }\sin x<0.}\end{align*}\)